Mathematics Curriculum

New York State Common Core Mathematics Curriculum

GEOMETRY • MODULE 5 Table of Contents1 Circles With and Without Coordinates

Module Overview ...... X Topic A: Central and Inscribed Angles (G-C.2, G-C.3) ...... X Lesson 1: Thales’ Theorem ...... X LessonDRAFT 2: Circles, Chords, Diameters, and Their Relationships ...... X Lesson 3: Rectangles Inscribed in Circles ...... X Lesson 4: Experiments with Inscribed Angles ...... X Lesson 5: Inscribed Angle Theorem and its Applications ...... X Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles ...... X Topic B: Arcs and Sectors (G-C.1, G-C.2, G-C.5) ...... X Lesson 7: The Angle Measure of an Arc...... X Lesson 8: Arcs and Chords ...... X Lesson 9: Arc Length and Areas of Sectors ...... X Lesson 10: Unknown Length and Area Problems ...... X Mid-Module Assessment and Rubric ...... X Topics A through B (assessment 1 day, return, remediation, or further applications 1 day) Topic C: Secants and Tangents (G-C.2, G-C.3, G-C.4) ...... X Lesson 11: Properties of Tangents ...... X Lesson 12: Tangent Segments ...... X Lesson 13: The Inscribed Angle Alternate a Tangent Angle ...... X Lesson 14: Secant Lines; Secant Lines that Meet Inside a Circle ...... X Lesson 15: Secant Angle Theorem, Exterior Case ...... X Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams ...... X Topic D: Equations for Circles and Their Tangents (G-GPE.1, G-GPE.4) ...... X Lesson 17: Writing the Equation for a Circle ...... X

1 Each lesson is ONE day, and ONE day is considered a 45-minute period.

Module 5: Circles With and Without Coordinates Date: 8/29/14 1

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Lesson 18: Recognizing Equations of Circles ...... X Lesson 19: Equations for Tangent Lines to Circles ...... X Topic E: Cyclic Quadrilaterals and Ptolemy’s Theorem (G-C.3) ...... X Lesson 20: Cyclic Quadrilaterals ...... X Lesson 21: Ptolemy’s Theorem ...... X End-of-Module Assessment and Rubric ...... X Topics C throughDRAFT D (assessment 1 day, return, remediation or further applications 1 day)

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Geometry • Module 5 Circles With and Without Coordinates

OVERVIEW With geometric intuition well established through Modules 1, 2, 3, and 4, students are now ready to explore the rich geometry of circles. This module brings together the ideas of similarity and congruence studied in Modules 1 and 2, the properties of length and area studied in Modules 3 and 4, and the work of geometric construction throughout the entire year. It also includes the specific properties of triangles, special quadrilaterals,DRAFT parallel lines and transversals, and rigid motions established and built upon throughout this mathematical story. This module’s focus is on the possible geometric relationships between a pair of intersecting lines and a circle drawn on the page. If the lines are perpendicular and one passes through the center of the circle, then the relationship encompasses the perpendicular bisectors of chords in a circle and the association between a tangent line and a radius drawn to the point of contact. If the lines meet at a point on the circle, then the relationship involves inscribed angles; and if they meet at the center of the circle, central angles. If the lines meet at a different point inside the circle or at a point outside the circle, the relationship includes the secant angle theorems and tangent angle theorems. Topic A, through a hands-on activity, leads students to Thales’ theorem (that an angle drawn from a diameter of a circle to a point on the circle is sure to be a right angle), to possible converses of Thales’ theorem, and eventually to the general inscribed-central angle theorem. Students use this result to solve unknown angle problems. Through this work, students construct triangles and rectangles inscribed in circles and study their properties (G-C.2, G-C.3). Topic B defines the measure of an arc and establishes results relating chord lengths and the measures of the arcs they subtend. Students build on their knowledge of circles from Module 2 and prove that all circles are similar. Students also develop a formula for arc length, as well as a formula for the area of a sector, and practice their skills solving unknown area problems (G-C.1, G-C.2, G-C.5). In Topic C students explore geometric relations in diagrams of two secant lines, or a secant and tangent line (and even two tangent lines) meeting a point inside or outside of a circle. They establish the secant angle theorems and tangent-secant angle theorems. By drawing auxiliary lines students also notice similar triangles and thereby discover l relationships between lengths of line segments appearing in these diagrams (G-C.2, G- C.3, G-C.4). Topic D brings in coordinate geometry to establish the equation of a circle. Students solve problems to find the equations of specific tangent lines or the coordinates of specific points of contact. They also express circles via analytic equations (G-GPE.1, G-GPE.4). The module concludes with Topic E focusing on the properties of quadrilaterals inscribed in circles and establishing Ptolemy’s theorem. This result codifies the Pythagorean theorem, curious facts about triangles, properties of the regular pentagon, and trigonometric relationships. It serves as a final unifying flourish for students’ year-long study of geometry (G-C.A.3).

Module 5: Circles With and Without Coordinates Date: 8/29/14 3

Focus Standards

Understand and apply theorems about circles. G-C.1 Prove2 that all circles are similar. G-C.2 Identify and describe relationships among inscribed angles, radii, and chords. Include3 the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle. G-C.3 Construct the inscribed and circumscribed circles of a triangle, and prove2 properties of angles for a quadrilateral inscribed in a circle. Find arc lengthsDRAFT and areas of sectors of circles. G-C.5 Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector.

Translate between the geometric description and the equation for a conic section. G-GPE.1 Derive the equation of a circle of given center and radius using the Pythagorean theorem; complete the square to find the center and radius of a circle given by an equation.

Use coordinates to prove simple geometric theorems algebraically. G-GPE.4 Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, 3) lies on the circle centered at the origin and containing the point (0, 2). √

Extension Standards

Understand and apply theorems about circles. G-C.4 (+) Construct a tangent line from a point outside a given circle to the circle.

Apply trigonometry to general triangles. G-SRT.9 (+) Derive the formula = ( )( ) ( ) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular1 to the opposite side 𝐴𝐴 2 𝑎𝑎 𝑏𝑏 𝑠𝑠𝑠𝑠𝑠𝑠 𝐶𝐶

2 Prove and apply (in preparation for Regents Exams). 3 Include angles formed by secants (in preparation for Regents Exams).

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Foundational Standards

Experiment with transformations in the plane. G-CO.3 Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. G-CO.5 Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.

Prove geometric theorems. G-CO.C.9 Prove theorems about lines and angles. Theorems include: Vertical angles are congruent; DRAFTwhen a transversal crosses parallel lines, alternate interior angles are congruent, and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. G-CO.C.10 Prove2 theorems about triangles. Theorems include: Measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. G-CO.C.11 Prove2 theorems about parallelograms. Theorems include: Opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.

Make geometric constructions. G-CO.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Prove theorems involving similarity. G-SRT.5 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.

Understand congruence and similarity using physical models, transparencies, or geometry software. 8.G.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles in real- world and mathematical problems in two and three dimensions. 8.G.8 Apply the Pythagorean theorem to find the distance between two points in a coordinate system.

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NYS COMMON CORE MATHEMATICS CURRICULUM Module Overview M5 GEOMETRY

Focus Standards for Mathematical Practice MP.1 Make sense of problems and persevere in solving them. Students solve a number of complex unknown angle and unknown area geometry problems, work to devise the geometric construction of given objects, and adapt established geometric results to new contexts and to new conclusions. MP.3 Construct viable arguments, and critique the reasoning of others. Students must provide justification for the steps in geometric constructions and the reasoning in geometric proofs, as well as create their own proofs of results and their extensions. MP.7 Look for and make use of structure. Students must identify features within complex diagrams (similar triangles, parallel chords, cyclic quadrilaterals, for example) which provide insight as to how to move forward with their thinking. DRAFT Terminology

New or Recently Introduced Terms . Chord (Given a circle C, and let P and Q be points on C. The segment PQ is called a chord of C.) . Tangent line (A tangent line to a circle is a line in the same plane that intersects the circle in one and �� only one point. This point is called the point of tangency.) . Secant line (A secant line to a circle is a line that intersects a circle in exactly two points.) . Inscribed angle (An inscribed angle is an angle whose vertex is on a circle, and each side of the angle intersects the circle in another point.) . Central angle (A central angle of a circle is an angle whose vertex is the center of a circle.) . Arc length (The length of an arc is the circular distance around the arc.) . Sector (Let arc AB be an arc of a circle with center O and radius r. The union of the segments OP, where P is any point on the arc AB, is called a sector. The arc AB is called the arc of the sector, and r is called its radius.� ) � � . Inscribed polygon (A polygon is inscribed in a circle if all vertices of the polygon lie on the circle.) . Cyclic quadrilateral (A quadrilateral inscribed in a circle is called a cyclic quadrilateral.)

Familiar Terms and Symbols4 . Circle . Diameter . Radius

4 These are terms and symbols students have seen previously.

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NYS COMMON CORE MATHEMATICS CURRICULUM Module Overview M5 GEOMETRY

Suggested Tools and Representations . Compass and straightedge . Geometer’s Sketchpad or Geogebra Software . White and colored paper, markers

Assessment Summary

Assessment Type Administered Format Standards Addressed

Mid-Module G-C.1, G-C.2, G-C.3, After Topic B Constructed response with rubric AssessmentDRAFT Task G-C.5. G-C.1, G-C.2, G-C.3, End-of-Module After Topic D Constructed response with rubric G-C.4, G-GPE.1, Assessment Task G-GPE.4.

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GEOMETRY • MODULE 5

Topic A: Central and Inscribed Angles G-C.A.2, G-DRAFTC.A.3 Focus Standard: G-C.A.2 Identify and describe relationships among inscribed angles, radii, and chords.

G-C.A.3 Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle. Instructional Days: 6 Lesson 1: Thales’ Theorem Lesson 2: Circles, Chords, Diameters and Their Relationships Lesson 3: Rectangles Inscribed in Circles Lesson 4: Experiments with Inscribed Angles Lesson 5: Inscribed Angles Theorem and Its Application Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles

The module begins with students exploring Thales’ theorem with a paper pushing discovery exercise where students push angles of triangles and trapezoids through a segment of fixed length to discover arcs of a circle (G-C.A.2). Students revisit the terms diameter and radius and are introduced to the terms central angle and inscribed angle. Lesson 2 begins the study of inscribed angles which is the focus of this module. In this lesson, students look at the relationships between circles and their diameters and chords, realizing through proofs (G-C.A.2) that the perpendicular bisector of a chord contains the center, the diameter is the longest chord, and congruent chords are equidistant from the center. Lesson 3 continues the study of inscribed angles by having students use a compass and straight edge to inscribe a rectangle in a circle (G-C.A.3) and then study similarities of circles and properties of other polygons that allow certain polygons to be inscribed in circles. Inscribed angles are compared to central angles in the same arcs in Lesson 4 with students using trapezoids and a paper pushing exercise similar to that of Lesson 1 to understand the difference between a major and minor arc. Students then explore inscribed and central angles and use the repeated patterns (MP. 7) to realize that the measure of a central angle is double the angle inscribed in the same arc. In Lesson 5, students are introduced to the inscribed angle theorem but only inscribed angles that are not obtuse. Students prove that inscribed angles are half the angle measure of the arcs they subtend. Lesson 6 completes the study of the inscribed angle theorem as students use their knowledge of chords, radii, diameters, central angles, and inscribed angles to persevere in solving a variety of unknown angle problems (MP.1). Throughout

Topic A: Central and Inscribed Angles Date: 8/29/14 1

NYS COMMON CORE MATHEMATICS CURRICULUM Topic C M5 GEOMETRY

this module students perform activities and constructions to enhance their understanding of the concepts studied. They pull the concepts together with proofs (MP.3), using concepts previously studied to arrive at new theorems and definitions.

DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Lesson 1: Thales’ Theorem

Student Outcomes . Using observations from a pushing puzzle, explore the converse of Thales’ theorem: If is a right triangle, then , , and are three distinct points on a circle with a diameter. △ 퐴퐵퐶 . Prove the statement of Thales’ theorem: If , , and are three different points on a circle with a 퐴 퐵 퐶 �퐴퐶�� diameter, then ∠ is a right angle. 퐴 퐵 퐶 �퐴퐶�� 퐴퐵퐶 Lesson Notes Every lesson inDRAFT this module is about an overlay of two intersecting lines and a circle. This will be pointed out to students later in the module, but keep this in mind as you are presenting lessons. In this lesson, students investigate what some say is the oldest recorded result, with proof, in the history of geometry – Thales’ theorem, attributed to Thales of Miletus (c. 624-c. 546 BCE), about 300 years before Euclid. Beginning with a simple experiment, students explore the converse of Thales’ theorem. This motivates the statement of Thales’ theorem, which students then prove using known properties of rectangles from Module 1.

Classwork Opening (15 minutes) Students explore the converse of Thales’s theorem with a pushing puzzle. Give each student a sheet of plain white paper, a sheet of colored cardstock, and a colored pen. Provide several minutes for the initial exploration before engaging students in a discussion of their observations and inferences.

Scaffolding: Opening Exercise (5 minutes) . For students with eye- 1. Mark points A and B on the sheet of white paper provided by your teacher. hand coordination or 2. Take the colored paper provided, and push that paper up between points A and B on the visualization problems, white sheet. model the Opening 3. Mark on the white paper the location of the corner of the colored paper, using a different Exercise as a class, and color than black. Mark that point C. See the example below. then provide students with a copy of the work to C complete the exploration.

. For advanced learners, A B explain the paper pushing puzzle, and let them come up with a hypothesis on 4. Do this again, pushing the corner of the colored paper up between the black points but at a different angle. Again, mark the location of the corner. Mark this point D. what they are creating and how they can prove it MP.7 5. Do this again and then again, multiple times. Continue to label the points. What curve do without seeing questions. & the colored points (C, D, …) seem to trace? – MP.7/MP.8 MP.8

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Discussion (8 minutes) . What curve do the colored points (C, D, …) seem to trace?  They seem to trace a semicircle. . If that is the case, where might the center of that semicircle be?  The midpoint of the line segment connecting points and on the white paper will be the center point of the semicircle. 퐴 퐵 . What would the radius of this semicircle be?  The radius is half the distance between points and (or the distance between point and the midpoint of the segment joining points and ). 퐴 퐵 퐴 . Can we prove that the marked points created by the corner of the colored paper do indeed lie on a circle? 퐴 퐵 What would we need to show? Have students do a 30-second Quick Write, and then share as a whole class.  We need to show that each marked point is the same distance from the midpoint of the line segment DRAFTconnecting the original points and . 퐴 퐵 Exploratory Challenge (12 minutes)

Allow students to come up with suggestions for how to prove that each marked point from the Opening Exercise is the same distance from the midpoint of the line segment connecting the original points and . Then offer the following approach. 퐴 퐵 . Have students draw the right triangle formed by the line segment between the two original points and and any one of the colored points ( , , …) created at the corner of the colored paper; then construct a rotated copy of that triangle underneath it. A sample drawing might be as follows: 퐴 퐵 퐶 퐷

A B A’ B’

C’

Allow students to read the question posed and have a few minutes to think independently and then share thoughts with an elbow partner. Lead students through the questions below. It may be helpful to have students construct the argument outlined in Steps 1–4 below several times for different points on the same diagram. The idea behind the proof is that no matter which colored point is chosen, the distance from that colored point to the midpoint of the segment between points and must be the same as the distance from any other colored point to that midpoint. 퐴 퐵

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Exploratory Challenge

Choose one of the colored points ( , , ...) that you marked. Draw the right triangle formed by the line segment connecting the original two points and and that colored point. Draw a rotated copy of the triangle underneath it. 푪 푫 Label the acute angles in the original푨 triangle푩 as and , and label the corresponding angles in the rotated triangle the same. 풙 풚 Todd says ’ is a rectangle. Maryam says ’ is a quadrilateral, but she’s not sure it’s a rectangle. Todd is right but doesn’t know how to explain himself to Maryam. Can you help him out? 푨푩푪푪 푨푩푪푪 1. What composite figure is formed by the two triangles? How would you prove it? A rectangle is formed. We need to show that all four angles measure °.

ퟗퟎ a. What is the sum of and ? Why? DRAFT90°; the sum of the acute풙 angles풚 in any right triangle is 90°.

b. How do we know that the figure whose vertices are the colored points ( , , …) and points and is a rectangle? 푪 푫 푨 푩 All four angles measure °. The colored points ( , , …) are constructed as right angles, and the angle at points and measures + , which is °. ퟗퟎ 푪 푫 푨 푩 풙 풚 ퟗퟎ 2. Draw the two diagonals of the rectangle. Where is the midpoint of the segment connecting the two original points and ? Why?

푨The midpoint푩 of the segment connecting points and is the intersection of the diagonals of the rectangle because the diagonals of a rectangle are congruent and bisect each other. 푨 푩

3. Label the intersection of the diagonals as point . How does the distance from point to a colored point ( , , …) compare to the distance from to points and ? 푷 푷 푪 푫 The distances from to each of푷 the points푨 are equal.푩

푷 4. Choose another colored point, and construct a rectangle using the same process you followed before. Draw the two diagonals of the new rectangle. How do the diagonals of the new and old rectangle compare? How do you know?

One diagonal is the same (the one between points and ), but the other is different since it is between the new colored point and its image under a rotation. The new diagonals intersect at the same point because diagonals of a rectangle intersect at their midpoints, and the midpoint푨 푩 of the segment connecting points and has not changed. The distance from to each colored point equals the distance from to each original푷 point and . By transitivity, the distance from to the first colored point, C, equals the distance from to the푨 second푩 colored point, D. 푷 푷 푨 푩 푷 푷

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

MP.3 5. How does your drawing demonstrate that all the colored points you marked do indeed lie on a circle? & For any colored point, we can construct a rectangle with that colored point and MP.7 the two original points, and , as vertices. The diagonals of this rectangle intersect at the same point because diagonals intersect at their midpoints, and the midpoint of the 푨diagonal푩 between points and is . The distance from to that colored point푷 equals the distance from to points and . By transitivity, the distance from to the first colored푨 point푩, , 푷equals the distance from 푷 to any other colored point. 푷 푨 푩 푷 푪 By definition,푷 a circle is the set of all points in the plane that are the same distance from a given center point. Therefore, each colored point on the drawing lies on the circle with center and a radius equal to half the length of the original line segment joining pints and . 푷 푨 푩 . Take a few minutes to write down what you have just discovered, and share that with your neighbor. . We haveDRAFT proven the following theorem: Theorem: Given two points and , let point be the midpoint between them. If is a point such that is right, then = = . 퐴 퐵 푃 퐶 In particular, that means that point is on a circle with center and diameter . ∠퐴퐶퐵 퐵푃 퐴푃 퐶푃

퐶 푃 �퐴퐵�� . This demonstrates the relationship between right triangles and circles. Theorem. If is a right triangle with ∠ the right angle, then , , and are three distinct points on a circle with a diameter.△ 퐴퐵퐶 퐶 퐴 퐵 퐶 Proof: If is a right angle, and is the midpoint �퐴퐵�� between points and , then = = implies that a circle with∠퐶 center and radius 푃 contains the points , , and . 퐴 퐵 퐵푃 퐴푃 퐶푃 푃 퐴푃 . This last theorem is the converse of Thales’ theorem, 퐴 퐵 퐶 which is discussed below in Example 1. . Review definitions previously encountered by students as stated in Relevant Vocabulary.

RELEVANT VOCABULARY

Circle. Given a point in the plane and a number > , the circle with center and radius is the set of all points in the plane that are distance from the point . 푪 풓 ퟎ 푪 풓 Radius. May refer either풓 to the line segment푪 joining the center of a circle with any point on that circle (a radius) or to the length of this line segment (the radius).

Diameter. May refer either to the segment that passes through the center of a circle whose endpoints lie on the circle (a diameter) or to the length of this line segment (the diameter).

Chord. Given a circle , and let and be points on . The segment is called a chord of .

Central Angle. A central푪 angle of푷 a circle푸 is an angle whose푪 vertex is the�푷푸� �center� of a circle. 푪

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Circle Radius Diameter Chord Central Angle

Example 1 (8 minutes)

Share with students that they have just recreated the converse of what some say is the oldest recorded result, with proof, in the historyDRAFT of geometry –Thales’ theorem, attributed to Thales of Miletus (c. 624- c. 546 BCE), some three centuries before Euclid! See Wikipedia, for example, on why the theorem might be attributed to Thales although it was clearly known before him. http://en.wikipedia.org/wiki/Thales%27_Theorem. Lead students through Examples1 and 2, and then let them struggle with a partner to determine a method to prove Thales’ theorem. If students are particularly struggling, give them the hint in the scaffold box. Once students have developed a strategy, lead the class through the remaining parts of this example.

Example 1

In the Exploratory Challenge, you proved the converse of a famous theorem in geometry. Thales’ theorem states: If , , and are three distinct points on a circle and segment is a diameter of the circle, then is right.

푨Notice푩 that,푪 in the proof in the Exploratory Challenge, you started푨푩 with a right angle (the corner∠ of푨푪푩 the colored paper) and created a circle. With Thales’ theorem, you must start with the circle, and then create a right angle.

Prove Thales’ theorem.

1. Draw circle with distinct points , , and on the circle and diameter . Prove that is a right angle. 푷 푨 푩 푪 푨푩 푨푪푩 ∠

2. Draw a third radius ( ). What types of triangles are and ? How do you know? 푷푪 △ 푨푷푪 △ 푩푷푪 They are isosceles triangles. Both sides of each triangle are radii of circle and are, therefore, of equal length. Scaffolding: 푷 . If students are struggling to develop a strategy to MP.1 3. Using the diagram that you just created, develop a strategy to prove Thales’ theorem. prove Thales’ Theorem, Look at each of the angle measures of the triangles, and see if we can prove is 90°. give them this hint:

∠푨푪푩 Draw a third radius, and 4. Label the base angles of as ° and the bases of as °. Express the measure use the result, also known of in terms of ° and °. to Thales, that the base △ 푨푷푪 풃 △ 푩푷푪 풂 angles of an isosceles The∠ measure푨푪푩 of 풂 is °풃 + °. triangle are congruent. 푨푪푩 풂 풃 ∠

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

5. How can the previous conclusion be used to prove that is a right angle?

+ = because the sum of the angle measures∠ in푨푪푩 a triangle is °. Then + = , so is a right angle. ퟐ풂 ퟐ풃 ퟏퟖퟎ ퟏퟖퟎ 풂 풃 ퟗퟎ ∠푨푪푩

Point out to students that and are examples of central angles.

∠푥 ∠푦 Exercises 1 – 2 (5 minutes) Allow students to do Exercises 1 – 2 individually and then compare answers with a neighbor. Use this as a means of informal assessment, and offer help where needed.

1. is a diameter of the circle shown. The radius is 12.5 cm, and AC = 7 cm. Find a. . �푨푩�� DRAFT풎∠푪°

ퟗퟎ b. AB 25 cm

c. BC

24 cm

2. In the circle shown, is a diameter. Find a. �푩푪�� 풎∠푫푨푩

ퟎ ퟏퟒퟒ b.

풎∠푩푨푬°

ퟏퟐퟖ c.

퐦∠°퐃퐀퐄 ퟖퟖ Closing (2 minutes) Give students a few minutes to explain the prompt to their neighbor, and then call the class together and share. Use this time to informally assess understanding and clear up misconceptions. . Explain to your neighbor the relationship that we have just discovered between a right triangle and a circle. Illustrate this with a picture.  If is a right triangle and the right angle is , , , and are distinct points on a circle and is the diameter of the circle. ∆퐴퐵퐶 ∠퐶 퐴 퐵 퐶 �퐴퐵��

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Lesson Summary

THEOREMS . Thales’ Theorem: If , , and are three different points on a circle with a diameter, then is a right angle. 퐴 퐵 퐶 �퐴퐵�� . Converse of Thales’∠퐴퐶퐵 Theorem: If is a right triangle with the right angle, then , , and are three distinct points on a circle with a diameter. ∆퐴퐵퐶 ∠퐶 . Therefore:퐴 퐵 퐶 Given distinct points , , and on a circle,�퐴퐵�� is a right triangle with the right angle if and only if is a diameter of the circle. 퐴 퐵 퐶 ∆퐴퐵퐶

. Given∠퐶 two points and , let point�퐴퐵�� be the midpoint between them. If is a point such that is right, then = = . 퐴 퐵 푃 퐶 RELEVANTDRAFT VOCABULARY∠퐴퐶퐵 퐵푃 퐴푃 퐶푃 . Circle. Given a point in the plane and a number > , the circle with center and radius is the set of all points in the plane that are distance

from the point . 푪 풓 ퟎ 푪 풓 풓

. Radius. May refer푪 either to the line segment joining the center of a circle with any point on that circle (a radius) or to the length of this line segment (the radius).

. Diameter. May refer either to the segment that passes through the center of a circle whose endpoints lie on the circle (a diameter) or to the length of this line segment (the diameter).

. Chord. Given a circle , and let and be points on . The segment

is called a chord of . �� 푪 푷 푸 푪 푷푸 . Central Angle. A central푪 angle of a circle is an angle whose vertex is the center of a circle.

Exit Ticket (5 minutes)

Lesson 1: Thales’ Theorem Date: 8/29/14 7

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Name Date

Lesson 1: Thales’ Theorem

Exit Ticket

Circle is shown below.

1. Draw퐴 two diameters of the circle. 2. Identify the shape defined by the endpoints of the two diameters. 3. Explain whyDRAFT this shape will always result.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

Exit Ticket Sample Solutions

Circle is shown below.

1. Draw푨 two diameters of the circle. 2. Identify the shape defined by the endpoints of the two diameters. 3. Explain why this shape will always result.

DRAFT

The shape defined by the endpoints of the two diameters will always form a rectangle. According to Thales’ theorem, whenever an angle is drawn from the diameter of a circle to a point on its circumference, then the angle formed is a right angle. All four endpoints represent angles drawn from the diameter of the circle to a point on its circumference; therefore, each of the four angles is a right angle. The resulting quadrilateral will, therefore, be a rectangle by definition of rectangle.

Problem Set Sample Solutions

1. A, B, and C are three points on a circle, and angle is a right angle. What’s wrong with the picture below? Explain your reasoning. 푨푩푪

Draw in three radii (from to each of the three triangle vertices), and label congruent base angles of each of the three resulting isosceles triangles. See diagram to see angle measures. In the “big” triangle ( ), we get + + = °. Using푶 the distributive property and division, we obtain ( + + ) = ° and + + = °. But we also have ° = = + . Substitution results in + + = △+푨푩푪, giving a value ퟐ풂of ° –ퟐ풃 a contradiction.ퟐ풄 ퟏퟖퟎ ퟐ 풂 풃 풄 ퟏퟖퟎ 풂 풃 풄 ퟗퟎ ퟗퟎ ∠푩 풃 풄 풂 풃 풄 풃 풄 풂 ퟎ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

2. Show that there is something mathematically wrong with the picture below.

Draw three radii ( , , and ). Label as and as . Also label as x and as x since is isosceles (both sides are radii). If is a right angle (as indicated on the drawing), then + = °. �� Since is isosceles,푶푨 푶푩 푶푪= + . ∠Similarly,푩푨푪 풂 ∠=푩푪푨+ .풄 Now adding∠ the푶푨푪 angles of ∠푶푪푨 results in △+푨푶푪+ + + + = °. Using the distributive∠푨푩푪 property and division, we obtain + + = 풂 °. 풄 ퟗퟎ Substitution△ 푨푶푩 takes us to ∠+푨푩푶= +풂 +풙 , which is a∠ c푪푩푶ontradiction.풄 풙 Therefore, the figure above△ is푨푩푪 mathematically DRAFTimpossible.풂 풂 풙 풄 풙 풄 ퟏퟖퟎ 풂 풄 풙 ퟗퟎ 풂 풄 풂 풄 풙

3. In the figure below, is the diameter of a circle of radius miles. If = miles, what is ?

푨푪�� ퟏퟕ 푩푪 ퟑퟎ 푨푪

16 miles.

4. In the figure below, is the center of the circle, and is a diameter.

푶 푨푫��

a. Find . ° 풎∠푨푶푩 ퟒퟖ b. If : = : , what is ? ° 풎 ∠푨푶푩 풎∠푪푶푫 ퟑ ퟒ 풎∠푩푶푪 ퟔퟒ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M5 T GEOMETRY

5. is a diameter of a circle, and is another point on the circle. The point lies on the line such that = . Show that = . (Hint: draw a picture to help you explain your thinking!) �푷푸�� 푴 푹 ⃖푴푸��⃗ Since푹푴 푴푸= (given),풎∠푷푹푴 풎∠= 푷푸푴 (both are right angles, by Thales’ theorem and by the angle addition postulate), and = (reflexive property), then by . It follows that 푹푴 푴푸 (corresponding풎∠푹푴푷 sides풎∠ of congruent푸푴푷 triangles) and that∠ 푸푴푷 = (by definition∠푹푴푷 of congruent angles). 푴푷 푴푷 △ 푷푹푴 ≅△ 푷푸푴 푺푨푺 ∠푷푹푴 ≅ ∠푷푸푴 풎∠푷푹푴 풎∠푷푸푴

6. Inscribe in a circle of diameter such that is a diameter. Explain why: a. ( ) = △ 푨푩푪 ퟏ �푨푪�� = ( ) 풔풊풏 is∠ the푨 hypotenuse푩푪 , and . Since sine is the ratio of the opposite side to the hypotenuse, will necessarily equal the length of the opposite side, that is, the length of . �� 푨푪 푨푪 ퟏ 풔풊풏 ∠푨 푩푪�� DRAFTb. ( ) = is the hypotenuse, and = . Since cosine is the ratio of the adjacent side to the hypotenuse, cos( ) 풄풐풔 ∠푨 푨푩 will necessarily equal the length of the adjacent side, that is, the length of . �� 푨푪 푨푪 ퟏ ∠푨 푨푩��

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Lesson 2: Circles, Chords, Diameters, and Their Relationships

Student Outcomes . Identify the relationships between the diameters of a circle and other chords of the circle.

Lesson Notes Students are askedDRAFT to construct the perpendicular bisector of a line segment and draw conclusions about points on that bisector and the endpoints of the segment. They relate the construction to the theorem stating that any perpendicular bisector of a chord must pass through the center of the circle.

Classwork Scaffolding: Opening Exercise (4 minutes) Post a diagram and display the steps to create a Construct the perpendicular bisector of line segment below (as you did in Module 1). perpendicular bisector used in Lesson 4 of Module 1. 푨푩�� . Label the endpoints of the segment A and B. . Draw circle A with center Draw another line that bisects but is not perpendicular to it. A and radius . . Draw circle B with center B List one similarity and one difference푨푩�� between the two bisectors. �퐴퐵�� and radius . . Label the points of Answers will vary. All points on the perpendicular bisector are equidistant from points and . �퐵퐴�� Points on the other bisector are not equidistant from points A and B. The perpendicular bisector intersection as C and D. meets at right angles. The other bisector meets at angles that are not congruent. 푨 푩 . Draw . �푨푩�� 퐶퐷⃖��⃗ You may wish to recall for students the definition of equidistant: Equidistant:. A point is said to be equidistant from two different points and if = .

Points and can be퐴 replaced in the definition above with other figures (lines,퐵 etc.)퐶 퐴퐵as long퐴퐶 as the distance to those figures is given meaning first. In this lesson, we will define the distance from the center of a circle to a chord. This definition퐵 will퐶 allow us to talk about the center of a circle as being equidistant from two chords.

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Discussion (12 minutes) Ask students independently or in groups to each draw chords and describe what they notice. Answers will vary depending on what each student drew. Lead students to relate the perpendicular bisector of a line segment to the points on a circle, guiding them toward seeing the relationship between the perpendicular bisector of a chord and the center of a circle. . Construct a circle of any radius, and identify the center as point . . Draw a chord, and label it . 푃 . Construct the perpendicular bisector of . �퐴퐵�� . What do you notice about the perpendicular bisector of ? �퐴퐵��  It passes through point P, the center of the circle. �퐴퐵�� . Draw DRAFTanother chord and label it CD. . Construct the perpendicular bisector�� of CD. . What do you notice about the perpendicular�� bisector of CD?  It passes through point , the center of the circle.�� . What can you say about the points on a circle in relation to the center of the circle? 푃  The center of the circle is equidistant from any two points on the circle. MP.3 . Look at the circles, chords, and perpendicular bisectors created by your neighbors. What statement can you & make about the perpendicular bisector of any chord of a circle? Why? (MP.3/MP.7) MP.7  It must contain the center of the circle. The center of the circle is equidistant from the two endpoints of the chord because they lie on the circle. Therefore, the center lies on the perpendicular bisector of the chord. That is, the perpendicular bisector contains the center. . How does this relate to the definition of the perpendicular bisector of a line segment? Scaffolding:  The set of all points equidistant from two given points (endpoints of a . Review the definition of line segment) is precisely the set of all points on the perpendicular central angle by posting a bisector of the line segment. visual guide. . A central angle of a circle

is an angle whose vertex is Exercises 1–6 (20 minutes) the center of a circle. Assign one proof to each group, and then jigsaw, share, and gallery walk as students present their work.

Exercises 1–6

1. Prove the theorem: If a diameter of a circle bisects a chord, then it must be perpendicular to the chord.

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 2

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Draw a diagram similar to that shown below.

Given: Circle with diameter , chord , and = .

Prove: 푪 �푫푬�� 푨푩�� 푨푭 푩푭 DRAFT= �푫푬�� ⏊�푨푩� �� Given 푨푭 = 푩푭 Reflexive property

푭푪 = 푭푪 Radii of the same circle are equal in measure

푨푪 푩푪 SSS

△ 푨푭푪 ≅△= 푩푭푪 Corresponding angles of congruent triangles are equal in measure

풎∠푨푭푪 and 풎∠푩푭푪 are right angles Equal angles that form a linear pair each measure °

∠푨푭푪 ∠푩푭푪 Definition of perpendicular lines ퟗퟎ

푫푬OR�� ⏊푨푩�� = Given

푨푭 = 푩푭 Radii of the same circle are equal in measure

푨푪 푩푪= Base angles of an isosceles are equal in measure

풎∠푭푨푪 풎∠푭푩푪 SAS

△ 푨푭푪 ≅=△ 푩푭푪 Corresponding angles of congruent triangles are equal in measure

풎∠푨푭푪 and 풎∠푩푭푪 are right angles Equal angles that form a linear pair each measure °

∠푨푭푪 ∠푩푭푪 Definition of perpendicular lines ퟗퟎ

푫푬��⏊푨푩�� 2. Prove the theorem: If a diameter of a circle is perpendicular to a chord, then it bisects the chord. Use a diagram similar to that in Exercise 1 above.

Given: Circle with diameter , chord , and

Prove: bisects푪 �푫푬�� 푨푩�� 푫푬��⏊�푨푩��

�푫푬� �� 푨푩�� Given

푫푬��⏊푨푩� �and�� are right angles Definition of perpendicular lines

∠푨푭푪 and∠ 푩푭푪 are right triangles Definition of right triangle

△ 푨푭푪 △ 푩푭푪 All right angles are congruent

∠푨푭푪= ≅ ∠ 푩푭푪 Reflexive property

푭푪 = 푭푪 Radii of the same circle are equal in measure

푨푪 푩푪

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△ 푨푭푪= ≅ △ 푩푭푪 Corresponding sides of congruent triangles are equal in length

푨푭 bisects푩푭 Definition of segment bisector

푫푬OR�� 푨푩�� Given

푫푬��⏊푨푩� �and�� are right angles Definition of perpendicular lines

∠푨푭푪 ∠푩푭푪 All right angles are congruent

∠푨푭푪= ≅ ∠ 푩푭푪 Radii of the same circle are equal in measure

푨푪 푩푪= Base angles of an isosceles triangle are congruent

풎∠푭푨푪 = 풎∠푭푩푪 Two angles of triangle are equal in measure, so third angles are equal

풎∠푨푪푭 풎∠푩푪푭 ASA DRAFT△ 푨푭푪= ≅△ 푩푭푪 Corresponding sides of congruent triangles are equal in length 푨푭 bisects푩푭 Definition of segment bisector

푫푬 �� 푨푩�� 3. The distance from the center of a circle to a chord is defined as the length of the perpendicular segment from the center to the chord. Note that, since this perpendicular segment may be extended to create a diameter of the circle, therefore, the segment also bisects the chord, as proved in Exercise 2 above. Prove the theorem: In a circle, if two chords are congruent, then the center is equidistant from the two chords.

Use the diagram below.

Given: Circle with chords and ; = ; is the midpoint of and is the midpoint of .

Prove: =푶 푨푩�� 푪푫�� 푨푩 푪푫 푭 �푨푩�� 푬 푪푫

= 푶푭 푶푬 Given.

푨푩 푪푫; If a diameter of a circle bisects a chord, then the diameter must be perpendicular to the chord. 푶푭��⏊�푨푩�� 푶푬��⏊�푪푫�� and are right angles Definition of perpendicular lines

∠푨푭푶 and∠ 푫푬푶 are right triangles Definition of right triangle

△E and푨푭푶 are midpoints△ 푫푬푶 of and Given

= 푭 푪푫�� 푨푩�� = and and are midpoints of and

푨푭 = 푫푬 푨푩All radii푪푫 of a circle푭 are푬 equal in measure �푨푩�� 푪푫��

푨푶 푫푶 HL

△ 푨푭푶= ≅△ 푫푬푶 Corresponding sides of congruent triangles are equal in length

푶푬 푶푭 4. Prove the theorem: In a circle, if the center is equidistant from two chords, then the two chords are congruent.

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 4

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Use the diagram below.

Given: Circle with chords and ; = ; is the midpoint of and is the midpoint of

Prove: AB= 푶 푨푩�� 푪푫�� 푶푭 푶푬 푭 �푨푩�� 푬 푪푫 DRAFT= 푪푫 Given 푶푭 푶푬; If a diameter of a circle bisects a chord, then it must be perpendicular to the chord. 푶푭��⏊�푨푩�� 푶푬��⏊�푪푫�� and are right angles Definition of perpendicular lines.

∠푨푭푶 and∠ 푫푬푶 are right triangles Definition of right triangle.

△ 푨푭푶= △ 푫푬푶 All radii of a circle are equal in measure.

푨푶 푫푶 HL

△AF푨푭푶 = ≅ △ 푫푬푶 Corresponding sides of congruent triangles are equal in length.

is the푫푬 midpoint of ,and is the midpoint of Given 푭 �푨푩�� 푬 AB = 푪푫 = and and are midpoints of and .

푪푫 푨푭 푫푬 푭 푬 �푨푩�� 푪푫�� 5. A central angle defined by a chord is an angle whose vertex is the center of the circle and whose rays intersect the circle. The points at which the angle’s rays intersect the circle form the endpoints of the chord defined by the central angle.

Prove the theorem: In a circle, congruent chords define central angles equal in measure.

Use the diagram below.

We are given that the two chords ( and ) are congruent. Since all radii of a circle are congruent, . Therefore, by SSS. since corresponding angles of congruent �� triangles are equal in measure. 푨푩 푪푫 푨푶 ≅ 푩푶 ≅ 푪푶�� ≅ �푫푶�� △ 푨푩푶 ≅ △ 푪푫푶 ∠푨푶푩 ≅ ∠푪푶푫

6. Prove the theorem: In a circle, if two chords define central angles equal in measure, then they are congruent.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Using the diagram from Exercise 5 above, we now are given that = . Since all radii of a circle are congruent, . Therefore, by SAS. because corresponding sides of congruent triangles are congruent 풎∠푨푶푩 풎 ∠푪푶푫 푨푶�� ≅ 푩푶�� ≅ �푪푶�� ≅ �푫푶�� △ 푨푩푶 ≅△ 푪푫푶 �푨푩�� ≅ �푪푫��

Closing (4 minutes) Have students write all they know to be true about the diagrams below. Bring the class together, go through the Lesson Summary, having students complete the list that they started, and discuss each point. A reproducible version of the graphic organizer shown is included at the end of the lesson.

Diagram Explanation of Diagram Theorem or Relationship

DRAFTIf a diameter of a circle bisects a chord, then it must be perpendicular to the chord. Diameter of a circle bisecting a chord If a diameter of a circle is perpendicular to a chord, then it bisects the chord.

If two chords are congruent, then the center is equidistant from the two chords. Two congruent chords equidistance If the center is equidistant from two from center chords, then the two chords are congruent.

Congruent chords define central angles equal in measure. If two chords define central angles Congruent chords equal in measure, then they are congruent.

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 6

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Lesson Summary

THEOREMS about chords and diameters in a circle and their converses:

• If a diameter of a circle bisects a chord, then it must be perpendicular to the chord.

• If a diameter of a circle is perpendicular to a chord, then it bisects the chord.

• If two chords are congruent, then the center is equidistant from the two chords.

• If the center is equidistant from two chords, then the two chords are congruent.

• Congruent chords define central angles equal in measure. DRAFT• If two chords define central angles equal in measure, then they are congruent. RELEVANT VOCABULARY • Equidistant. A point is said to be equidistant from two different points and if = . 퐴 퐵 퐶 퐴퐵 퐴퐶

Exit Ticket (5 minutes)

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 7

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Name Date

Lesson 2: Circles, Chords, Diameters, and Their Relationships

Exit Ticket

1. Given circle A shown, AF = AG and BC = 22. Find DE.

DRAFT

2. In the figure, circle has a radius of 10. .

푷 �푨푩�� ⊥ �푫푬�� a. If = , what is the length of AC?

푨푩 ퟖ

b. If = , what is the length of ?

푫푪 ퟐ 푨푩

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 8

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

Exit Ticket Sample Solutions

1. Given circle shown, = and = . Find .

22 푨 푨푭 푨푮 푩푪 ퟐퟐ 푫

2. In the figure, circle has a radius of 10. . a. If = , what is the length of AC? 퐏 �퐀퐁�� ⊥ 퐃퐄�� 4푨푩 ퟖ

b. If = , what is the length of ?

12푫푪 ퟐ 푨푩 DRAFT

Problem Set Sample Solutions 1. In this drawing, = , = , and = . What is ?

푨푩 ퟑퟎ 푶푴 ퟐퟎ 푶푵 ퟏퟖ 푪

(~ . )

ퟐ √ퟑퟎퟏ ퟑퟒ ퟕ 2. In the figure to the right, and ; = . Find AC. �푨푪��⏊�푩푮�� 푫푭��⏊�푬푮�� 푬푭 ퟏퟐ 24

3. In the figure, = , = . Find EG. Explain your work. 푨푪 ퟐퟒ 풂풏풅 푫푮 ퟏퟑ 5, is a right triangle with hypotenuse = radius = 13 and = , so = by Pythagorean theorem. = = . ∆푨푩푮 푨푩 ퟏퟐ 푩푮 ퟓ 푩푮 푮푬 ퟓ

Lesson 2: Circles, Chords, Diameters, and Their Relationships Date: 8/29/14 9

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

4. In the figure, = , = . Find . 푨푩 ퟏퟎ 푨푪 ퟏퟔ 4 푫푬

5. In the figure, = , and the two concentric circles have radii of and . Find . 푪푭 ퟖ ퟏퟎ ퟏퟕ 9 푫푬

DRAFT

6. In the figure, the two circles have equal radii and intersect at points and . and are centers of the circles. = , and the radius of each circle is . . 푩 푫 푨 푪 푨푪 ퟖ ퟓ Find . Explain your work. �푩푫��⏊�푨푪�� 6 푩푫 = = (radii)

푩푨 = 푩푪 = ퟓ

푨푮 = 푮푪 (Pythagoreanퟒ theorem) 푩푮 = ퟑ

푩푫 ퟔ

7. In the figure, the two concentric circles have radii of and . Chord of the larger circle intersects the smaller circle at and . = . . ퟔ ퟏퟒ �푩푭�� a. Find . 푪 푬 푪푬 ퟖ �푨푫��⏊��푩푭�� 푨푫

b. Findퟐ√ퟓ .

푩푭

ퟖ √ퟏퟏ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 T GEOMETRY

8. In the figure, is the center of the circle, and = .

Prove that 푨 bisects . 푪푩 푪푫 Let and푨푪� �� be perpendiculars∠퐁퐂퐃 from to and , respectively.

�푨푬=�� 푨푭�� Given 푨 �푪푩�� 푪푫��

푪푩 = 푪푫 If two chords are congruent, then the center is equidistant from the two chords. 푨푬 푨푭 = Reflexive property

푨푪 푨푪= = ° A line joining the center and the midpoint of a chord is perpendicular to the chord.

∠푪푬푨 ∠푪푭푨 ퟗퟎ HL

△ 푪푬푨=≅△ 푪푭푨 Corresponding angles of congruent triangles are equal in measure.

∠푬푪푨 bisects∠ 푭푪푨 Definition of angle bisector DRAFT푨푪 �� ∠푩푪푫 9. In class, we proved: Congruent chords define central angles equal in measure. a. Give another proof of this theorem based on the properties of rotations. Use the figure from Exercise 5.

We are given that the two chords ( and ) are congruent. Therefore, a rigid motion exists that carries to . The same rotation that carries to also carries to and to . The angle of �� rotation is the measure of , and푨푩 the rotation푪푫 is clockwise. �푨푩�� 푪푫�� 푨푩�� 푪푫�� 푨푶�� 푪푶�� 푩푶�� 푫푶�� ∠푨푶푪 b. Give a rotation proof of the converse: If two chords define central angles of the same measure, then they must be congruent.

Using the same diagram, we are given that . Therefore, a rigid motion (a rotation) carries to . This same rotation carries to and to . The angle of rotation is the measure of , and the rotation is clockwise. ∠푨푶푩 ≅ ∠푪푶푫 ∠푨푶푩 ∠푪푶푫 �푨푶�� 푪푶�� 푩푶�� 푫푶�� ∠푨푶푪

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Graphic Organizer on Circles

Diagram Explanation of Diagram Theorem or Relationship DRAFT

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Lesson 3: Rectangles Inscribed in Circles

Student Outcomes . Inscribe a rectangle in a circle. . Understand the symmetries of inscribed rectangles across a diameter.

Lesson Notes Have students use a compass and straightedge to locate the center of the circle provided. If necessary, remind students of their work in Module 1 on constructing a perpendicular to a segment and of their work in Lesson 1 in this module on Thales’ theorem.DRAFT Standards addressed with this lesson are G-C.A.2 and G-C.A.3.

Classwork Scaffolding: . Display steps to construct Opening Exercise (9 minutes) a perpendicular line at a Students will follow the steps provided and use a compass and straightedge to find the point. center of a circle. This exercise reminds students about constructions previously studied . Draw a segment through that will be needed in this lesson and later in this module. the point, and using a compass mark a point Using only a compass and straightedge, find the location of the center of the circle below. Follow the steps provided. equidistant on each side of the point.

Draw chord . . Label the endpoints of the Construct a chord perpendicular to at endpoint B. segment and . �� Mark the point푨푩 of intersection of the perpendicular �� . Draw circle with center chord and the circle as point C. 푨푩 퐴 퐵 and radius . is a diameter of the circle. 퐴 Construct a second diameter in the same way. . Draw circle with center �푨푪�� 퐴 �퐴퐵�� Where the two diameters meet is the center of the and radius . circle. 퐵 . Label the points of 퐵 �퐵퐴�� intersection as and . . Draw . 퐶 퐷 . For students struggling ⃖퐶퐷��⃗ with constructions due to eye-hand coordination or fine motor difficulties, provide set squares to construct perpendicular lines and segments.

. For advanced learners, give directions without steps and have them construct from memory.

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Explain why the steps of this construction work.

The center is equidistant from all points on the circle. Since the diameter goes through the center, the intersection of any two diameters is a point on both diameters and must be the center.

Exploratory Challenge (10 minutes)

Guide students in constructing a rectangle inscribed in a circle by constructing a right triangle (as in the Opening Exercise) and rotating the triangle about the center of the circle. Have students explore an alternate method, such as drawing a single chord, then constructing perpendicular chords three times. Review relevant vocabulary. . MP.7 How can you use a right triangle (such as the one you constructed in the Opening Exercise above) to produce a rectangle whose four vertices lie on the circle? (MP.1)  DRAFTWe can rotate the triangle 180° around the center of the circle (or around the midpoint of the diameter, which is the same thing).

Exploratory Challenge

Construct a rectangle such that all four vertices of the rectangle lie on the circle below.

. Suppose we wanted to construct a rectangle with vertices on the circle, but we didn't want to use a triangle. Is there a way we could do this? Explain.  We can construct a chord anywhere on the circle, then construct the perpendicular to one of its endpoints, and then repeat this twice more to construct our rectangle. . How can you be sure that the figure in the second construction is a rectangle?  We know it is a rectangle because all four angles are right angles.

RELEVANT VOCABULARY

Inscribed Polygon. A polygon is inscribed in a circle if all vertices of the polygon lie on the circle.

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Exercises 1–5 (20 minutes) For each exercise, ask students to explain why the construction is certain to produce the requested figure and to explain the symmetry across the diameter of each inscribed figure. Before students begin the exercises, ask the class, “What is symmetry?” Have a discussion, and let the students explain symmetry in their own words. They should describe symmetry as a reflection across an axis so that a figure lies on itself. Exercise 5 is a challenge exercise and can either be assigned to advanced learners or covered as a teacher-led example. In Exercise 5, students prove the converse of Thales’ theorem that they studied in Lesson 1.

Exercises 1–5

1. Construct a kite inscribed in the circle below, and explain the construction using symmetry.

DRAFT

Construct as before, but this time reflect it across the diameter. It is a kite because, by reflection, there are two opposite pairs of congruent adjacent sides. △ 푨푩푪

2. Given a circle and a rectangle, what must be true about the rectangle for it to be possible to inscribe a congruent copy of it in the circle?

The diagonals of the rectangle must be the length of the diameter of the circle.

3. The figure below shows a rectangle inscribed in a circle.

a. List the properties of a rectangle. Opposite sides parallel and congruent, four right angles, diagonals congruent, and bisect each other.

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b. List all the symmetries this diagram possesses. Opposite sides are congruent, all four angles are congruent, diagonals are congruent, the figure may be reflected onto itself across the perpendicular bisector of the sides of the rectangle, the figure may be rotated onto itself with either a ° or a ° rotation either clockwise or counterclockwise.

ퟏퟖퟎ ퟑퟔퟎ c. List the properties of a square. Opposite sides parallel, all sides congruent, 4 right angles, diagonals congruent, bisect each other, and are perpendicular.

d. List all the symmetries of the diagram of a square inscribed in a circle. In addition to the symmetries listed above, all four sides are congruent, the figure may be reflected onto itself across the diagonals of the square, the figure may be rotated onto itself with either a ° or a ° rotation either clockwise or counterclockwise. ퟗퟎ ퟐퟕퟎ DRAFT 4. A rectangle is inscribed into a circle. The rectangle is cut along one of its diagonals and reflected across that diagonal to form a kite. Draw the kite and its diagonals. Find all the angles in this new diagram, given that the acute angle between the diagonals of the rectangle in the original diagram was °.

Draw diagrams such as the following: ퟒퟎ

Given = °. Then = °; = = °; = = °; = °; = °. 풎∠푨푫푩 ퟒퟎ

풎∠푩푫푬 ퟒퟎ 풎∠푩푨푫 풎∠푩푬푫 ퟗퟎ 풎∠푨푩푫 풎∠푬푩푫 ퟓퟎ 풎∠푨푩푬 ퟏퟎퟎ 풎∠푨푫푬 ퟖퟎ 5. CHALLENGE: Show that the 3 vertices of a right triangle are equidistant from the midpoint of the hypotenuse by showing that the perpendicular bisectors of the legs pass through the midpoint of the hypotenuse. (This is called the side-splitter theorem.) a. Draw the perpendicular bisectors of . b. Label the point where they meet P. What is point ? �푨푩�� 퐚퐧퐝 �푨푪�� The center of the circle. 푷

c. What can be said about the distance from P to each vertex of the triangle? What is the relationship between the circle and the triangle?

Point P is equidistant from the three vertices of the triangle , , and , so the circle is circumscribed about . 푨 푩 푪 ∆푨푩푪

d. Repeat this process, this time sliding to another place on the circle and call it ’. What do you notice?

You get the푩 same center. e. Using what you have learned about angles, chords, and their

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relationships, what does the position of point depend on? Why?

The position of P depends only on , not푷 the position of as long as and ’ are on the same side of . and have the same measure because they are inscribed in the same arc. The center of the circle in both cases is P. 풎∠푨푩푪 푩 푩 푩 푨푪�� ∠푨푩푪 ∠푨푩′푪

Closing (1 minute) Have students discuss the question with a neighbor or in groups of 3. Call the class back together and review the definition below. . Explain how the symmetry of a rectangle across the diameter of a circle helps inscribe a rectangle in a circle.  Since the rectangle is composed of two right triangles with the diameter as the hypotenuse, it is DRAFTpossible to construct one right triangle and then reflect it across the diameter.

Lesson Summary

RELEVANT VOCABULARY

• Inscribed Polygon. A polygon is inscribed in a circle if all vertices of the polygon lie on the circle.

Exit Ticket (5 minutes)

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Name Date

Lesson 3: Rectangles Inscribed in Circles

Exit Ticket

Rectangle is inscribed in circle . Boris says that diagonal could pass through the center, but it doesn’t have to pass through the center. Is Boris correct? Explain your answer in words, or draw a picture to help you explain your thinking. 퐴퐵퐶퐷 푃 퐴퐶

DRAFT

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Exit Ticket Sample Solutions

Yes, the diagonal must pass through the center of the circle. Since each vertex of the rectangle is a right angle, the hypotenuse of the right triangle formed by each angle and the diagonal of the rectangle must be the diameter of the circle (by the work done in Lesson 1 of this module). The diameter of the circle passes through the center of the circle; therefore, the diagonal passes through the center.

Problem Set Sample Solutions

1. DRAFTUsing only a piece of . copy paper and a pencil, find the location of the center of the circle below.

ퟖ ퟓ풙ퟏퟏ

Lay the paper across the circle so that its corner lies on the circle. The points where the two edges of the paper cross the circle are the endpoints of a diameter. Mark those points, and draw the diameter using the edge of the paper as a straightedge. Repeat to get a second diameter. The intersection of the two diameters is the center of the circle.

2. Is it possible to inscribe a parallelogram that is not a rectangle in a circle? No, although it is possible to construct an inscribed polygon with one pair of parallel sides (i.e., a trapezoid); a parallelogram requires that both pairs of opposite sides be parallel and both pairs of opposite angles be congruent. A parallelogram is symmetric by 180 degree rotation about its center and has NO other symmetry unless it is a rectangle. Two parallel lines and a circle create a figure that is symmetric by a reflection across the line through the center of the circle that is perpendicular to the two lines. If a trapezoid is formed with vertices where the parallel lines meet the circle, the trapezoid has reflectional symmetry. Therefore, it cannot be a parallelogram-—-UNLESS it is a rectangle.

3. In the figure, is a rectangle inscribed in circle . = ; = . Find . 푩푪푫푬 푨 푫푬 ퟖ 푩푬 ퟏퟐ 푨푬

ퟐ√ퟏퟑ

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4. Given the figure, = = and = . Find the radius of the circle. 푩푪 푪푫 ퟖ 푨푫 ퟏퟑ Mark the midpoint of as point . = = , so = . is a right triangle, so by the Pythagorean theorem, = . Using the Pythagorean �푩푪�� 푬 푩푬 푬푪 ퟒ 푬푫 ퟏퟐ ∆푬푨푫 theorem again givens = . 푬푨 ퟓ 푨푪 √ퟒퟏ

5. In the figure, and are parallel chords cm apart. = cm, = cm, and . 푫푭�� 푩푮�� ퟏퟒ 푫푭 ퟏퟐ 푨푩 ퟏퟎ Find . �푬푯��⏊�푩푮�� Draw푩푮 . = °, = , = . By Pythagorean theorem, DRAFT= . ∆푫푬푨 풎∠푫푬푨 ퟗퟎ 푫푬 ퟔ 푫푨 ퟏퟎ 푬푨In ퟖ , = °, = , = , so = . This means = .

∆ cm푨푩푯 풎∠푨푯푩 ퟗퟎ 푨푩 ퟏퟎ 푨푯 ퟔ 푩푯 ퟖ 푩푮 ퟏퟔ

ퟏퟔ 6. Use perpendicular bisectors of the sides of a triangle to construct a circle that circumscribes the triangle. (Students did a construction similar to this in G10-M1 Lesson 4).

Draw any triangle.

Construct the perpendicular bisector of the sides.

The perpendicular bisectors meet at the circumcenter. Using the center and the distance to one vertex as a radius, draw the circle.

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Lesson 4: Experiments with Inscribed Angles

Student Outcomes . Explore the relationship between inscribed angles and central angles and their intercepted arcs.

Lesson Notes As with Lesson 1 in this module, students use simple materials to explore the relationship between different types of angles in circles. In Lesson 1, the exploration was limited to angles inscribed in diameters; in this lesson, we extend the concept to includeDRAFT all inscribed angles. This lesson sets up concepts taught in Lessons 5 – 7. Problem 6 of the Problem Set is particularly important in setting up Lesson 5. Problem 7 of the Problem Set is an extension and will be revisited in Lesson 7.

Classwork Have available for each student (or group) a straight edge, white paper, and trapezoidal paper cutouts, created by slicing standard colored 8.5 x 11 sheets of paper or cardstock from edge to edge using a paper cutter. There should be a variety of trapezoids with different acute angles available.

Opening Exercise (5 minutes) Project the circle shown on the board. Have students identify the central angle, inscribed angle, minor arc, major arc, and intercepted arc of an angle. Have students write the definition of each in their own words, and then discuss the formal definitions. This vocabulary could be introduced with a series of prompts such as: • is a minor arc. is a major arc. Explain the difference between a major arc and minor arc. 퐵퐸� 퐸퐷퐵� • is an inscribed angle. is a central angle. Explain the difference between an inscribed angle and a ∠central퐵퐷퐶 angle. ∠퐵퐴퐶 • and both intercept arc . Explain what you think it means for an angle to intercept an arc. ∠ 퐶퐷퐵 ∠퐶퐴퐵 퐵퐶�

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RELEVANT VOCABULARY

ARC. An arc is a portion of the circumference of a circle.

MINOR AND MAJOR ARC. Let be a circle with center , and let and be different points that lie on but are not the endpoints of the same diameter. The minor arc is the set containing , , and all points of that are in the interior of . The major arc is the set푪 containing , , and all푶 points of 푨 that 푩lie in the exterior of . Examples:푪 Minor Arc , . Major Arc , . Answers will vary. 푨 푩 푪 ∠푨푶푩 푨 푩 푪 ∠푨푶푩 푩푬INSCRIBED� 푬푫� ANGLE. 푬푫푩An� inscribed푫푪푬� angle is an angle whose vertex is on a circle and each side of the angle intersects the circle in another point. Examples: , . Answers will vary.

CENTRAL ANGLE. A central angle∠푩푫푪 of a∠ circle푬푪푫 is an angle whose vertex is the center of a circle. Examples: , . Answers will vary. ∠푪푨푩 ∠푩푨푬 INTERCEPTED ARC OF AN ANGLE. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of the arc. Examples: , . Answers will vary. DRAFT� � 푬푫 푪푭

Your teacher will provide you with a straight edge, a sheet of colored paper in the shape of a trapezoid, and a sheet of plain white paper.

. Draw 2 points no more than inches apart in the middle of the plain white paper, and Scaffolding: label them and . . If students are struggling ퟑ D . Use the acute푨 angle푩 of your colored trapezoid to plot C with acute and obtuse a point on the white sheet by placing the colored angles of a trapezoid being cutout so that the points and are on the edges of supplementary, have them the acute angle and then plotting the position of the 푨 푩 confirm by folding or vertex of the angle. Label that vertex . A B tearing the trapezoid into . Repeat several times. Name the points푪 , , …. segments containing the angles and putting them 푫 푬 together as they did in Exploratory Challenge (10 minutes) Grade 5 Module 6. . The students’ task is as appears below: Display the definition of supplementary angles.

D C

B A

. As students receive their materials, ask them to label the acute angle of the trapezoid. What is the relationship between the acute angle and the obtuse angle?  They are supplementary. . As students complete the point-plotting, ask, “What shape do the plotted points form?”  The points seem to be the major arc of a circle. MP.3 . How can you find the minor arc of the circle? Explain how you know. &  MP.7 We can find the minor arc of the circle by pushing the supplementary angle of the trapezoid through

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the two original points from above. If the acute angle creates a major arc, the supplementary angle would produce a smaller (minor) arc. . How does this relate to the work we did on Thales’ theorem in Lesson 1?  In Lesson 1, we showed that a triangle created by connecting the endpoints of a diameter with any other point on a circle is a right triangle. We used a right angle (a corner of a plain piece of paper) to create our original semicircle. Here, we are using the acute and obtuse angles of a trapezoid to create major and minor arcs of a circle.

Example 1 (10 minutes)

Have students further explore the angles formed by connecting points and in their drawing with any one of the points they marked at the vertex ( , , …) as it was moved through points and . 퐴 퐵 . WhenDRAFT you trace over the 퐶angles퐷 퐸 formed by points and and the 퐴vertex 퐵point ( , , …) you marked, what do you notice about the measures of the angles you drew? 퐴 퐵 퐶 퐷 퐸  All angles drawn with a vertex on the major arc have the same measure – the measure of the acute angle of the trapezoid. . What happens when you trace over the angles formed by points and and the vertex of the obtuse angle?  All angles drawn with a vertex on the minor arc have the same measure – the measure of the obtuse 퐴 퐵 angle of the trapezoid.

Example 1

1. Draw several of the angles formed by connecting points and on your paper with any of the additional points you marked as the acute angle was pushed through the points ( , , ,…). What do you notice about the measures of these angles? 푨 푩 푪 푫 푬 MP.8 All angles have the same measure – the measure of the acute angle on the trapezoid.

2. Draw several of the angles formed by connecting points A and B on your paper with any of the additional points you marked as the obtuse angle was pushed through the points from above. What do you notice about the measures of these angles?

All angles have the same measure – the measure of the obtuse angle on the trapezoid.

Example 2 (10 minutes)

Continue the exploration, providing each student with several copies of the circle below, a straightedge, and scissors. They will select a point on the circle and create an inscribed angle. Each student will cut out his or her angle and compare it to the angle of several neighbors. All students started with the same arc thus, all inscribed angles will have the same measure. This can also be confirmed using protractors to measure the angles instead of cutting the angles out or modeled by the teacher.

Example 2

1. In the circle below, draw a point on the circle, and label it . Create angle .

퐃 ∠퐁퐃퐂

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2. is called an inscribed angle. Can you explain why?

∠The퐁퐃퐂 vertex is on the circle, and the sides of the angle pass through points that are also on the circle.

3. Arc is called the intercepted arc. Can you explain why?

It is 퐁퐂the� arc cut in the circle by the inscribed angle.

4. DRAFTCarefully cut out the inscribed angle, and compare it to the angles of several of your neighbors. What appears to be true about each of the angles you drew?

All appear to have the same measure.

5. Draw another point on a second circle, and label it point . Create angle , and cut it out. Compare and . What appears to be true about the two angles? 퐄 ∠퐁퐄퐂 ∠퐁퐃퐂 ∠All퐁퐄퐂 appear to have the same measure.

6. What conclusion may be drawn from this? Will all angles inscribed in the circle from these two points have the same measure?

All angles inscribed in the circle from these two points will have the same measure.

Explain to your neighbor what you have just discovered.

Example 3 (3 minutes)

Extend the exploration, having students draw a circle, select two points on the circle, and use those two points as endpoints of an intercepted arc for a central angle.

Example 3

1. In the circle below, draw the angle formed by connecting points and to the center of the circle.

퐁 퐂

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Is and inscribed angle? Explain.

∠퐁퐀퐂No. The vertex is not on the circle; the vertex is the center of the circle.

2. Is it appropriate to call this the central angle? Why or why not? The acute angle ( ) is formed by connecting points and to the center point ( ), so it would be appropriate to call this a central angle. ∠푩푨푪 푩 푪 푨

3. What is the intercepted arc?

The intercepted arc is .

푩푪� 4. Is the measure of the same as the measure of one of the inscribed angles in Example 2?

No, the measure of∠ 퐁퐀퐂 is greater. DRAFT ∠푩푨푪 5. Can you make a prediction about the relationship between the inscribed angle and the central angle? The inscribed angle is about half the central angle. The central angle is double the inscribed angle.

Closing (2 minutes) Have students explain to a partner the answer to the prompt below, and then call the class together to review the Lesson Summary. . What is the difference between an inscribed angle and a central angle?

Lesson Summary • All inscribed angles from the same intercepted arc have the same measure. RELEVANT VOCABULARY • ARC. An arc is a portion of the circumference of a circle. • MINOR AND MAJOR ARC. Let be a circle with center , and let and be different points that lie on but are not the endpoints of the same diameter. The minor arc is the 푪set containing , , and 푶all points of푨 that푩 are in the interior of . The푪 major arc is the set containing , , and all points of that lie in the exterior of 푨. 푩 푪 ∠푨푶푩 푨 푩 • INSCRIBED ANGLE.푪 An inscribed angle is∠ 푨푶푩an angle whose vertex is on a circle, and each side of the angle intersects the circle in another point.

• CENTRAL ANGLE. A central angle of a circle is an angle whose vertex is the center of a circle. • INTERCEPTED ARC OF AN ANGLE. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of

the arc.

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Exit Ticket (5 minutes) DRAFT

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Name Date

Lesson 4: Experiments with Inscribed Angles

Exit Ticket

Joey marks two points on a piece of paper, as we did in the Opening Exercise, and labels them and . Using the trapezoid shown below, he pushes the acute angle through points and from below several times so that the sides of the angle touch points and , marking the location of the vertex each time. Joey claims that퐴 the shape퐵 he forms by doing this is the minor arc of a circle and that he can form the major퐴 arc by퐵 pushing the obtuse angle through points and from above. “The퐴 obtuse퐵 angle has the greater measure, so it will form the greater arc,” states Joey. DRAFT 퐴 Ebony퐵 disagrees, saying that Joey has it backwards. “The acute angle will trace the major arc,” claims Ebony.

C D C

B A

1. Who is correct, Joey or Ebony? Why?

2. How are the acute and obtuse angles of the trapezoid related?

3. If Joey pushes one of the right angles through the two points, what type of figure is created? How does this relate to the major and minor arcs created above?

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Exit Ticket Sample Solutions

Joey marks two points on a piece of paper, as we did in the Opening Exercise, and labels them and . Using the trapezoid shown below, he pushes the acute angle through points and from below several times so that the sides of the angle touch points and , marking the location of the vertex each time. Joey claims that푨 the shape푩 he forms by doing this is the minor arc of a circle and that he can form the major푨 arc by푩 pushing the obtuse angle through points and from above. “The obtuse푨 angle푩 has the greater measure, so it will form the greater arc,” states Joey. 푨 Ebony푩 disagrees, saying that Joey has it backwards. “The acute angle will trace the major arc,” claims Ebony.

D C

B A DRAFT 1. Who is correct, Joey or Ebony? Why? Ebony is correct. The acute angle vertex traces out the major arc of the circle.

2. How are the acute and obtuse angles of the trapezoid related? They are supplementary.

3. If Joey pushes one of the right angles through the two points, what type of figure is created? How does this relate to the major and minor arcs created above?

A semicircle is created. Both arcs will be the same measure ( °).

ퟏퟖퟎ

Problem Set Sample Solutions

1. Using a protractor, measure both the inscribed angle and the central angle shown on the circle below.

= ______= ______°; ° 풎∠푩푪푫 풎∠푩푨푫

ퟗퟎ ퟏퟖퟎ

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2. Using a protractor, measure both the inscribed angle and the central angle shown on the circle below.

DRAFT = ______= ______풎∠°;푩푫푪° 풎∠푩푨푪

ퟑퟎ ퟔퟎ

3. Using a protractor, measure both the inscribed angle and the central angle shown on the circle below.

= ______= ______

풎∠°;푩푫푪° 풎∠푩푨푪

ퟓퟎ ퟏퟎퟎ 4. What relationship between the measure of the inscribed angle and the measure of the central angle that intercept the same arc is illustrated by these examples?

The measure of the inscribed angle appears to be half the measure of the central angle that intercepts the same arc.

5. Is your conjecture at least true for inscribed angles that measure °?

Yes, according to Thales’ theorem, if , , and are points on a circleퟗퟎ where is a diameter of the circle, then is a right angle. Since a diameter represents an ° angle, our conjecture is always true for angles that �� measure °. 푨 푩 푪 푨푪 ∠푨푩푪 ퟏퟖퟎ ퟗퟎ

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6. Prove that = in the diagram below.

풚 ퟐ풙

is an isosceles triangle since all radii of a circle are congruent. Therefore, = = . In addition, = + since the measure of an exterior angle of a triangle is equal to the sum of the measures of the opposite△ 푨푩푪 interior angles. By substitution, = + , or = . 풎∠푩 풎∠푪 풙 풚 풎∠푩 풎∠푪 풚 풙 풙 풚 ퟐ풙 7. Red ( ) and blue ( ) lighthouses are located on the coast of the ocean. Ships traveling are in safe waters as long as the angle from the ship ( ) to the two lighthouses ( ) is always less than or equal to some angle called the DRAFT푹 푩 “danger angle.” What happens to as the ship gets closer to shore and moves away from shore? Why do you think a larger angle is dangerous?푺 ∠푹푺푩 Ө Ө The closer the boat is to the shore, the larger will be, and as the boat moves away from shore, gets smaller. smaller means the ship is in deeper water, which is safer for ships. Ө Ө 푨 Ө

Red (R)

Blue (B)

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Lesson 5: Inscribed Angle Theorem and its Applications

Student Outcomes . Prove the inscribed angle theorem: The measure of a central angle is twice the measure of any inscribed angle that intercepts the same arc as the central angle. . Recognize and use different cases of the inscribed angle theorem embedded in diagrams. This includes recognizing and using the result that inscribed angles that intersect the same arc are equal in measure.

Lesson Notes Lesson 5 introducesDRAFT but does not finish the inscribed angle theorem. The statement of the inscribed angle theorem in this lesson should be only in terms of the measures of central angles and inscribed angles, not the angle measures of intercepted arcs. The measure of the inscribed angle is deduced and the central angle is given, not the other way around in Lesson 5. This lesson only includes inscribed angles and central angles that are acute or right. Obtuse angles will not be studied until Lesson 7. The Opening Exercise and Examples 1 and 2 are the complete proof of the inscribed angle theorem (central angle version).

Classwork Opening Lead students through a discussion of the Opening Exercise (an adaptation of Lesson 4 Scaffolding: Problem Set 6), and review terminology, especially intercepted arc. Knowing the definition of intercepted arc is critical for understanding this and future lessons. The goal is for . Include a diagram in which students to understand why the Opening Exercise supports but is not a complete proof of point B and angle ABC are the inscribed angle theorem and then to make diagrams of the remaining cases, which are already drawn. addressed in Examples 1-2 and Exercise 1. Opening Exercise (7 minutes)

and are points on a circle with center .

푨1. Draw푪 a point on the circle so that 푶 is a diameter. Then draw the angle . 푩 푨푩�� 2. What angle in your diagram is an inscribed angle? ∠푨푩푪

∠ 푨푩푪 3. What angle in your diagram is a central angle?

∠푨푶푪

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4. What is the intercepted arc of angle ?

Minor arc ∠푨푩푪

푨푪� 5. What is the intercepted arc of ?

Minor arc ∠푨푶푪

푨푪� 6. The measure of the inscribed angle is and the measure of the central angle is . Find in terms of ? 풙 풚 We are풎∠ given푪푨푩 = , and풙 we know that = = , so is an isosceles triangle, meaning is also x. The sum of the angles of a triangle is , so 풎∠푫 풙 푨푩 푨푪 푨푫 ∆푪푨푫 = – . and are supplementary meaning that 풎∠푨푪푫 ퟏퟖퟎ = – ( – ) = ; therefore, = . 풎∠푪푨푫 ퟏퟖퟎ ퟐ풙 ∠푪푨푫 ∠푪푨푩 풎∠푪푨푩 ퟏퟖퟎ ퟏퟖퟎ ퟐ풙 ퟐ풙 풚 ퟐ풙 RELEVANTDRAFT VOCABULARY Inscribed angle theorem (as it will be stated in Lesson 7). The measure of an inscribed angle is half the angle measure of its intercepted arc.

Inscribed angle. An inscribed angle is an angle whose vertex is on a circle, and each side of the angle intersects the circle in another point.

Arc intercepted by an angle. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of the arc. An angle inscribed in a circle intercepts exactly one arc; in particular, the arc intercepted by a right angle is the semicircle in the interior of the angle.

Exploratory Challenge (10 minutes) Scaffolding: Review the definition of intercepted arc, inscribed angle, and central angle and then state . Post drawings of each case the inscribed angle theorem. Then highlight the fact that we have proved one case but as they are studied in the not all cases of the inscribed angle theorem (the case in which a side of the angle passes class as well as the through the center of the circle). Sketch drawings of various cases to set up; for instance, definitions of inscribed Example 1 could be the inside case and Example 2 could be the outside case. You may angles, central angles, and want to sketch them in a place where you can refer to them throughout the class. intercepted arcs.

MP.7 . What do you notice that is the same or different about each of these pictures?  Answers will vary.

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. What arc is intercepted by ?  The minor arc ∠퐴퐵퐶 . How do you know this arc is intercepted by ? 퐴퐶�  The endpoints of the arc ( and ) lie on the angle. ∠퐴퐵퐶  All other points of the arc are inside the angle. 퐴 퐶  Each side of the angle contains one endpoint of the arc. . Today we’re going to talk about the inscribed angle theorem. It says the following: The measure of a central angle is twice the measure of any inscribed angle that intercepts the same arc as the central angle. Does the Opening Exercise satisfy the conditions of the inscribed angle theorem?  Yes. is an inscribed angle and is a central angle both of which intercept the same arc. . What did we show about the inscribed angle and central angle with the same intersected arc in the Opening ∠퐴퐵퐶 ∠퐴푂퐶 Exercise?  DRAFTThat the measure of is twice the measure of .  This is because is an isosceles triangle whose legs are radii. The base angles satisfy ∠퐴푂퐶 ∠퐴퐵퐶 = = . So, = 2 because the exterior angle of a triangle is equal to the sum of the measures of the△ opposite푂퐵퐶 interior angles. 푚∠퐵 푚∠퐶 푥 푦 푥 . Does the conclusion of the Opening Exercise match the conclusion of the inscribed angle theorem?  Yes. . Does this mean we have proven the inscribed angle theorem?  No. The conditions of the inscribed angle theorem say that could be any inscribed angle. The vertex of the angle could be elsewhere on the circle. ∠퐴퐵퐶 . How else could the diagram look?  Students may say that diagram could have the center of the circle be inside, outside, or on the inscribed angle. The Opening Exercise only shows the case when the center is on the inscribed angle. (Discuss with students that the precise way to say “inside the angle” is to say “in the interior of the angle.”)  We still need to show the cases when the center is inside and outside the inscribed angle.

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. There is one more case, when is on the minor arc between and instead of the major arc.  If students ask where the x and y are in this diagram, say we’ll find out in Lesson 7. 퐵 퐴 퐶

DRAFT Examples 1-2 (10 minutes)

These examples prove the second and third case scenarios – the case when the center of the circle is inside or outside the inscribed angle and the inscribed angle is acute. Both use similar computations based on the Opening Exercise. Example 1 is easier to see than Example 2. For Example 2, you may want to let students figure out the diagram on their own, but then go through the proof as a class. Go over proofs of Examples 1-2 with the case when angle is acute. If a student draws so that angle is obtuse, save the diagram for later when doing Lesson 7. ∠퐴퐵퐶 퐵 ∠퐴퐵퐶 Note that the diagrams for the Opening Exercise as well as Examples 1-2 have been labeled so that in each diagram, is the center, is an inscribed angle, and is a central angle. This consistency highlights parallels between the computations in the three cases. 푂 ∠퐴퐵퐶 ∠퐴푂퐶

Scaffolding: Example 1 . Before doing examples 1 and are points on a circle with center . and 2, or instead of going 1. What is the intercepted arc of ? Color it red. 푨 푪 푶 through the examples, Minor arc ∠푪푶푨 teachers could do an exploratory activity using 푨푪� the triangles shown in 2. Draw triangle . What type of triangle is it? Why? Exercise 3 parts (a), (b), An isosceles triangle푨푶푪 because = (they are radii of the same circle). and (c). Have students

푶푪 푶푨 measure the central and inscribed angles, 3. What can you conclude about and ? Why? intercepting the same arc They are equal because base angles풎∠푶푪푨 of an isosceles풎∠푶푨푪 triangle are equal in measure. with a protractor, then log the measurements in a table, and look for a pattern.

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4. Draw a point B on the circle so that is in the interior of the inscribed angle .

푶 ∠푨푩푪

The diagram should resemble the inside case of the discussion diagrams.

5. What is the intercepted arc of angle ? Color it green. DRAFTMinor arc ∠푨푩푪 푨푪�

6. What do you notice about arc ?

It is the same arc that was intercepted푨푪� by the central angle.

7. Let the measure of be and the measure of be . Can you prove that = ? (Hint: Draw the diameter that contains point .) ∠푨푩푪 풙 ∠푨푶푪 풚 Let풚 ퟐ풙 be a diameter. Let , , , and be the measures푩 of , , , and , respectively. We can express and in terms of these measures: �� ퟏ ퟏ ퟐ ퟐ =푩푫 + , and = +풙 .풚 By풙 the opening풚 exercise, ∠푪푩푫 ∠푪푶푫 ∠푨푩푫 =∠푨푶푫 and = . Thus = . 풙 풚 풙 풙ퟏ 풙ퟐ 풚 풚ퟏ 풚ퟐ 풚ퟏ ퟐ풙ퟏ 풚ퟐ ퟐ풙ퟐ 풚 ퟐ풙 8. Does your conclusion support the inscribed angle theorem? Yes, even when the center of the circle is in the interior of the inscribed angle, the measure of the inscribed angle is equal to half the measure of the central angle that intercepts the same arc.

9. If we combine the Opening Exercise and this proof, have we finished proving the inscribed angle theorem? No. We still have to prove the case where the center is outside the inscribed angle.

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Example 2

and are points on a circle with center . 1. Draw a point on the circle so that is in the exterior of the inscribed angle . 푨 푪 푶 푩 푶 ∠푨푩푪

DRAFTThe diagram should resemble the outside case of the discussion diagrams.

2. What is the intercepted arc of angle ? Color it yellow.

Minor arc ∠푨푩푪

푨푪� 3. Let the measure of be , and the measure of be . Can you prove that = ? (Hint: Draw the diameter that contains point .) ∠푨푩푪 풙 ∠푨푶푪 풚 Let 풚 beퟐ풙 a diameter. Let , , , and be the measures푩 of , , , and , respectively. We can express and in terms �� ퟏ ퟏ ퟐ ퟐ of these푩푫 measures: 풙 풚 풙 풚 ∠푪푩푫= ∠푪푶푫 an∠d 푨푩푫= ∠푨푶푫 By the Opening Exercise, = 풙 and 풚 = . Thus, = . 풙 풙ퟐ − 풙ퟏ 풚 풚ퟐ − 풚ퟏ 풚ퟏ ퟐ풙ퟏ 풚ퟐ ퟐ풙ퟐ

풚 ퟐ풙 4. Does your conclusion support the inscribed angle theorem? Yes, even when the center of the circle is in the exterior of the inscribed angle, the measure of the inscribed angle is equal to half the measure of the central angle that intercepts the same arc.

5. Have we finished proving the inscribed angle theorem? We have shown all cases of the inscribed angle theorem (central angle version). We do have one more case to study in Lesson 7, but it is ok not to mention it here. The last case is when the location of is on the minor arc between and . 푩 푨

푪 Ask students to summarize the results of these theorems to each other before moving on. Exercises 1–5 (10 minutes) Exercises are listed in order of complexity. Students do not have to do all problems. Problems can be specifically assigned to students based on ability.

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Exercises 1 – 5

1. Find the measure of the angle with measure . a. = ° b. = ° c. = ° 풙

풎∠푫 ퟐퟓ 풎∠푫 ퟏퟓ 풎∠푩푨푪 ퟗퟎ

x x

DRAFT

° ° °

ퟓퟎ ퟏퟓퟎ ퟒퟓ

d. = ° e. f. = °

풎∠푩 ퟑퟐ 풎∠푫 ퟏퟗ

° = = ° °

ퟓퟖ 풙 풘 ퟔퟎ ퟕퟏ

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2. Toby says is a right triangle because = °. Is he correct? Justify your answer. ∆푩푬푨 풎∠푩푬푨 ퟗퟎ Toby is not correct. The = °. is inscribed in the same arc as the central angle, so it has a measure of °. This means that = ° because the sum of the angles풎∠푩푬푨 of a triangleퟗퟓ ∠ is푩푪푫 °. = since they are vertical angles, so the triangle isퟑퟓ not right. 풎∠푫푬푪 ퟗퟓ ퟏퟖퟎ 풎∠푩푬푨 풎∠푫푬푪

3. Let’s look at relationships between inscribed angles. DRAFTa. Examine the inscribed polygon below. Express in terms of and in terms of . Are the opposite angles in any quadrilateral inscribed in a circle supplementary? Explain. 풙 풚 풚 풙

= ° ; = ° . The angles are supplementary.

풙 ퟏퟖퟎ − 풚 풚 ퟏퟖퟎ − 풙 b. Examine the diagram below. How many angles have the same measure, and what are their measures in terms of ?

풙

Let and be the points on the circle that both the original angles contain. All the angles intercepting the minor arc between and have measure x, and the angles intercepting the major arc between and measure푩 푫 ° . 푩 푫 푩 푫 ퟏퟖퟎ − 풙

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4. Find the measures of the labeled angles.

a. b. .

= , = =

c. 풙 ퟐퟖ 풚 ퟓퟎ d. 풚 ퟒퟖ DRAFT

= = , =

풙 ퟑퟐ 풙 ퟑퟔ 풚 ퟏퟐퟎ e. f.

= = 풙 ퟒퟎ

풙 ퟑퟎ Closing (3 minutes) . With a partner, do a 30-second Quick Write of everything that we learned today about the inscribed angle theorem.

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 Today we began by revisiting the Problem Set from yesterday as the key to the proof of a new theorem, the inscribed angle theorem. The practice we had with different cases of the proof allowed us to recognize the many ways that the inscribed angle theorem can show up in unknown angle problems. We then solved some unknown angle problems using the inscribed angle theorem combined with other facts we knew before. . Have students add the theorems in the Lesson Summary to their graphic organizer on circles started in Lesson 2 with corresponding diagrams.

Lesson Summary

THEOREMS DRAFT• The inscribed angle theorem: The measure of a central angle is twice the measure of any inscribed angle that intercepts the same arc as the central angle.

• Consequence of inscribed angle theorem: Inscribed angles that intercept the same arc are equal in measure. RELEVANT VOCABULARY • INSCRIBED ANGLE. An inscribed angle is an angle whose vertex is on a circle, and each side of the angle intersects the circle in another point. • INTERCEPTED ARC. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle,

and each side of the angle contains an endpoint of the arc. An angle inscribed in a circle intercepts exactly one arc; in particular, the arc intercepted by a right angle is the semicircle in the interior of the angle.

Exit Ticket (5 minutes)

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Name Date

Lesson 5: Inscribed Angle Theorem and its Application

Exit Ticket

The center of the circle below is . If angle has a measure of 15 degrees, find the values of and . Explain how you know. 푂 퐵 푥 푦

DRAFT

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Exit Ticket Sample Solutions

The center of the circle below is . If angle B has a measure of degrees, find the values of and . Explain how you know. 푶 ퟏퟓ 풙 풚 = . Triangle is isosceles, so base angles are congruent. = = . = . is a central angle inscribed in the same arc as inscribed angle . So = . 풚 ퟏퟓ 푪푶푩 ∠푶푪푩 풂풏풅 ∠푶푩푪 풎∠푶푩푪 ퟏퟓ 풎∠푶푪푩 풙 ퟑퟎ ∠푪푶푨 ∠푪푩푨 풎∠푪푶푨 ퟐ풎∠푪푩푨

Problem Set Sample Solutions Problems 1-2 are intended to strengthen students’ understanding of the proof of the inscribed angle theorem. The other problems are applications of the inscribed angle theorem. Problems 3-5 are the most straightforward of these, followed by Problem 6, then Problems 7-9, which combine use of the inscribed angle theorem with facts about triangles, angles, and polygons.DRAFT Finally Problem 10 combines all the above with the use of auxiliary lines in its proof.

Find the value of x in each exercise.

1. 2.

= =

풙 ퟑퟒ 풙 ퟗ

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3. 4.

DRAFT

= =

풙 ퟑퟎ 풙 ퟒퟎ

5. 6.

= =

풙 ퟔퟎ 풙 ퟔ

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7. 8.

DRAFT

= =

풙 ퟐퟎ 풙 ퟒퟔ 9. a. The two circles shown intersect at . The center of the larger circle, , lies on the circumference of the smaller circle. If a chord of the larger circle, , cuts the smaller circle at , find and . 푬 풂풏풅 푭 푫

�푭푮�� 푯 풙 풚

= , =

풙 ퟏퟎퟎ 풚 ퟓ

b. How does this problem confirm the inscribed angle theorem? is a central angle of the larger circle and is double , the inscribed angle of the larger circle. is inscribed in the smaller circle and equal in measure to , also inscribed in the smaller circle. ∠푭푫푬 ∠푭푮푬 ∠푭푫푬 ∠푭푯푬

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10. In the figure below, and intersect at point E.

�� Prove: + 푬푫 푩푪= ( )

풎∠푫푨푩 풎∠푬푨푪 ퟐ 풎∠푩푭푫

Proof: Join .

�푩푬�� = ( ______) ퟏ

풎∠푩푬푫 ퟐ 풎∠

= ( ______) ퟏ 풎∠푬푩푪 ퟐ 풎∠ In ,

DRAFT∆푬푩푭 + = ______

풎∠푩푬푭 풎∠푬푩푭 풎∠

( ______) + ( ______) = ______ퟏ ퟏ ퟐ 풎∠ ퟐ 풎∠ 풎∠ + = ( )

∴ 풎∠푫푨푩 풎∠푬푨푪 ퟐ 풎∠푩푭푫 ; ; ; , ,

푩푨푫 푬푨푪 푩푭푫 푫푨푩 푬푨푪 푩푭푫

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Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles

Student Outcomes . Use the inscribed angle theorem to find the measures of unknown angles. . Prove relationships between inscribed angles and central angles.

Lesson Notes Lesson 6 continuesDRAFT the work of Lesson 5 on the inscribed angle theorem. Many of the problems in Lesson 6 have chords that meet outside of the circle, and students are looking at relationships between triangles formed within circles and finding angles using their knowledge of the inscribed angle theorem and Thales’ theorem. When working on unknown angle problems, present them as puzzles to be solved. Students are to use what is known to find missing pieces of the puzzle until they find the piece asked for in the problem. Calling these puzzles instead of problems will encourage students to persevere in their work and see it more as a fun activity. Scaffolding: Classwork . Create a Geometry Axiom/Theorem wall, Opening Exercise (10 minutes) similar to a Word Wall, so students will have easy Allow students to work in pairs or groups of three and work through the proof below, reference. Allow students writing their work on large paper. Some groups may need more guidance, and others may to create colorful designs need you to model this problem. Call students back together as a class, and have groups and display their work. present their work. Use this as an informal assessment of student understanding. For example, a student Compare work and clear up misconceptions. Also, talk about different strategies groups draws a picture of an used. inscribed angle and a central angle intercepting the same arc and color In a circle, a chord and a diameter are extended outside of the circle to meet at point C. If codes it with the angle = °, and = °, find . 푫푬�� 푨푩�� relationship between the 풎∠푫푨푬 ퟒퟔ 풎∠푫푪푨 ퟑퟐ 풎∠푫푬푨 two noted. Students could be assigned axioms, theorems, or terms to illustrate so that all students would have work displayed. . For advanced learners, present the problem from the Opening Exercise and Let = , ask them to construct the = proof without the guided 풎∠푫푬푨 풚 steps. 풎∠푬푨푬 풙

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In , = Reason angles inscribed in same arc are congruent

∆푨푩푫 풎∠푫푩푨 풚 = 90° Reason angle inscribed in semicircle

풎∠푨푫푩 + + + = 180 Reason sum of angles of triangle = 180°

∴ ퟒퟔ 풙 풚 ퟗퟎ + = 44

풙 풚 MP.3 In , = + Reason Ext. angle of a triangle is equal to the sum of the remote interior angles. ∆푨푪푬 풚 풙 ퟑퟐ + + = 44 Reason substitution

풙 풙 ퟑퟐ = 6

풙 DRAFT = 38

풚 = ° 풎∠푫푬푨 ퟑퟖ Exploratory Challenge (15 minutes)

Display the theorem below for the class to see. Have the students state the theorem in their own words. Lead students through the first part of the proof of the theorem with leading questions, and then divide the class into partner groups. Have half of the groups prove why ’ cannot be outside of the circle and half of the class prove why ’ cannot be inside of the circle, then as a whole class, have groups present their work and discuss. 퐵 퐵 Do the following as a whole class: THEOREM: If , , ’, and are four points with and ’ on the same side of line , and angles and are congruent, then , , ’, and all lie on the 퐴 퐵 퐵 퐶 퐵 퐵 same circle. ⃖퐴퐶��⃗ ∠퐴퐵퐶 ∠퐴퐵′퐶 퐴 퐵 퐵 퐶 . State this theorem in your own words, and write it on a piece of paper. Share it with a neighbor.  If we have 2 points on a circle ( and ), and two points between those two points on the same side ( and ’), and if we draw two angles that are congruent ( 퐴 ), then퐶 all of the points ( , , ’, and ) lie on the same circle. 퐵 퐵 ∠퐴퐵퐶 푎푛푑 ∠퐴퐵′퐶 퐴 퐵 퐵 . Let’s start with points , , and . Draw a circle containing points , , and . 퐶  Students draw a circle with points , , and on the circle. 퐴 퐵 퐶 퐴 퐵 퐶 . Draw . 퐴 퐵 퐶  Students draw . ∠퐴퐵퐶

∠퐴퐵퐶

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles Date: 8/29/14 2

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

. Do we know the measure of ?  No. If students want to measure it, remind them that all circles drawn by their classmates are different, ∠퐴퐵퐶 so we are finding a general case. . Since we don’t know the measure of this angle, assign it to be the variable , and label your drawing.  Students label diagram. 푥 . In the theorem, we are told that there is another point ’. What else are we told about ’?

 ’ lies on the same side of as . 퐵 퐵  퐵 ⃖퐴퐶��⃗ 퐵 . What are we trying to prove? ∠퐴퐵퐶 ≈ ∠퐴퐵′퐶  ’ lies on the circle too. Assign half the class this investigation. Let them work in pairs, and 퐵 provide leadingDRAFT questions as needed. . Let’s look at a case where ’ is not on the circle. Where could ’ lie? 퐵  Outside of the circle or inside the circle. 퐵 . Let’s look at the case where it lies outside of the circle first. Draw B’ outside of your circle and draw .  Students draw ’ and . ∠퐴퐵′퐶

퐵 ∠퐴퐵′퐶 . What is mathematically wrong with this picture?  Answers will vary. We want students to see that the inscribed angle formed where intersects the circle has a measure of since it is inscribed in the same arc �� as not . See diagram.퐴퐵′ 푥 . To further clarify, have students draw the triangle with ∠퐴퐵퐶 ∠퐴퐵′퐶 the inscribed segment as shown. Further discuss what is mathematically incorrect with the angles marked ∆ in퐴퐵 the′퐶 triangle. 푥 . What can we conclude about ’?  ’ cannot lie outside of the circle. 퐵 Assign the other half of the class this investigation. Let them work in 퐵 pairs and provide leading questions as needed. . Where else could ’ lie?  In the circle or on the circle. 퐵 . With a partner, prove that ’ cannot lie inside the circle using the steps above to guide you. 퐵 . Circle around as groups are working, and help where necessary, leading some groups if required. . Call the class back together, and allow groups to present their findings. Discuss both cases as a class.

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles Date: 8/29/14 3

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

. Have students do a 30-second Quick Write on what they just discovered.  If , , ’, and are 4 points with and ’ on the same side of the line , and angles and are congruent, then , , ’, and all lie on the same circle. 퐴 퐵 퐵 퐶 퐵 퐵 ⃖퐴퐶��⃗ ∠퐴퐵퐶 ′ ∠퐴퐵 퐶 퐴 퐵 퐵 퐶 Exercises 1 - 4 (13 minutes) Have students work through the problems (puzzles) below in pairs or homogeneous groups of three. Some groups may need one-on-one guidance. As students complete problems, have them summarize the steps that they took to solve each problem, then post solutions at 5-minute intervals. This will give groups that are stuck hints and show different methods for solving.

Find the value in each figure below, and describe how you arrived at the answer. DRAFT 풙 1. Hint: Thales’ theorem 2.

= ° inscribed in a semicircle = = °

풎∠푩푬푪 = ퟗퟎ = ° base angles of an isosceles 풎∠corresponding푪푩푬 풎∠ angles푪푨푫 areퟑퟒ congruent

풎∠triangle푬푩푪 are 풎∠congruent푬푪푩 andퟒퟓ sum of angles of a triangle = ° linear pair with

= ° 풎∠푩푨푫 ퟏퟒퟔ = ∠=푪푨푫° inscribed ퟏ ퟏퟖퟎ = = ° angles inscribed in the same arc angle풎∠푨푫푬 is halfퟐ of풎∠ measure푩푨푫 ofퟕퟑ central angle

are풎∠ 푬푩푪congruent풎∠ 푬푫푪 ퟒퟓ intercepting same arc = =

풙 ퟒퟓ 풙 ퟕퟑ

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles Date: 8/29/14 4

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

3. 4. DE = EF = FC

DRAFT

= = = ° inscribed angles are half ퟏ the measure of the central angle intercepting the same arc 풎∠푩푬푪 풎∠푪푭푩 ퟐ 풎∠푩푨푪 ퟓퟐ Draw center of circle O. = linear pair with = = ° central angle doubleퟎ 풎∠푫푬푮 = ퟏퟐퟖ° linear pair with ∠푩푬푪 풎∠measure푬푶푭 of inscribedퟐ풎∠푬푪푭 angleퟔퟎ intercepting same arc 풎∠푮푭푫 = ퟏퟐퟖ° sum of angles of a∠ quadrilateral푪푭푩 = = = ° 풎∠푬푮푭is 360° ퟕퟒ 풎∠angles푫푶푬 inscribed풎∠푬푶푭 in 풎∠congruent푭푶푪 chordsퟔퟎ are ≅ x = 74 = ° sum of angles of a circle = 360° 풎∠=푫푶푪 angleퟏퟖퟎ inscribed in a semicircle 풙 ퟗퟎ

Closing (2 minutes) Have students do a 30-second Quick Write of what they have learned about the inscribed angle theorem. Bring the class back together and debrief. Use this as a time to informally assess student understanding and clear up misconceptions. . Write all that you have learned about the inscribed angle theorem.  The measure of the central angle is double the measure of any inscribed angle that intercepts the same arc.  Inscribed angles that intercept the same arc are congruent.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

Lesson Summary:

THEOREMS

• The inscribed angle theorem: The measure of a central angle is twice the measure DRAFTof any inscribed angle that intercepts the same arc as the central angle

• Consequence of inscribed angle theorem: Inscribed angles that intercept the same arc are equal in measure.

• If A, B, B’, and C are four points with B and B’ on the same side of line , and angles and are congruent, then A, B, B’, and C all lie on the same circle. 퐴퐶⃖��⃗ RELEVANT∠퐴퐵퐶 VOCABULARY∠퐴퐵′퐶 • CENTRAL ANGLE. A central angle of a circle is an angle whose vertex is the center of a circle.

• INSCRIBED ANGLE. An inscribed angle is an angle whose vertex is on a

circle, and each side of the angle intersects the circle in another point.

• INTERCEPTED ARC. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of the arc. An angle inscribed in a circle intercepts exactly one arc; in particular, the arc intercepted by a right angle is the semicircle in the interior of the angle.

Exit Ticket (5 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

Name Date

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles

Exit Ticket

Find the measure of angles and . Explain the relationships and theorems used.

푥 푦 DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

Exit Ticket Sample Solutions

1. Find the measures of angles and . Explain the relationships and theorems used.

= (linear pair with풙 풚 ). = = ° (inscribed angle is half measure of central ퟏ angle with sameퟎ intercepted arc). = = = ° (corresponding angles are congruent) 풎∠푬푨푪 ퟒퟐ ∠푩푨푬 풎∠푬푭푪 ퟐ 풎∠푬푨푪 ퟐퟏ = 풙 ퟐퟏ 풎∠푨푩푫 풎∠푬푨푪 ퟒퟐ

풚 ퟒퟐ

Problem Set Sample Solutions The first two problems are easier and require straightforward use of the inscribed angle theorem. The rest of the problems vary in difficulty but could be time consuming. Consider allowing students to choose the problems that they do and assigning a number of problems to be completed. You may want everyone to do Problem 8, as it is a proof with some parts of stepsDRAFT given as in the Opening Exercise.

In Problems 1–5, find the value .

풙 1.

ퟒퟎ ퟓ 2.

ퟓퟕ

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles Date: 8/29/14 8

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

DRAFT

ퟏퟓ

ퟑퟏ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

DRAFT

ퟗퟎ 6. If = , express y in terms of .

푩푭 푭푪 풙

풚 ퟐ풙

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

7. a. Find the value .

풙

DRAFT

풙 ퟗퟎ b. Suppose the = °. Prove that = °.

= (alt.angles풎∠푪 congruent),풂 =풎∠ 푫푬푩 (inscribedퟑ풂 angles half the central angle, + + = (angles of triangle = 180), = – , + = (angles form line), – + 푫 풂 ∠푨 ퟐ풂 풂 ퟐ풂 ∠푨푬푫 ퟏퟖퟎ = (substitution), = ∠푨푬푫 ퟏퟖퟎ ퟑ풂 ∠푨푬푫 ∠푩푬푫 ퟏퟖퟎ ퟏퟖퟎ ퟑ풂 ∠푩푬푫 ퟏퟖퟎ 푩푬푫 ퟑ풂

8. In the figure below, three identical circles meet at B, F and C, E respectively. = . , , and , , lie on straight lines. 푩푭 푪푬 푨 푩 푪 푭 푬 푫

Prove is a parallelogram.

푨푪푫푭

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M5 GEOMETRY

Proof:

DRAFT

Join and .

푩푬 푪푭 = Reason: ______

푩푭 푪푬 a = ______= ______= ______= d Reason: ______

______= ______

Alternate angles are equal.

푨푪 ��‖푭푫�� ______= ______

Corresponding angles are equal.

푨푭 ��‖푩푬�� ______= ______

Corresponding angles are equal.

푩푬 ��‖푪푫��

푨푭 ��‖푩푬��‖푪푫�� is a parallelogram.

푨푪푫푭 Given; b, f, e, angles inscribed in congruent arcs are congruent; = ; = ; =

∠푪푩푬 ∠푭푬푩 ∠푨 ∠푪푩푬 ∠푫 ∠푩푭

Lesson 6: Unknown Angle Problems with Inscribed Angles in Circles Date: 8/29/14 12

GEOMETRY • MODULE 5

Topic B: Arcs and Sectors

G-C.A.1, G-C.A.2, G-C.B.5 Focus Standard:DRAFT G-C.A.1 Prove that all circles are similar. G-C.A.2 Identify and describe relationships among inscribed angles, radii, and chords. G-C.B.5 Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Instructional Days: 4 Lesson 7: The Angle Measure of an Arc Lesson 8: Arcs and Chords Lesson 9: Arc Length and Areas of Sectors Lesson 10: Unknown Length and Area Problems

In Topic B, students continue studying the relationships between chords, diameters, and angles and extend that work to arcs, arc length, and areas of sectors. Students use previous knowledge of inscribed and central angles and use that structure and repeated reasoning to develop an understanding of circles, secant lines, and tangent lines (MP.7). The module begins with Lesson 7 as students revisit the inscribed angle theorem, this time stating it in terms of inscribed arcs (G-C.A.2). This concept is extended to studying similar arcs which leads students to understand that all circles are similar (G-C.A.1). Students then look at the relationships between chords and subtended arcs and prove congruent chords lie in congruent arcs. They also prove that arcs between parallel lines are congruent using transformations (G-C.A.2). Lessons 9 and 10 switch the focus from angles to arc length and areas of sectors. Students use formulas learned previously about the area and circumference of circles and combine those with concepts learned in this module to determine arc length, areas of sectors, and similar triangles (G-C.B.5). In Lesson 9, students are introduced to radians as the ratio of arc length to the radius of a circle. Lesson 10 reinforces these concepts with problems involving unknown length and area. Topic B requires that students use and apply previous knowledge to see the structure in new applications and see the repeated patterns in these problems to arrive at theorems relating chords, arcs, angles, secant lines, and tangent lines to circles (MP.7). For example, students know that an inscribed angle has a measure of half the central angle intercepting the same arc. When they discover that the measure of a central angle is equal to the angle measure of its intersected arc, they conclude that the measure of an inscribed angle is half the angle of its intercepted arc. Students then conclude that congruent arcs have congruent chords and arcs between parallel chords are congruent by applying previously studied concepts.

Topic B: Arcs and Sectors Date: 8/29/14 1

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Lesson 7: The Angle Measure of an Arc

Student Outcomes . Define the angle measure of arcs, and understand that arcs of equal angle measure are similar. . Restate and understand the inscribed angle theorem in terms of arcs: The measure of an inscribed angle is half the angle measure of its intercepted arc. . Explain and understand that all circles are similar.

Lesson NotesDRAFT Lesson 7 introduces the angle measure of an arc and finishes the inscribed angle theorem. Only in this lesson is the inscribed angle theorem stated in full: “The measure of an inscribed angle is half the angle measure of its intercepted arc.” When we state the theorem in terms of the intercepted arc, the requirement that the intercepted arc is in the interior of the angle that intercepts it guarantees the measure of the inscribed angle is half the measure of the central angle. In Lesson 7, we will also calculate the measure of angles inscribed in obtuse angles. Lastly, we will address G- C.A.1 and show that all circles are similar, a topic previously covered in Module 2, Lesson 14.

Classwork Opening Exercise (5 minutes) This Opening Exercise reviews the relationship between inscribed angles and central angles, concepts that need to be solidified before we introduce the last part of the inscribed angle theorem (arc measures). Have students complete the exercise individually and compare answers with a partner, then pull the class together to discuss. Be sure students can identify inscribed and central angles.

If the measure of is °, name 3 other angles that have the same measure and explain why. ∠푮푩푭 ퟏퟕ Answers will vary. , , , , they are all congruent because they are inscribed in the same arc. ∠푮푯푭 ∠푮푪푭 ∠푮푫푭 ∠푮푬푭 MP.3 What is the measure of ? Explain.

°; it is the central angle∠푮푨푭 with an inscribed arc of °. The measure of the central angle is double the measure of the inscribedퟑퟒ angle of the same arc. ퟏퟕ

Can you find the measure of ? Explain.

°; and are vertical∠푩푨푫 angles and are congruent. ퟑퟒ ∠푩푨푫 ∠푮푨푭

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Discussion (15 minutes)

This lesson begins with a full class discussion that ties what students know about central Scaffolding: angles and inscribed angles to relating these angles to the arcs they are inscribed in, which . Provide students a copy of is a new concept. In this discussion, we will define some properties of arcs. As properties the picture if they have are defined, list them on a board or large paper so students can see them. difficulty creating the . We have studied the relationship between central angles and inscribed angles. drawing. Can you help define an inscribed angle? . As definitions are given and as new terms are  An angle is inscribed in an arc if the sides of the angle contain the added, have students endpoints of the arc; the vertex of the angle is a point on the circle, but repeat each definition and not an endpoint on the arc. term aloud. . Can you help me define a central angle? . Create a definition wall.  Answers will vary. A central angle for a circle is an angle whose vertex is DRAFTat the center of the circle. . Let’s draw a circle with an acute central angle.  Students draw a circle with an acute central angle. . Display the picture below.

Scaffolding:

. Provide students a graphic organizer for arcs (major and minor). . For ELL students, do choral repetition of major and

minor arcs using hand gestures (arms wide for major and close together . How many arcs does this central angle divide this circle into? for minor).  2 . What do you notice about the two arcs?  One is longer than the other is. One arc is contained in the angle and its interior, and one arc is contained in the angle and its exterior. (Students might say “inside the angle” or “outside the angle.” Help them to state it precisely.) . In a circle with center , let and be different points that lie on the circle but are not the endpoints of a diameter. The minor arc between and is the set cont푂aining퐴 , , 퐵and all points of the circle that are in the interior of . 퐴 퐵 퐴 퐵 . Explain to your neighbor what a minor arc is, and write the definition. ∠퐴푂퐵

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. The way we show a minor arc using mathematical symbols is ( with an arc over them). Write this on your drawing. 퐴퐵� 퐴퐵 . Can you predict what we call the larger arc?  The major arc. . Now, let’s write the definition of a major arc.  In a circle with center , let and be different points that lie on the circle but are not the endpoints of a diameter. The major arc is the set containing , , and all points of the circle that lie in the exterior of . 푂 퐴 퐵 퐴 퐵 . Can we call it ? ∠퐴푂퐵  No, because we already called the minor arc . 퐴퐵� . We would write the major arc as where X is any point on the circle outside of the central angle. Label the 퐴퐵� major arc. 퐴푋퐵� . Can youDRAFT define a semicircle in terms of arc?  In a circle, let and be the endpoints of a diameter. A semicircle is the set containing , , and all points of the circle that lie in a given half-plane of the line determined by the diameter. 퐴 퐵 퐴 퐵 . If I know the measure of , what do you think the angle measure of is?  The same measure. ∠퐴푂퐵 퐴퐵� . Let’s say that statement.  The angle measure of a minor arc is the measure of the corresponding central angle. . What do you think the angle measure of a semicircle is? Why?  180°. It is half a circle, and a circle measures 360°. . Now let’s look at . If the angle measure of is 20°, what do you think the angle measure of would be? Explain. 퐴푋퐵� 퐴퐵� 퐴푋퐵�  340° because it is the other part of the circle not included in the 20°. Since a full circle is 360°, the part not included in the 20° would equal 340°. . Discuss what we have just learned about minor and major arcs and semicircles and their measures with a partner. . Look at the diagram. If is 20°, can you find the angle measure of and ? Explain. 퐴퐵�  All are 20° because they all have the same 퐶퐷� 퐸퐹� central angle. The circles are dilations of each other, so angle measurement is conserved and the arcs are similar. . We are discussing angle measure of the arcs, not length of the arcs. Angle measure is only the amount of turning that the arc represents, not how long the arc is. Arcs of different lengths can have the same angle measure. Two arcs (of possibly different circles) are similar if they have the same angle measure. Two arcs in the same or congruent circles are congruent if they have the same angle measure. . Explain these statements to your neighbor. Can you prove it?

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

. In this diagram, I can say that and are adjacent. Can you write a definition of adjacent arcs? 퐵퐶� 퐶퐷�  Two arcs on the same circle are adjacent if they share exactly one endpoint. . If = 25° and = 35°, what is the angle measure of ? Explain. 퐵퐶� 퐶퐷� 퐵퐷�  60°. Since they were adjacent, together they create a larger arc whose angle measures can be added together. Or, calculate the measure of the arc as 360–(25 + 35), as the representation could be from to without going through point C. 퐵 . This is a parallel to the 180 protractor axiom (angles add). If AB 퐷 and BCDRAFT are adjacent arcs, then = + . . Central angles and inscribed angles intercept arcs on a circle. An angle intercepts an arc if the endpoints of the 푚퐴퐶� 푚퐴퐵� 푚퐵퐶� arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of the arc. . Draw a circle and an angle that intercepts an arc and an angle that does not. Explain your drawing to your neighbor.  Answers will vary.

. Tell your neighbor the relationship between the measure of a central angle and the measure of the inscribed angle intercepting the same arc.  The measure of the central angle is double the measure of any inscribed angle that intercepts the same arc. . Using what we have learned today, can you state this in terms of the measure of the intercepted arc?  The measure of an inscribed angle is half the angle measure of its intercepted arc. The measure of a central angle is equal to the angle measure of its intercepted arc.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Example 1 (8 minutes)

This example extends the inscribed angle theorem to obtuse angles; it also shows the relationship between the measure of the intercepted arc and the inscribed angle. Students will need a protractor.

Example 1: What if we started with an angle inscribed in the minor arc between and ?

푨 푪

DRAFT

. Draw a point on the minor arc between and .  Students draw point B. 푩 푨 푪 . Draw the arc intercepted by ? Make it red in your diagram. ∠푨푩푪  Students draw the arc and color it red. . In your diagram, do you think the measure of an arc between and is half of the measure of the inscribed angle? Why or why not? 푨 푪  The phrasing and explanations can vary. However, there is one answer; the measure of the inscribed arc is twice the measure of the inscribed angle. . Using your protractor, measure . Write your answer on your diagram. ∠퐴퐵퐶  Answers will vary. . Now measure the arc in degrees. Students may struggle with this, so ask… . Can you think of an easier way to measure this arc in degrees?  We could measure , and then subtract that measure from 360° . Write the measure of the arc in degrees on your diagram. ∠퐴푂퐶 . Do your measurements support the inscribed angle theorem? Why or why not?  Yes, the measure of the inscribed angle is half the measure of its intercepted arc. . Compare your answer and diagram to your neighbor’s answer and diagram. . Restate the inscribed angle theorem in terms of intercepted arcs.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

 The measure of an inscribed angle is half the angle measure of its intercepted arc.

Exercises 1–4 (5 minutes) Scaffolding: : : : = : : : 1. In circle , . Find . Help students see the a. 푨 푩푪� 푪푬� 푬푫� �푫푩� ퟏ ퟐ ퟑ ퟒ connection between the 36° 풎∠푩푨푪 central angles and inscribed angle by color-

coding. b. . Outline central angle in 108° 풎∠ 푫푨푬 red. . Outline the inscribed angle

that has the same c. intercepted arc as the DRAFT144° 풎 푫푩� central angle in blue.

180° 풎푪푬푫�

2. In circle B, AB = CD. Find

° 풎푪푫� ퟔퟎ b.

° 풎 푪푨푫� ퟑퟎퟎ c.

° 풎 푨푫� ퟏퟖퟎ 3. In circle A, is a diameter and = °. If = , find

a. 푩푪�� 풎∠푫푨푪 ퟏퟎퟎ 풎푬푪� ퟐ풎푩푫�

° 풎∠푩푨푬 ퟐퟎ b.

° 풎 푬푪� ퟏퟔퟎ c.

° 풎 푫푬푪� ퟐퟔퟎ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

4. Given circle A with = °, find

a. 풎∠푪푨푫 ퟑퟕ

° 풎 푪푩푫� ퟑퟐퟑ b.

. °풎∠ 푪푩푫 ퟏퟖ ퟓ c.

. 풎∠° 푪푬푫 DRAFTퟏퟔퟏ ퟓ

Example 2 (4 minutes)

In this example, students will ponder the question, “Are all circles similar?” This is intuitive but easy to show, as the ratio of the circumference of two circles is equal to the ratio of the diameters and the ratio of the radii. . Project the circle at right on the board. . What is the circumference of this circle in terms of radius, r?  2

. What if we휋푟 double the radius, what is the circumference?  2 (2 ) = 4

. What if we휋 triple푟 the 휋original radius, what is the circumference?  2 (3) = 6

. What determines휋 푟 the휋 circumference of a circle? Explain.  The radius. The only variable in the formula is r (radius), so as radius changes, the size of the circle changes. . Does the shape of the circle change? Explain.  All circles have the same shape; they are just different sizes depending on the length of the radius. . What does this mean is true of all circles?  All circles are similar.

Closing (3 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Call class together, and show the diagram. . Express the measure of the central angle and the inscribed angle in terms of the angle measure .  The central angle has a measure of °. ∘ 푥 °  The inscribed angle∠ 퐶퐴퐷 has a measure of푥 . 1 . State the inscribed angle theorem∠퐶퐵퐷 to your neighbor. 2 푥  The measure of an inscribed angle is half the angle measure of its intercepted arc.

Lesson Summary

THEOREMS DRAFT• INSCRIBED ANGLE THEOREM: The measure of an inscribed angle is half the measure of its intercepted arc. • Two arcs (of possibly different circles) are similar if they have the same angle measure. Two arcs in the same or congruent circles are congruent if they have the same angle measure.

• All circles are similar. RELEVANT VOCABULARY • ARC. An arc is a portion of the circumference of a circle.

• MINOR AND MAJOR ARC. Let be a circle with center , and let and be different points that lie on but are not the endpoints of the same diameter. The minor arc is the 푪set containing , , and 푶all points of푨 that푩 푪 are in the interior of . The major arc is the set containing , , and all points of that lie in the exterior of 푨. 푩 푪 ∠푨푶푩 푨 푩 • SEMICIRCLE.푪 In a circle, let and be ∠the푨푶푩 endpoints of a diameter. A semicircle is the set containing , , and all points of the circle that lie in a given half-plane of the line determined by the diameter퐴 . 퐵 퐴 퐵 • INSCRIBED ANGLE. An inscribed angle is an angle whose vertex is on a circle and each side of the angle intersects the circle in another point.

• CENTRAL ANGLE. A central angle of a circle is an angle whose vertex is the center of a circle.

• INTERCEPTED ARC OF AN ANGLE. An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the

interior of the angle, and each side of the angle contains an endpoint of the arc. Exit Ticket (5 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Name Date

Lesson 7: Properties of Arcs

Exit Ticket

1. Given circle A with diameters and and = 56°. �퐵퐶�� 퐷퐸��

푚퐶퐷� a. Name a central angle. DRAFT b. Name an inscribed angle.

c. Name a chord that is not a diameter.

d. What is the measure of ?

∠퐶퐴퐷 e. What is the measure of ?

∠퐶퐵퐷 f. Name 3 angles of equal measure.

g. What is the degree measure of ?

퐶퐷퐵�

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

Exit Ticket Sample Solutions

1. Given circle A with diameters and and = °

�� a. Name a central angle. 푩푪 푫푬 풎푪푫 ퟓퟔ

∠푪푨푫 b. Name an inscribed angle. Answers will vary. .

∠푪푬푫 c. Name a chord that is not a diameter. Answers will vary. . DRAFT 푪푬�� d. What is the measure of ?

° ∠푪푨푫

ퟓퟔ e. What is the measure of ?

° ∠푪푩푫

ퟐퟖ⁰ f. Name 3 angles of equal measure. = =

풎∠푪푬푫 풎∠푪푭푫 풎∠푪푩푫 g. What is the degree measure of ?

° 푪푫푩� ퟏퟖퟎ Problem Set Sample Solutions The first two problems are easier and require straightforward use of the inscribed angle theorem. The rest of the problems vary in difficulty, but could be time consuming. Consider allowing students to choose the problems that they do and assigning a number of problems to be completed. You may want everyone to do Problem 8, as it is a proof with some parts of steps given as in the Opening Exercise.

1. Given circle A with = ,

풎∠푪푨푫 ퟓퟎ⁰ a. Name a central angle.

∠b.푪푨푫 Name an inscribed angle.

∠푪푩푫

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

c. Name a chord. Answers will vary.

d. Name a minor푩푫�� arc.��

Answers will vary.

푪푫� e. Name a major arc.

푪푩푫� f. Find ,

° 풎푪푫�

ퟓퟎ g. Find . DRAFT° � 풎푪푩푫

ퟑퟏퟎ h. Find .

° 풎∠푪푩푫

ퟐퟓ 2. Given circle A, find the measure of each minor arc.

= °

풎푩푪� = ퟔퟒ°

풎푪푫� = ퟔퟒ °

풎푪푬� =ퟏퟏퟔ °

풎푩푫� ퟏퟏퟔ

3. Given circle A, find a.

° 풎∠푩푨푫 ퟏퟎퟎ b.

° 풎∠푪푨푩 ퟖퟎ c.

° 풎푩푪� ퟖퟎ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

° 풎푩푫� ퟏퟎퟎ e.

° 풎푩푪푫� ퟐퟔퟎ 4. Find the angle measure of angle .

풙

ퟑퟑ

DRAFT

5. In the figure, = ° and = °. Find .

풎∠푩푨푪 ퟏퟐퟔ 풎∠푩푬푫 ퟑퟐ 풎∠푫푬푪

°°

ퟖퟓ

6. In the figure = °, and = . K is the midpoint of and J is the midpoint of . Find and .ퟎ 풎∠푩푪푫 ퟕퟒ 풎∠푩푫푪 ퟒퟐ 푪푩�

푩푫� 풎∠푲푩푫 풎∠푪푲푱 Solution: Join , , , , , and .

푩푲 푲푪 푲푫 푲푱 푱푪 푱푫 = ______

풎푩푲� 풎푲푪� = = ° ______ퟒퟐ 풎∠푲푫푪 ퟐ ퟐퟏ = ______

풂 In , = ______

∆푩푪푫 풃

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 7 M5 GEOMETRY

= ______

풄 = ______

풎푩푱� 풎푱푫� = ______

풎∠푱푪푫 = ______

풅 = + = ______

풎∠푲푩푫 풂 풃 = + = ______

풎∠푪푲푱 풄 풅 Midpoint forms congruent arcs; angle bisector; °, congruent angles inscribed in same arc; °, sum of angles of DRAFTtriangle = °; °, congruent angles inscribed in same arc; °, angle bisector; °, congruent angles inscribed in same arc; °; °. ퟐퟏ ퟔퟒ ퟏퟖퟎ ퟔퟒ ퟑퟕ ퟑퟕ ퟖퟓ ퟗퟏ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

Lesson 8: Arcs and Chords

Student Outcomes . Congruent chords have congruent arcs, and the converse is true. . Arcs between parallel chords are congruent.

Lesson Notes In this lesson, students use concepts studied earlier in this module to prove three new concepts: Congruent chords have congruent arcs; congruent arcs have congruent chords; arcs between parallel chords are congruent. The proofs are designed for studentsDRAFT to be able to begin independently, so this is a great lesson to allow students the freedom to try a proof with little help getting started. This lesson that highlights MP.7 as students study different circle relationships and draw auxiliary lines and segments. MP.1 and MP.3 are also highlighted as students attempt a series of proofs without initial help from the teacher. Classwork Opening Exercise (5 minutes) The Opening Exercise reminds students of our work in Lesson 2 relating circles, chords, and radii. It sets the stage for Lesson 8. Have students try this exercise on their own, then compare answers with a neighbor, particularly the explanation of their work. Bring the class back together, and have a couple of students present their work and do a quick review.

Given circle with , = , and = . Find and . Explain your work.

= , 푨= . � 푩푪�� ⊥ 푫푬�� 푭푨 ퟔ 푨푪 ퟏퟎ 푩푭 푫푬 MP.7 , = – = & 푩푭 is aퟒ radius푫푬 withퟏퟔ a measure of . If = then . MP.1 Connect푨푩 and . In , ퟏퟎ = 푭푨 = ퟔ . Both푩푭 ퟏퟎ andퟔ ퟒ are right triangles and congruent, so by the Pythagorean theorem, 푨푫 푨푬 ∆푫푨푬 푨푫 푨푬 ퟏퟎ ∆푫푭푨 ∆푬푭푨 = = , making = .

푫푭 푭푬 ퟖ 푫푬 ퟏퟔ

Exploratory Challenge (12 minutes) Scaffolding: In this example, students will use what they have learned about the relationships between . Display the theorems, chords, radii, and arcs to prove that congruent chords have congruent arcs and congruent axioms, and definitions minor arcs have congruent chords. They will then extend this to include major arcs. We that we have studied in are presenting the task and then letting students think through the first proof. Give this module on a word students time to struggle and talk with their groups. This is not a difficult proof and can be wall. done with concepts from Lesson 2 that they are familiar with or by using rotation. Once . Chorally repeat the groups finish and talk about the first proof, they then do two more proofs that are similar. definitions as a class.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

Walk around, and give help where needed, but not too quickly. Display the picture below to the class. Scaffolding:

. If groups are struggling with the proof, give them the following leading questions and steps: . Draw a picture of the problem. . Draw two triangles, one joining the center to each chord.

. What is true about the sides of the triangle DRAFTconnected to the center of

the circle? . Are the triangles congruent? How? . Tell me what you see in this diagram. . What does that mean  A circle, a chord, a minor arc, a major arc. about the central angles? . What do you notice about the chord and the minor arc? . If we say the central angle has a measure of x°, what  They have the same endpoints. is the measure of each . We say that arc is subtended by chord . Can you repeat that with me? chord? . Think about what we know  Arc is� subtended by chord . �� 퐴퐵 퐴퐵 about rotating figures, will . What do you think we mean by the word “subtended”? 퐴퐵� �퐴퐵�� that help us?  The chord cuts the circle and forms the arc. The chord and arc have the . Try drawing a picture. same endpoints. . Display circle at right. What can we say about arc ?  Arc is subtended by chord . 퐶퐷� . If AB = CD, what do you think would be true about and ? 퐶퐷� �퐶퐷��  They are equal (congruent). 푚퐴퐵� 푚퐶퐷� Put students in heterogeneous groups of three, and present the task. Set up a 5-minute check to be sure that groups are on the right path and to give ideas to groups who are struggling. Have groups show their work on large paper or poster board and display work, then have a whole class discussion showing the various ways to achieve the proof. . With your group, prove that if the chords are congruent, the arcs subtended by those chords are congruent.  Some groups will use rotations and others similar triangles similar to the work that was done in Lesson 2. Both ways are valid and sharing will expose students to each method. . Now prove that in a circle congruent minor arcs have congruent chords.

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 Students should easily see that the process is almost the same, and that it is indeed true. . Do congruent major arcs have congruent chords too?  Since major arcs are the part of the circle not included in the minor arc, if minor arcs are congruent, 360° minus the measure of the minor arc will also be the same.

Exercise 1 (5 minutes) Have students try Exercise 1 individually, and then do a pair-share. Wrap up with a quick whole class discussion.

1. Given circle with m(BC) ̂= ° and , find . Explain your work. 푨 ퟓퟒ ∠푪푫푩 ≅ ∠푫푩푬 풎푫푬� = ° . = ° because the central angle has the same DRAFTmeasure as its subtended arc. = ° because an inscribed angle has � 풎half푫푬 the measureퟓퟒ 풎∠ of푪푨푩 the centralퟓퟒ angle with the same inscribed arc. Since , = ° because풎∠푪푫푩 it is ퟐퟕdouble the angle inscribed in it.

∠ 푪푫푩 ≅ ∠푫푩푬 풎푫푬� ퟓퟒ

Example 1 (5 minutes)

In this example, students prove that arcs between parallel chords are congruent. This is a teacher led example. Students will need a compass and Scaffolding: straight edge to construct a diameter and a copy of the circle below. . For advanced learners, display the picture below,

and ask them to prove the Display the picture at right to the class. theorem without the provided questions, and . What do you see in this diagram? then present their proofs  A circle, two arcs, a pair of parallel chords in class. . What looks to be true about the arcs?  They appear to be congruent. . This is true, and here is the theorem: In a circle, arcs between parallel chords are congruent. . Repeat that with me.  In a circle, arcs between parallel chords are congruent. . Let’s prove this together. Construct a diameter perpendicular to the parallel chords.  Students construct the perpendicular diameter. . What does this diameter do to each chord?  The diameter bisects each chord. . Reflect across the diameter (or fold on the diameter). What happens to the endpoints?

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

 The reflection takes the endpoints on one side to the endpoints on the other side. It therefore takes arc to arc. Distances from the center are preserved. . What have we proven?  Arcs between parallel chords are congruent. . Draw . Can you think of another way to prove this theorem using properties of angles formed by parallel lines? �퐶퐷��  = because alternate interior angles are congruent. This means m = m both have inscribed angles of the same measure, so the arc angle measures are congruent and twice the � � measurem∠BCD of m∠EDCtheir inscribed angles. CE BD

Exercise 2 (5 minutes) Have students DRAFTwork on Exercise 2 in pairs. This exercise requires use of all concepts Scaffolding: studied today. Pull the class back together to share solutions. Use this as a way to assess . Review the properties of student understanding. quadrilaterals. . For students who can’t 2. If two arcs in a circle have the same measure, what can you say about the quadrilateral picture the figure, provide formed by the four endpoints? Explain. them with

If the arcs are congruent, their endpoints can be joined to form chords that are parallel ( .).

푩푪��‖푫푬�� The chords subtending the congruent arcs are congruent ( ).

A quadrilateral with one pair of opposite sides parallel and�푩푫� �the��‖푪푬� other�� pair of sides congruent is an isosceles trapezoid.

Exercises 3 - 5 (5 minutes)

3. Find the angle measure of and .

= , = ° 푪푫� 푬푫� ퟎ 퐦퐂퐃� ퟏퟑퟎ 퐦퐄퐃� ퟓퟎ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

4. = and : = : . Find a. 풎푪푩 � 풎푬푫� 풎푩푫� 풎푬푪� ퟐ ퟑ 풎∠24° 푩푪푭

풎∠144°푬푫푭

168풎∠°푪푭푬 5. DRAFT is a diameter of circle A. : : = : : . Find �푩푪a.�� 풎푩푫� 풎푫푬� 풎푬푪� ퟏ ퟑ ퟓ 풎20푩푫°�

160풎푫푪�°

280풎푬푩�°

Closing (3 minutes) Have students do a 30-second Quick Write on what they have learned in this lesson about chords and arcs. Pull the class together to review, and have them add these to the circle graphic organizer started in Lesson 2. . Congruent chords have congruent arcs. . Congruent arcs have congruent chords. . Arcs between parallel chords are congruent.

Lesson Summary

THEOREMS

. Congruent chords have congruent arcs. . Congruent arcs have congruent chords. . Arcs between parallel chords are congruent.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

Exit Ticket (5 minutes) DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

Name Date

Lesson 8: Arcs and Chords

Exit Ticket

1. Given circle with radius 10, prove = .

퐴 퐵퐸 퐷퐶

DRAFT

2. Given the circle at right, find .

푚퐵퐷�

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

Exit Ticket Sample Solutions

1. Given circle A with radius 10, prove = .

= (vertical angles푩푬 are congruent)푫푪

풎∠푩푨푬= 풎∠ 푫푨푪(arcs are equal in degree measure to their inscribed central angles)

풎푩푬�= 풎 푫푪(chords� are equal in length if they subtend congruent arcs)

푩푬 푫푪

2. Given the circle at right, find .

° 풎푩푫�

ퟔퟎ DRAFT Problem Set Sample Solutions Problems 1–3 are straightforward and easy entry. Problems 5–7 are proofs and may be challenging for some students. You may consider only assigning some problems or allowing student choice while requiring some problems of all students.

1. Find

풎푪푬�° ퟕퟎ b.

풎푩푫�° ퟕퟎ c.

풎푪푫�° ퟒퟎ

2. In circle A, is a diameter, = , and = °. 푩푪�� 풎푪푬� 풎푬푫� 풎∠푪푨푬

ퟑퟐ a. Find .

° 풎∠푪푨푫

ퟑퟐ b. Find .

° 풎∠푨푫푪

ퟓퟖ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

3. In circle A, is a diameter, 2 = , and . Find . 푩푪�� 풎푪푬� 풎푬푫� 푩푪��‖푫푬�� . ° 풎∠푪푫푬

ퟐퟐ ퟓ

DRAFT

4. In circle A, is a diameter and = ° .

푩푪�� 푪푬� ퟔퟖ a. Find . ° 풎푪푫�

ퟔퟖ

b. Find . ° 풎∠푫푩푬

ퟔퟖ

c. Find ° 풎∠푫푪푬

ퟏퟏퟐ

5. In the circle given, . Prove

Join . 푩푪� ≅ 푬푫� 푩푬�� ≅ �푫푪�� BC = ED (congruent arcs have chords equal in length) �푪푬�� = (angles inscribed in same arc are equal in measure) = (angles inscribed in congruent arcs are equal in measure풎∠푪푩푬) 풎∠푬푫푪 풎∠푩푪푬 풎∠푫푬푪 (AAS)

∆푩푪푬 ≅ ∆ (corresponding푫푬푪 sides of congruent triangles are congruent) 푩푬 �� ≅ �푫푪��

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M5 GEOMETRY

6. Given circle A with , show .

Join , , 푨푫�� =�� ‖푪푬�� = = 푩푫� (radii)≅ 푫푬� , , (base angles of isosceles triangles are �� congruent푩푫 푫푬) 푨푬푨푪 푨푬 푨푫(Alternate푨푩 interior ∠angles푨푬푪 ≅are ∠ 푨푪푬congruent)∠푨푬푫 ≅ ∠푨푫푬 ∠푨푫푩+ ≅ ∠푨푩푫+ = (sum of angles of ∆) = (substitution)∠푨푬푪 ≅ ∠푬푨푫 = ퟎ (SAS) BD풎∠ =푨푫푬ퟎ DE (corresponding풎∠푫푬푨 풎∠ parts푬푨푫 of congruentퟏퟖퟎ ퟎ triangles) ퟑ풎 (arcs∠푨푬푫 ퟏퟖsubtendedퟎ by congruent풎∠ 푨푬푫chords) ퟔퟎ ∆푩푨푫 ≅ ∆푫푨푬 ≅ ∆푬푨푪 푩푫� ≅ 푫푬�

DRAFT 7. In circle A, is a radius and and = °. Find . Complete the proof.

= 푨푩� �� 푩푪� ≅ ______푩푫� 풎∠푪푨푫 ퟓퟒ 풎∠ 푨푩푪

푩푪 푩푫 _____ = ______

풎∠ 풎∠ + + = ______

풎∠ 푩푨푪 풎∠푪푨푫 풎∠푩푨푫 _____ + ° = ° ______

ퟐ풎∠ ퟓퟒ ퟑퟔퟎ = ______

풎∠푩푨푪 = ______

푨푩 푨푪 _____ = ______

풎∠ 풎∠ + = ______

ퟐ풎 ∠푨푩푪 풎∠푩푨푪 = _____

풎∠푨푩푪 Chords of congruent arcs; BAC, BAD, angles inscribed in congruent arcs are congruent; 360°, circle; BAC; 153°; radii; ABC , ACB, base angles of isosceles; 180°, angles of triangle equal 180°; 13.5°

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 9 M5 GEOMETRY

Lesson 9: Arc Length and Areas of Sectors

Student Outcomes . When students are provided with the angle measure of the arc and the length of the radius of the circle, they understand how to determine the length of an arc and the area of a sector...

Lesson Notes This lesson explores the following geometric definitions: ARC: An arc isDRAFT any of the following three figures—a minor arc, a major arc, or a semicircle. 1 LENGTH OF AN ARC: The length of an arc is the circular distance around the arc. MINOR AND MAJOR ARC: In a circle with center , let and be different points that lie on the circle but are not the endpoints of a diameter. The minor arc between and is the set containing , , and all points of the circle that are in the interior of . The푂 major퐴 arc is퐵 the set containing , , and all points of the circle that lie in the exterior of . 퐴 퐵 퐴 퐵 ∠퐴푂퐵 퐴 퐵 RADIAN: A radian is the measure∠퐴푂퐵 of the central angle of a sector of a circle with arc length of one radius length.

SECTOR. Let arc be an arc of a circle with center and radius . The union of the segments , where is any point on the arc , is called a sector. The arc is called the arc of the sector, and is called its radius. � 퐴퐵 푂 푟 푂푃 푃 SEMICIRCLE: In a circle,퐴퐵� let and be the endpoints퐴퐵� of a diameter. A semicircle is the set 푟containing , , and all points of the circle that lie in a given half-plane of the line determined by the diameter. 퐴 퐵 퐴 퐵

Classwork Opening (2 minutes) . In Lesson 7, we studied arcs in the context of the degree measure of arcs and how those measures are determined. . Today we examine the actual length of the arc, or arc length. Think of arc length in the following way: If we laid a piece of string along a given arc and then measured it against a ruler, this length would be the arc length.

Example 1 (12 minutes)

. Discuss the following exercise with a partner.

1 This definition uses the undefined term distance around a circular arc (G-CO.1). In grade 4, student might use wire or string to find the length of an arc.

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Example 1:

MP.1 a. What is the length of the arc of degree measure in a circle of radius ?–MP.1

ퟔퟎ˚ ퟏퟎ 풄풎

Scaffolding: . Prompts to help struggling students along: . If we can describe arc length as the length of the

string that is laid along an = ( × ) arc, what is the length of DRAFTstring laid around the ퟏ 푨풓풄 풍풆풏품풕풉 ퟐ흅 ퟏퟎ = ퟔ entire circle? (The ퟏퟎ흅 circumference, 2 ) 푨풓풄 풍풆풏품풕풉 The marked arc lengthퟑ is . . What portion of the entire ퟏퟎ흅 circle is the arc measure휋푟 ퟑ 풄풎 ? ( = ; the arc Encourage students to articulate why their computation works. Students should be able measure60 tells1 us that the 60˚ 360 6 to describe that the arc length is a fraction of the entire circumference of the circle, and arc length is of the length that fractional value is determined by the arc degree measure divided by . This will 1 of the entire circle.) help them generalize the formula for calculating the arc length of a circle with arc degree 6 measure and radius . 360˚

푥˚ 푟 b. Given the concentric circles with center A and with = , calculate the arc length intercepted by on each circle. The inner circle has a radius of 10 and each circle has a radius 10 units greater than the previous풎∠푨 circle. ퟔퟎ⁰ ∠푨 Arc length of circle with radius = ( )( ) = ퟔퟎ ퟏퟎ흅 �� ퟑퟔퟎ ퟑ Arc length of circle with radius 푨푩 = � �(ퟐ흅)(ퟏퟎ) = ퟔퟎ ퟐퟎ흅 �� ퟑퟔퟎ ퟑ Arc length of circle with radius 푨푪 = � � (ퟐ흅)(ퟐퟎ) = = ퟔퟎ ퟑퟎ흅 푨푫�� ퟑퟔퟎ� ퟐ흅 ퟑퟎ ퟑ ퟏퟎ흅 c. An arc, again of degree measure , has an arc length of . What is the radius of the circle on which the arc sits? ퟔퟎ˚ ퟓ흅 풄풎 ( × ) = ퟏ ퟐ흅 풓 ퟓ흅 ퟔ = ퟐ흅풓= ퟑퟎ흅 The radius of the circle on which the arc sits is 풓. ퟏퟓ ퟏퟓ 풄풎 . Notice that provided any two of the following three pieces of information–the radius, the central angle (or arc degree measure), or the arc length–we can determine the third piece of information.

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d. Give a general formula for the length of an arc of degree measure on a circle of radius .–MP.8

Arc length = ( ) 풙˚ 풓 풙 ퟑퟔퟎ ퟐ흅풓 e. Is the length of an arc intercepted by an angle proportional to the radius? Explain.

Yes, the arc angle length is a constant times the radius when x is a constant angle measure, so it is proportional ퟐ흅풙 to the radius of an arc intercepted by an angle. ퟑퟔퟎ

Support parts (a)–(d) with these follow up questions regarding arc lengths. Draw the corresponding figure with each question as you pose the question to the class. . From the belief that for any number between 0 and 360, there is an angle of that measure, it follows that for any length between 0 and 2pir, there is an arc of that length on a circle of radius . . Additionally, we drew a parallel with the protractor axiom (“angles add”) in Lesson 7 with respect to arcs. DRAFT푟 For example, if we have arcs and as in the following figure, what can we conclude about ? 180˚

퐴퐵� 퐵퐶� 푚퐴퐵퐶�

 = + . We can draw the same parallel with arc lengths. With respect to the same figure, we can say: 푚퐴퐶� 푚퐴퐵� 푚퐵퐶� = +

. Then, given any minor arc푎푟푐, such푙푒푛푔푡 asℎ� minor퐴퐶� � arc푎푟푐 푙푒푛푔푡, whatℎ� must퐴퐵� � the푎푟푐 sum푙푒푛푔푡 of a ℎ�minor퐵퐶� � arc and its corresponding major arc (in this example major arc ) sum to? 퐴퐵�

퐴푋퐵�

 The sum of their arc lengths is the entire circumference of the circle, or 2 . . What is the possible range of lengths of any arc length? Can an arc length have a length of 0? Why or why not? 휋푟  No, an arc has, by definition, two different endpoints. Hence, its arc length is always greater than zero. . Can an arc length have the length of the circumference, 2 ?

휋푟

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Students may disagree about this. Confirm that an arc length refers to a portion of a full circle. Therefore, arc lengths fall between 0 and 2 ; 0 < < 2 .

휋푟 푎푟푐 푙푒푛푔푡ℎ 휋푟 . In part (a), the arc length is . Look at part (b). 10휋 Have students calculate the arc length as the 3 central angle stays the same, but the radius of the circle changes. If students write out the calculations, they will see the relationship and constant of proportionality that we are trying to discover through the similarity of the circles. . What variable is determining arc length as the central angle remains constant? Why?  The radius determines the size of the circle because all circles are DRAFTsimilar. . Is the length of an arc intercepted by an angle proportional to the radius? If so, what is the constant of proportionality?  Yes, , where x is a constant angle measure in degree, the constant of proportionality is . 2휋푥 휋푥 흅 . What is the arc360 length표푟 180 if the central angle has a measure of 1⁰? 180  휋 . What we180 have just shown is that for every circle, regardless of the radius length, a central angle of 1⁰ produces an arc of length . Repeat that with me. 휋  For every180 circle, regardless of the radius length, a central angle of 1⁰ produces an arc of length . 휋 . Mathematicians have used this relationship to define a new angle measure, a radian. A radian is the measure180 of the central angle of a sector of a circle with arc length of one radius length. Say that with me.  A radian is the measure of the central angle of a sector of a circle with arc length of one radius length. . So 1 = . What does 180⁰ equal in radian measure? 0 휋  180 radians.푟푎푑푖푎푛푠 . What does 360⁰ or a rotation through a full circle equal in radian measure? 휋  2π radians. . Notice, this is consistent with what we found above.

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Exercise 1 (5 minutes)

Exercise 1

The radius of the following circle is , and the = .

a. What is the arc length of ? ퟑퟔ 풄풎 풎∠푨푩푪 ퟔퟎ˚

푨푪�

DRAFT

The degree measure of arc is . Then the arc length of is � � = ( × 푨푪) ퟏퟐퟎ˚ 푨푪 ퟏ 푨풓풄 풍풆풏품풕풉 ퟐ흅 ퟑퟔ = ퟑ The푨풓풄 arc풍풆풏품풕풉 length ofퟐퟒ흅 is .

푨푪� ퟐퟒ흅 풄풎 b. What is the radian measure of the central angle? Arc length = ( )( )

=풂풏품풍풆 ( 풎풆풂풔풖풓풆 풐풇 풄풆풏풕풓풂풍 풂풏품풍풆 풊풏 풓풂풅풊풂풏풔 )(풓풂풅풊풖풔)

Arc푨풓풄 length풍풆풏품풕풉 = 풂풏품풍풆( ) 풎풆풔풖풓풆 풐풇 풄풆풏풕풓풂풍 풂풏품풍풆 풊풏 풓풂풅풊풂풏풔 ퟑퟔ 흅풙 = ( �ퟏퟖퟎ� ퟑퟔ )

Theퟐퟒ흅 measureퟑퟔ 풂풏품풍풆 of the풎풆풂풔풖풓풆 central angle풐풇 풄풆풏풕풓풂풍in radians풂풏품풍풆 = =풊풏 풓풂풅풊풂풏풔 radians. ퟐퟒ흅 ퟐ흅 ퟑퟔ ퟑ

Discussion (5 minutes) Discuss what a sector is and how to find the area of a sector. . A sector can be thought of as the portion of a disk defined by an arc.

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Sector. Let be an arc of a circle with center and radius . The union of all segments , where is any point of , is called a sector. 푨푩� 푶 풓 푶푷�� 푷 푨푩 �

Example 2 (8DRAFT minutes) Scaffolding: Allow students to work in partners or small groups on the questions before offering . We calculated arc length prompts. by determining the portion of the circumference the a. Circle has a radius of . What is the area of the circle? Write the formula. arc spanned. How can we = ( ( ) ) = 푶 ퟏퟎ 풄풎 use this idea to find the ퟐ ퟐ 푨풓풆풂 흅 ퟏퟎ ퟏퟎퟎ흅 풄풎 area of a sector in terms of b. What is the area of half of the circle? Write and explain the formula. area of the entire disk? = ( ( ) ) = . 10 is the radius of the circle, and = , which is the ퟏ ퟏ ퟏퟖퟎ fraction of the circle.ퟐ ퟐ 푨풓풆풂 ퟐ 흅 ퟏퟎ ퟓퟎ흅 풄풎 ퟐ ퟑퟔퟎ

c. What is the area of a quarter of the circle? Write and explain the formula. = ( ( ) ) = . 10 is the radius of the circle, and = , which is the fraction of the circle. ퟏ ퟐ ퟐ ퟏ ퟗퟎ 푨풓풆풂 ퟒ 흅 ퟏퟎ ퟐퟓ흅 풄풎 ퟒ ퟑퟔퟎ MP.7 d. Make a conjecture about how to determine the area of a sector defined by an arc measuring 60 degrees.

( ) = ( ( ) ) = ( ( ) ); the area of the circle times the arc measure divided by 360 ퟔퟎ ퟐ ퟏ ퟐ 푨풓풆풂 풔풆풄풕풐풓 푨푶푩 ퟑퟔퟎ 흅 ퟏퟎ ퟔ 흅 ퟏퟎ ( ) = ퟓퟎ흅 푨풓풆풂 풔풆풄풕풐풓 푨푶푩 The area of the sector ퟑ is . ퟓퟎ흅 ퟐ 푨푶푩 ퟑ 풄풎 Again, as with Example 1 Part (a), encourage students to articulate why the computation works.

e. Circle has a minor arc with an angle measure of . Sector has an area of . What is the radius of circle ? 푶 푨푩� ퟔퟎ˚ 푨푶푩 ퟐퟒ흅 =푶 ( ( ) ) ퟏ ퟐ ퟐퟒ흅 ( 흅 풓 ) =ퟔ ( ) ퟐ ퟏퟒퟒ흅= 흅 풓 풓 ퟏퟐ

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The radius has a length of .

ퟏퟐ 풖풏풊풕풔 f. Give a general formula for the area of a sector defined by arc of angle measure on a circle of radius ?

Area of sector = ( ) 풙˚ 풓 풙 ퟐ ퟑퟔퟎ 흅풓

Exercises 2–3 (7 minutes)

Exercise 2

The area of sector in the following image is . Find the measurement of the central angle labeled . 푨푶푩 ퟐퟖ흅 = ( 풙˚( ) ) 풙 ퟐ ퟐퟖ흅= DRAFT 흅 ퟏퟐ ퟑퟔퟎ 풙The centralퟕퟎ angle has a measurement of .

ퟕퟎ˚ Exercise 3

In the following figure, circle has a radius of , = and = = . Find: 푶 ퟖ 풄풎 풎∠푨푶푪 ퟏퟎퟖ˚ 푨푩� 푨푪� ퟏퟎa. 풄풎

∠푶푨푩

ퟑퟔ˚ b.

푩푪�

ퟏퟒퟒ˚ c. Area of sector

( 푩푶푪) = ( ( ) ) ퟏퟒퟒ ퟐ 푨풓풆풂 풔풆풄풕풐풓 푩푶푪 흅 ퟖ ( ) ퟑퟔퟎ. The푨풓풆풂 area풔풆풄풕풐풓 of sector푩푶푪 ≈ ퟖퟎis ퟒ. . ퟐ

푩푶푪 ퟖퟎ ퟒ 풄풎

Closing (1 minute) Present the following questions to the entire class, and have a discussion. . What is the formula to find the arc length of a circle provided the radius and an arc of angle measure ?  = ( )(2 ) 푟 푥˚ 푥 . What is 퐴푟푐the formula푙푒푛푔푡ℎ to find360 the휋푟 area of a sector of a circle provided the radius and an arc of angle measure ? 푟  = ( )( ) 푥˚ 푥 2 . What is 퐴푟푒푎a radian?표푓 푠푒푐푡표푟 360 휋푟

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 The measure of the central angle of a sector of a circle with arc length of one radius length.

Lesson Summary

RELEVANT VOCABULARY • ARC: An arc is any of the following three figures—a minor arc, a major arc, or a semicircle. 1 • LENGTH OF AN ARC: The length of an arc is the circular distance around the arc. • MINOR AND MAJOR ARC: In a circle with center , let and be different points that lie on the circle but are not the endpoints of a diameter. The minor arc between and is the set containing푂 , 퐴, and퐵 all points of the circle that are in the interior of . The major arc is the set containing ,

DRAFT, and all points of the퐴 circle퐵 that lie in the exterior퐴 of퐵 . ∠퐴푂퐵 퐴 • RADIAN: A radian is the measure of the central angle of a sector of a circle 퐵 ∠퐴푂퐵 with arc length of one radius length.

• SECTOR: Let arc be an arc of a circle with center and radius . The union of the segments , where is any point on the arc , is called a sector.

The arc is called퐴퐵� the arc of the sector, and is called푂 its radius.푟 푂푃 푃 퐴퐵� • SEMICIRCLE퐴퐵�: In a circle, let and be the endpoints푟 of a diameter. A semicircle is the set containing , , and all points of the circle that lie in a given half-plane of the line퐴 determined퐵 by the diameter. 퐴 퐵

Exit Ticket (5 minutes)

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Name Date

Lesson 9: Arc Length and Areas of Sectors

Exit Ticket

DRAFT

1. Find the arc length of .

푃푄푅�

2. Find the area of sector .

푃푂푅

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Exit Ticket Sample Solutions

1. Find the arc length of .

( ) = 푷푸푹�( × ) ퟏퟔퟐ 푨풓풄 풍풆풏품풕풉 푷푹� ퟐ흅 ퟏퟓ ( ) = ퟑퟔퟎ. 푨풓풄 풍풆풏품풕풉 푷푹� ( ퟏퟑ ퟓ흅 ) = 푪풊풓풄풖풎풇풆풓풆풏풄풆The arc length of 풄풊풓풄풍풆 is ( 푶 ퟑퟎ흅. ) or . .

푷푸푹� ퟑퟎ흅 − ퟏퟑ ퟓ흅 풄풎 ퟏퟔ ퟓ흅 풄풎 2. Find the area of sector .

( ) =푷푶푹 ( ( ) ) ퟏퟔퟐ ퟐ 푨풓풆풂(풔풆풄풕풐풓 푷푶푹) 흅 ퟏퟓ DRAFT = ퟑퟔퟎ. 푨풓The풆풂 area풔풆풄풕풐풓 of sector푷푶푹 ퟏퟎퟏis ퟐퟓ흅. . ퟐ 푷푶푹 ퟏퟎퟏ ퟐퟓ흅 풄풎

Problem Set Sample Solutions

1. and are points on the circle of radius and the measure of arc is . Find, to one decimal place each of the following: 푷 푸 ퟓ 풄풎 푷푸� ퟕퟐ˚ a. The length of arc � ( )푷푸 = ( × ) ퟕퟐ 푨풓풄 풍풆풏품풕풉 푷푸� ퟐ흅 ퟓ ( ) = ퟑퟔퟎ 푨풓풄The arc풍풆풏품풕풉 length푷푸� of ퟐ흅is or approximately . ..

푷푸� ퟐ흅 풄풎 ퟔ ퟑ 풄풎 b. Find the ratio of the arc length to the radius of the circle. = 흅 ퟐ흅 ퟏퟖퟎ ∙ ퟕퟐ ퟓ 풓풂풅풊풂풏풔 c. The length of chord

The length of is twice푷푸 the value of in .

= 푷푸 풙 △ 푶푸푹

풙 =ퟓ 풔풊풏=ퟑퟔ 푷푸Chord ퟐ풙 hasퟏퟎ a 풔풊풏lengthퟑퟔ of or approximately . . 푷푸 ퟏퟎ 풔풊풏 ퟑퟔ 풄풎 ퟓ ퟗ 풄풎 d. The distance of the chord from the center of the circle.

The distance of chord from푷푸 the center of the circle is labeled as in .

= 푷푸 풚 △ 푶푸푹

The풚 distanceퟓ 풄풐풔 ퟑퟔ of chord from the center of the circle is or approximately . 푷푸 ퟓ 풄풐풔 ퟑퟔ 풄풎 ퟒ 풄풎

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e. The perimeter of sector .

( ) =푷푶푸+ +

푷풆풓풊풎풆풕풆풓(풔풆풄풕풐풓) = ퟓ +ퟓ ퟐ흅 푷풆풓풊풎풆풕풆풓The perimeter풔풆풄풕풐풓 of sector ퟏퟎ isퟐ흅 ( + ) or approximately . . 푷푶푸 ퟏퟎ ퟐ흅 풄풎 ퟏퟔ ퟑ 풄풎 f. The area of the wedge between the chord and the arc

( ) = ( ) (푷푸 ) 푷푸�

푨풓풆풂(풘풆풅품풆) = 푨풓풆풂( 풔풆풄풕풐풓)( − 푨풓풆풂) △ 푷푶푸 ퟏ 푨풓풆풂 △ 푷푶푸 ퟏퟎ 풔풊풏 ퟑퟔ ퟓ 풄풐풔 ퟑퟔ ( ퟐ) = ( ( ) ) ퟕퟐ ퟐ 푨풓풆풂 풔풆풄풕풐풓 푷푶푹 흅 ퟓ ( ) = (ퟑퟔퟎ( ) ) ( )( ) ퟕퟐ ퟐ ퟏ DRAFTThe푨풓풆풂 area풘풆풅품풆 of sector is흅 approximatelyퟓ − ퟏퟎ 풔풊풏 . ퟑퟔ ퟓ. 풄풐풔 ퟑퟔ ퟑퟔퟎ ퟐ ퟐ 푷푶푸 ퟑ ퟖ 풄풎 g. The perimeter of this wedge ( ) = +

푷풆풓풊풎풆풕풆풓The perimeter풘풆풅품풆 of the wedgeퟐ흅 is approximatelyퟏퟎ 풔풊풏 ퟑퟔ . . ퟏퟐ ퟐ 풄풎 2. What is the radius of a circle if the length of a arc is ?

= ( ) ퟒퟓ˚ ퟗ흅 ퟒퟓ ퟗ흅 ퟐ흅풓 = ퟑퟔퟎ The풓 radiusퟑퟔ of the circle is units. ퟑퟔ

3. Arcs and both have an angle measure of , but their arc lengths are not the same. = and = .

a. 푨푩�What are퐂퐃� the arc lengths of arcs and ퟑퟎ?˚ �푶푩�� ퟒ �푩푫�� ퟐ � � = ( ×푨푩 ) 푪푫 ퟑퟎ 푨풓풄 풍풆풏품풕풉�푨푩� � ퟐ흅 ퟒ = ퟑퟔퟎ ퟐ 푨풓풄 풍풆풏품풕풉�푨푩� � 흅 The arc length of ퟑis units. ퟐ 푨푩� ퟑ 흅 = ( × ) ퟑퟎ 푨풓풄 풍풆풏품풕풉�푪푫� � ퟐ흅 ퟔ = ퟑퟔퟎ The푨풓풄 arc풍풆풏품풕풉 length�푪푫� of � 흅is units.

푪푫� 흅

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b. What is the ratio of the arc length to the radius for all of these arcs? Explain.

= radians, the angle is constant, so the ratio of arc length to radius will be the angle measure, 30⁰ x . ퟑퟎ흅 흅 흅

ퟏퟖퟎ ퟔ ퟏퟖퟎ c. What are the areas of the sectors and ?

( ) = ( ( )푨푶푩) 푪푶푫 ퟑퟎ ퟐ 푨풓풆풂 풔풆풄풕풐풓 푨푶푩 흅 ퟒ ( ) = ퟑퟔퟎ ퟒ 푨풓풆풂 풔풆풄풕풐풓 푨푶푩 흅 The area of the sector ퟑ is . ퟒ ퟐ 푨푶푩 ퟑ 흅 풖풏풊풕풔 ( ) = ( ( ) ) ퟑퟎ ퟐ 푨풓풆풂 풔풆풄풕풐풓 푪푶푫 흅 ퟔ The area of the sector ퟑퟔퟎ is . ퟐ DRAFT 푨푶푩 ퟑ흅 풖풏풊풕풔 4. In the circles shown, find the value of x. The circles shown have central angles that are equal in measure. a. b.

= radians = ퟐ흅 풙 ퟑ 풙 ퟑퟎ c. d.

= = ퟏퟖ ퟐ흅 풙 ퟓ 풙 ퟒퟓ

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5. The concentric circles all have center . The measure of the central angle is 45⁰. The arc lengths are given. a. Find the radius of each circle. 푨 Radius of inner circle: = , = 흅 ퟒퟓ흅 Radius of middle circle:ퟐ ퟏퟖퟎ= 풓 풓 , ퟐ= ퟓ흅 ퟒퟓ흅 Radius of outer circle: ퟒ= ퟏퟖퟎ ,풓 풓= ퟓ ퟗ흅 ퟒퟓ흅 ퟒ ퟏퟖퟎ 풓 풓 ퟗ b. Determine the ratio of the arc length to the radius of each circle, and interpret its meaning.

is the ratio of the arc length to the radius of each circle. It is the 흅 measure of the central angle in radians. ퟒ

DRAFT= 6. In the figure, if , find the length of arc ?

Since is of 푷푸� , thenퟏퟎ 풄풎the arc length of is of푸푹� ; the arc length of is ퟏ ퟏ

.ퟔ˚ ퟏퟓ ퟗퟎ˚ 푸푹� ퟏퟓ ퟏퟎ 풄풎 푸푹� ퟐ ퟑ 풄풎

7. Find, to one decimal place, the areas of the shaded regions. a.

= (or ( ) + ) ퟑ 푺풉풂풅풆풅 푨풓풆풂 푨풓풆풂 풐풇 풔풆풄풕풐풓 − 푨풓풆풂 풐풇 푻풓풊풂풏품풍풆 ퟒ 푨풓풆풂 풐풇 풄풊풓풄풍풆 푨풓풆풂 풐풇 풕풓풊풂풏품풍풆 = ( ( ) ) ( )( ) ퟗퟎ ퟐ ퟏ 푺풉풂풅풆풅 푨풓풆풂 흅 ퟓ − ퟓ ퟓ = ퟑퟔퟎ. . ퟐ The푺풉풂풅풆풅 shaded푨풓풆풂 area isퟔ approximatelyퟐퟓ흅 − ퟏퟐ ퟓ . . ퟐ ퟎ 풖풏풊풕

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 9 M5 GEOMETRY

b. The following circle has a radius of .

ퟐ

= ( ) + ퟑ 푺풉풂풅풆풅 푨풓풆풂 푨풓풆풂 풐풇 풄풊풓풄풍풆 푨풓풆풂 풐풇 풕풓풊풂풏품풍풆 Note: The triangle is a triangleퟒ with legs of length 2 (the legs are comprised by the radii, like DRAFTthe triangle in the previous question). ퟒퟓ⁰ − ퟒퟓ⁰ − ퟗퟎ⁰ = ( ( ) ) + ( )( ) ퟑ ퟐ ퟏ 푺풉풂풅풆풅 푨풓풆풂 흅 ퟐ ퟐ ퟐ = ퟒ + ퟐ 푺풉풂풅풆풅The shaded푨풓풆풂 area isퟑ흅 approximatelyퟐ 11. . ퟐ ퟒ 풖풏풊풕풔

= ( ) + ( )

푺풉풂풅풆풅 푨풓풆풂 푨풓풆풂 풐풇 ퟐ 풔풆풄풕풐풓풔 푨풓풆풂 풐풇 ퟐ 풕풓풊풂풏품풍풆풔 = + ( ) ퟗퟖ ퟒퟗ√ퟑ 푺풉풂풅풆풅 푨풓풆풂 ퟐ � 흅� ퟒ = ퟑ + ퟐ ퟏퟗퟔ 푺풉풂풅풆풅 푨풓풆풂 흅 ퟗퟖ√ퟑ The shaded area is approximatelyퟑ . . ퟐ ퟕퟒ ퟗퟗ 풖풏풊풕풔

Lesson 9: Arc Length and Areas of Sectors Date: 8/29/14 14

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

Lesson 10: Unknown Length and Area Problems

Student Outcomes . Students apply their understanding of arc length and area of sectors to solve problems of unknown area and length.

Lesson Notes This lesson continues the work started in Lesson 9 as students solve problems on arc length and area of sectors. The lesson is intended to be 45 minutes of problem solving with a partner. Problems vary in level of difficulty and can be assigned specificallyDRAFT based on student understanding. The problem set can be used in class for some students or assigned as homework. Students who need to focus on a small number of problems could finish the other problems at home. Teachers may choose to model two or three problems with the entire class. Exercise 4 is a modeling problem highlighting G-MG.A.1 and MP.4.

Scaffolding: Classwork . Post area of sector and arc Begin with a quick whole class discussion of an annulus. Project the figure on the right on length formulas for easy the board. reference. . A review of compound Opening Exercise (3 minutes) figures may be required before this lesson. In the following figure, a cylinder is carved out from within another cylinder of the same height; . Scaffold the task by asking the bases of both cylinders share the same center. students to compute the 1. Sketch a cross section of the figure parallel to the base. area of the circle with radius r, then the circle with radius s, and then ask how the shaded region is related to the two circles.

. Use an example with numerical values for s and r on the coordinate plane,

and ask students to estimate the area first (see

example below).

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

Confirm that students’ sketch is correct before allowing them to proceed to #2.

2. Mark and label the shorter of the two radii as and the longer of the two radii . Show how to calculate the area of the shaded region and explain the parts of the MP.2 풓 풔 & expression. MP.7 ( ) = ( ) ퟐ ퟐ 푨풓풆풂s = radius풔풉풂풅풆풅 of outer circle흅 풔 − 풓 r = radius of inner circle

The figure you sketched in #2 is called an annulus; it is a ring shaped region or the region lying between two concentric circles. In Latin, annulus means “little ring.” DRAFT Exercises (35 minutes)

1. Find the area of the following annulus.

( ) = ( . ) ퟐ ퟐ 푨풓풆풂(풔풉풂풅풆풅) = 흅 ퟗ. −ퟕ ퟓ 푨풓풆풂The area풔풉풂풅풆풅 of the annulusퟐퟒ ퟕퟓ흅 is . . ퟐ ퟐퟒ ퟕퟓ흅 풖풏풊풕풔 2. The larger circle of an annulus has a diameter of , and the smaller circle has a diameter of . . What is the area of the annulus? ퟏퟎ 풄풎 ퟕ ퟔ 풄풎 The radius of the larger circle is , and the radius of the smaller circle is . .

( ) = ( . ퟓ) 풄풎 ퟑ ퟖ 풄풎 ퟐ ퟐ 푨풓풆풂(풔풉풂풅풆풅) = 흅 ퟓ. −ퟑ ퟖ 푨풓풆풂The area풔풉풂풅풆풅 of the annulusퟏퟎ ퟓퟔ흅 is . . ퟐ ퟏퟎ ퟓퟔ흅 풄풎

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

3. In the following annulus, the radius of the larger circle is twice the radius of the smaller circle. If the area of the following annulus is , what is the radius of the larger circle? ퟐ ퟏퟐ흅 풖풏풊풕풔

( ):

푨풓풆풂 풔풉풂풅풆풅 (( ) ) = ퟐ ퟐ DRAFT흅 ퟐ풓 −= 풓 ퟏퟐ흅 ퟐ ퟑ풓 = ퟏퟐ The radius of the larger circle is twice the radius of the smaller풓 ퟐ circle or ( ) = ; the radius of the larger circle is . ퟐ ퟐ ퟒ ퟒ 풖풏풊풕풔

MP.4 4. An ice cream shop wants to design a super straw to serve with their extra thick milkshakes that is double the width and thickness of a standard straw. A standard straw is 8 mm in diameter and 0.5 mm thick. a. What is the cross-sectional (parallel to the base) area of the new straw (round to the nearest hundredth)? . mm2

ퟒퟕ ퟏퟐ b. If the new straw is 23 mm long, what is the maximum volume of milkshake that can be in the straw at one time (round to the nearest hundredth)?

, . mm3

ퟑ ퟓퟒퟎ ퟓퟕ c. A large milkshake is 32 ounces (approximately 950 mL). If Corbin withdraws the full capacity of a straw 10 times a minute, what is the minimum amount of time that it will take him to drink the milkshake (round to the nearest minute)?

minutes

ퟐퟕ 5. In the circle given, is the diameter and is perpendicular to chord . DF = 8 cm and FE = 2 cm. Find AC, BC, , the arc length of , and the area of sector (round to the nearest �� hundredth, if necessary푬푫 ). 퐂퐁 풎∠푪푨푩 퐂퐄퐁� 퐂퐄퐁� = cm, = cm, = ( . ) = . , arc length = . cm, area = . cm2 푨푪 ퟓ 푩푪 ퟖ 풎∠푪푨푩 ퟐ ퟓퟑ ퟏퟑ ퟏퟎퟔ ퟐퟔ⁰ ퟗ ퟐퟕ ퟐퟑ ퟏퟖ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

6. Given circle A with , find the following (round to the nearest hundredth, if necessary):

a. ∠퐁퐀퐂 ≅ ∠퐁퐀퐃 풎푪푫� ퟒퟓ˚ b.

풎푪푩푫� ퟑퟏퟓ˚ c.

풎푩푪푫�. ퟐퟎퟐ ퟓ˚ d. Arc length

. yds 푪푫� DRAFTퟗ ퟒퟐ e. Arc length

. yds 푪푩푫�

ퟔퟓ ퟗퟕ f. Arc length

. yds 푩푪푫�

ퟒퟐ ퟒퟏ g. Area of sector

2 . yds 푪푫�

ퟓퟔ ퟓퟓ h. Area of sector

2 . yds 푪푩푫�

ퟑퟗퟓ ퟖퟒ i. Area of sector

. yds2 푩푪푫�

ퟐퟓퟒ ퟒퟕ 7. Given circle A, find the following (round to the nearest hundredth, if necessary): a. Circumference of circle A yds

ퟗퟔ b. Radius of circle A . yds

ퟏퟓ ퟐퟖ c. Area of sector

2 . yds 푪푫�

ퟗퟏ ퟔퟗ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

8. Given circle A, find the following (round to the nearest hundredth, if necessary): a.

풎∠. 푪푨푫

ퟒퟕ ퟕퟒ⁰ b. Area of sector

푪푫�

ퟔퟎ

9. Find the area of the shaded region (round to the nearest hundredth). DRAFT .

ퟑퟓ ퟕퟑ

10. Many large cities are building or have built mega Ferris wheels. One is 600 feet in diameter and has 48 cars each seating up to 20 people. Each time the Ferris wheel turns Ө degrees, a car is in a position to load. a. How far does a car move with each rotation of Ө degrees (round to the nearest whole number)? feet

ퟑퟗ b. What is the value of Ө in degrees? .

ퟕ ퟓퟎ˚

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

11. ∆ABC is an equilateral triangle with edge length 20 cm. D, E, and F are midpoints of the sides. The vertices of the triangle are the centers of the circles creating the arcs shown. Find the following (round to the nearest hundredth): a. The area of the sector with center A. . cm2

ퟓퟐ ퟑퟔ b. The area of triangle ABC. . cm2

ퟏퟕퟑ ퟐퟏ c. The area of the shaded region. . cm2

ퟏퟔ ퟏퟑ d. The perimeter of the shaded region. DRAFT. cm

ퟑퟏ ퟒퟐ

12. In the figure shown, AC = BF = 5 cm, GH = 2 cm, and = . Find the area in the rectangle, but outside of the circles (round to the nearest퐦∠퐇퐁퐈 hundredthퟑퟎ⁰ ).

. cm2 ퟏퟑ ퟕퟐ

13. This is a picture of a piece of a mosaic tile. If the radius of each smaller circle is 1 inch, find the area of red section, the white section, and the blue section (round to the nearest hundredth).

Red = . in2, White = . in2, Blue = . in2

ퟐ ퟐퟖ ퟏퟎ ퟐퟖ ퟐ ퟐퟖ

Closing (2 minutes) Present the questions to the class, and have a discussion, or have students answer individually in writing. Use this as a method of informal assessment. . Explain how to find the area of a sector of a circle if you know the measure of the arc in degrees.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

o Find the fraction of the circumference by dividing the measure of the arc in degrees by 360, and then multiply by the circumference 2 . . Explain how to find the arc length of an arc if you know the central angle. 휋푟 o Find the fraction of the area by dividing the measure of the central angle in degrees by 360, then multiply by the circumference . 2 휋푟 Exit Ticket (5 minutes)

DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

Name Date

Lesson 10: Unknown Length and Area Problems

Exit Ticket

1. Given circle A, find the following: (Round to the nearest hundredth.) a. The in degrees. b. The area of sector . 푚퐵퐶�

퐵퐶� DRAFT

2. Find the shaded area (round to the nearest hundredth).

62⁰

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

Exit Ticket Sample Solutions

1. Given circle A, find the following:

a. The in degrees.

.풎푩푪�

ퟔퟖ ퟕퟓ⁰ b. The area of sector .

퐁퐂� ퟐ ퟏퟑퟓ 풇풕 2. Find the shaded area. . DRAFTퟐ ퟏퟓ ퟏퟓ 풄풎 Problem Set Sample Solutions Students should continue the work they began in class for homework.

1. Find the area of the shaded region if the diameter is 32 inches (round to the nearest hundredth).

214.47 in2

2. Find the area of the entire circle given the area of the sector.

500 in2

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY

3. are arcs of concentric circles with lying on the radii of the larger circle. Find the area of the region (round to the nearest hundredth). 퐃퐅� 퐚퐧퐝 퐁퐆� 퐁퐃�� 퐚퐧퐝 �퐅퐆��

108.00 cm2

4. Find the radius of the circle, x, y, and z (round to the nearest hundredth).

Radius = 7.64 cm, x = 74.99⁰, y = 134.99⁰, z = 16 cm DRAFT

5. In the figure, the radii of two concentric circles are 24 cm and 12 cm. = . If a chord of the larger circle intersects the smaller circle only at C, find the area of the shaded region in terms of π. 퐦퐃퐀퐄� ퟏퟐퟎ⁰ 퐃퐄��

ퟐ ퟐퟖퟖ흅 풄풎

Lesson 10: Unknown Length and Area Problems Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

Name Date

1. Consider a right triangle with sides of lengths 3, 4, and 5 centimeters drawn on a page.

a. Describe a sequence of straightedge and compass constructions that allow you to draw the circle that circumscribes the triangle. Explain why your construction steps successfully accomplish this task.

DRAFT

b. What is the distance of the side of the right triangle of length 3 from the center of the circle that circumscribes the triangle?

c. What is the area of the inscribed circle for the triangle?

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

2. A five-pointed star with vertices , , , , and is inscribed in a circle as shown. Chords and intersect at point . 𝐴𝐴 𝑀𝑀 𝐵𝐵 𝑁𝑁 𝐶𝐶 𝐴𝐴��𝐴𝐴� 𝑀𝑀��𝑀𝑀� 𝑃𝑃 DRAFT

a. What is the value of + + + + , the sum of the measures of the angles in the points of the star? Explain your answer. 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 𝑚𝑚∠𝑁𝑁𝑁𝑁𝑁𝑁 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 𝑚𝑚∠𝑀𝑀𝑀𝑀𝑀𝑀

b. Suppose is the midpoint of the arc and is the midpoint of arc , and = . What is , and why? 1 𝑀𝑀 𝐴𝐴𝐴𝐴 𝑁𝑁 𝐵𝐵𝐵𝐵 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 2 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

3. Two chords and in a circle with center intersect at right angles at point . equals the radius of the circle. �𝐴𝐴��𝐴𝐴� 𝐵𝐵𝐵𝐵�� 𝑂𝑂 𝑃𝑃 𝐴𝐴𝐴𝐴 DRAFT

a. What is the measure of the arc ?

𝐴𝐴𝐴𝐴

b. What is the value of the ratio ? Explain how you obtain your answer. 𝐷𝐷𝐷𝐷 𝐴𝐴𝐴𝐴

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

4. a. An arc of a circle has length equal to the diameter of the circle. What is the measure of that arc in radian? Explain your answer.

b. Two circles have a common center . Two rays from intercept the circles at points , , , and as shown. 𝑂𝑂 𝑂𝑂 Suppose 𝐴𝐴 =𝐵𝐵 2𝐶𝐶 5 and𝐷𝐷 that the area of the sector given by , , andDRAFT is 10 square centimeters. 𝑂𝑂𝑂𝑂 ∶ 𝑂𝑂𝑂𝑂 ∶ 𝐴𝐴 𝑂𝑂i. What𝐷𝐷 is the ratio of the measure of the arc to the measure of the arc ? 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵

ii. What is the area of the shaded region given by the points , , , and ?

𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷

iii. What is the ratio of the length of the arc to the length of the arc ?

𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

5. In this diagram, the points , , and are collinear and are the centers of three congruent circles. is the point of contact of two circles that are externally tangent. The remaining points at which two circles intersect are labeled , , 𝑃𝑃, and𝑄𝑄 as𝑅𝑅 shown. 𝑄𝑄

𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷

DRAFT

a. Segment is extended until it meets the circle with center at a point . Explain, in detail, why , , and are collinear. 𝐴𝐴��𝐴𝐴� 𝑃𝑃 𝑋𝑋 𝑋𝑋 𝑃𝑃 𝐷𝐷

b. In the diagram a section is shaded. What percent of the full area of the circle with center is shaded? 𝑄𝑄

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

A Progression Toward Mastery STEP 1 STEP 2 STEP 3 STEP 4 Assessment Missing or incorrect Missing or incorrect A correct answer A correct answer Task Item answer and little answer but with some evidence supported by evidence of evidence of some of reasoning or substantial reasoning or reasoning or application of evidence of solid application of application of mathematics to reasoning or mathematics to mathematics to solve the problem, application of solve the problem. solve the problem. or an incorrect mathematics to answer with solve the problem. substantial evidence of solid reasoning or DRAFTapplication of mathematics to solve the problem. 1 G-C.1 Student does not Student describes an Student accurately Student accurately G-C.2 describe an accurate accurate construction describes the describes the construction procedure procedure but doesn’t construction procedure, construction procedure, and does not find the fully justify its validity. justifies its validity, and justifies its validity, and radii, and hence areas, of Student makes some accurately determines accurately determines the circumscribed and progress towards finding the radius and area of the radii and areas of the inscribed circles. the radii, and hence the circumscribed circle. circumscribed and areas, of the Some progress towards inscribed circles. circumscribed and finding the radius of the inscribed circles. inscribed circle is made. 2 G-C.2 Student struggles to find Student makes some Student makes correct Student makes correct a relevant geometric progress applying the use of inscribed angle use of inscribed angle result that aids in inscribed angles results to answer the results to fully answer answering this question. theorem and some questions, but the questions with clear progress towards justifications might not justifications. answering the questions. be fully explained. 3 G-C.2 Student does not Student makes some Student provides correct Student takes full care to identify relevant central progress to identifying solutions but does not identify inscribed angles angles to the problem the relevant central give complete details as and their matching and makes little progress angles for the problem to how those solutions central angle measures towards finding and some steps towards are obtained. and provides clear, solutions. identifying angles. detailed solutions. 4 G-C.5 The student does not Student exhibits some Student exhibits some Student gives complete distinguish between arc understanding of the arc clear understanding of and correct length and arc measure measure, arc length, and the arc measure, arc explanations, exhibiting and does not make use radius relationships, and length, and radius a clear understanding of of the arc length/radius makes some progress to relationships, and makes the arc measure, arc proportionality. providing answers. good progress to length, and radius providing answers. relationships.

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

5 G-C.2 Student does not Student makes some Student provides clear Student gives complete G-C.3 articulate a clear and attempt to provide a explanations and and thorough correct explanation and clear explanation and develops a correct plan explanations and G-C.5 does not develop a plan makes some movement for computing the provides the correct final for computing the towards computing the desired area. solution. desired area. desired area.

DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

Name Date

1. Consider a right triangle with sides of lengths 3, 4, and 5 centimeters drawn on a page.

Label the vertices of the right triangle A, B, and C as shown. Because ABC is a right angle, AC will be the diameter of the circle. So the ∠midpoint of AC is the center.�� We first need to construct that midpoint.DRAFT �� 1. Set a compass at point A with its width equal to AC. Draw a circle. 2. Set the compass at C with the same width AC. Draw a circle. 3. Connect the two points of intersection of these circles with a line segment. This line segment intersects AC at its midpoint. Call it M. 4. Set the compass at ��M�� with width MA. Draw the circle. (This is the circumscribing circle for the triangle.

(Note that we could also locate the center of the circumscribing circle by constructing the perpendicular bisectors of any two sides of the triangle. Their point of intersection is the center.)

b. What is the distance of the side of the right triangle of length 3 from the center of the circle that circumscribes the triangle?

The distance d we seek is the length of the line segment connecting the center of the circumscribing circle to the midpoint of the side of the triangle of length 3. This segment is perpendicular to that side. Thus we see two similar right triangles, with scale factor 2. It 1 follows that d = 4 = 2. 2 ⋅

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NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

c. What is the area of the inscribed circle for the triangle?

Draw in three radii for the inscribed circle as shown. They meet the sides of the triangle at right angles. Each radius can be thought of as the height of a small triangle within the larger 3-4-5 triangle. If the radius of the inscribed circle is r, then the area of the whole 3-4-5 triangle is 1 1 1 3 r + 4 r 5 r = 6r. 2 2 2 1 But� ⋅ the⋅ �area� ⋅of ⋅the� �triangle⋅ ⋅ � is 3 4 = 6, and so r = 1, and the area of 2 2 the inscribed circle is π1 = π square⋅ ⋅ centimeters.

DRAFT

Module 5: Circles With and Without Coordinates Date: 8/29/14 9

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

2. A five-pointed star with vertices , , , , and is inscribed in a circle as shown. Chords and intersect at point . 𝐴𝐴 𝑀𝑀 𝐵𝐵 𝑁𝑁 𝐶𝐶 𝐴𝐴��𝐴𝐴� 𝑀𝑀��𝑀𝑀� 𝑃𝑃

DRAFT

a. What is the value of + + + + , the sum of the measures of the angles in the points of the star? Explain your answer. 𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 𝑚𝑚∠𝑁𝑁𝑁𝑁𝑁𝑁 𝑚𝑚∠𝐶𝐶𝐶𝐶𝐶𝐶 𝑚𝑚∠𝐴𝐴𝐴𝐴𝐴𝐴 𝑚𝑚∠𝑀𝑀𝑀𝑀𝑀𝑀 The measure of BAN is half the measure of the arc BN, similarly for the four remaining angles. Thus, the∠ sum of all five angle measures is the half the sum of the measures of ALL the arcs in the circle.

1 The sum of angle measures is 360° = 180°. 2 ⋅

The angles marked a are congruent because they are inscribed angles from congruent arcs. This is similar for the angles marked b. We are also told that m CBA is double m BAN.

∠ ∠

From part (a), we have 4a + 2b = 180°, and so 2a + b = 90°.

Using the fact that angles in BPC sum to 180° we get

m BPC = 180° 2a b = 90△°. ∠ − −

Module 5: Circles With and Without Coordinates Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

a. What is the measure of the arc ?

𝐴𝐴𝐴𝐴

Draw in the central angle to chord AB. We see an equilateral triangle. The central angle, and hence arc AB,�� has measure 60°.

b. What is the value of the ratio ? Explain how you obtain your answer. 𝐷𝐷𝐷𝐷 𝐴𝐴𝐴𝐴 Draw chord BC. By the inscribed/central angle theorem m ACB=30°�.� � ��In PBC it then follows that m PBC=60°, and so arc∠ DC has a measure△ of 120°. Let’s draw this∠ central angle.

Draw the line segment OX perpendicular to DC as shown. We r see that DXO is half an�� equilateral� triangle,� �and�� so XO = . 2 △

r 2 3r DC Thus, DX = r2 = , and so DC = 3r, and = 3. 2 √2 AB � − � � √ √

Module 5: Circles With and Without Coordinates Date: 8/29/14 11

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

4. a. An arc of a circle has length equal to the diameter of the circle. What is the measure of that arc in radian? Explain your answer.

An arc of length one radius of the circle represents one radian of turning. Thus, an arc of length two radii, a diameter, has a measure of 2 radian.

b. Two circles have a common center . Two rays from intercept the circles at points , , , and as shown. 𝑂𝑂 𝑂𝑂 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷 Suppose = 2 5 and that the area of the sector given by , , andDRAFT is 10 square centimeters. 𝑂𝑂𝑂𝑂 ∶ 𝑂𝑂𝑂𝑂 ∶ 𝐴𝐴 𝑂𝑂 𝐷𝐷 i. What is the ratio of the measure of the arc to the measure of the arc ? 𝐴𝐴𝐴𝐴 Both arcs𝐵𝐵𝐵𝐵 represent the same amount of turning (that is, they have the same central angle), and so have the same measure. This ratio is 1.

ii. What is the area of the shaded region given by the points , , , and ?

Let θ be the measure of AOD in radian. Since OA𝐴𝐴 : 𝐵𝐵OB𝐶𝐶 = 2 :𝐷𝐷 5, we have that OA = 2x, and OB = 5x for∠ some value x.

θ Now the area of sector AOD is 10 = π(2x)2 = 2θx2, so θx2 = 5. 2π

θ 2 ⋅25 2 125 The area of sector BOC is π(5x) = θx = . 2π 2 2 125 Thus, the area of the shaded −region is 10 = 52.5 square centimeters. 2 5 (Alternatively: The sectors AOD and BOC− are similar with scale factor . Thus, the 2 5 2 125 area of sector BOC is 10 × = , and the area of the shaded region is 2 2 125 10 = 52.5 square centimeters.) 2 � � − iii. What is the ratio of the length of the arc to the length of the arc ?

Since the radii of the circles come in𝐴𝐴 a𝐴𝐴 2 to 5 ratio, the same𝐵𝐵 𝐵𝐵is true for these arclengths.

Module 5: Circles With and Without Coordinates Date: 8/29/14 12

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷

DRAFT a. Segment is extended until it meets the circle with center at a point . Explain, in detail, why , , and are collinear. 𝐴𝐴��𝐴𝐴� 𝑃𝑃 𝑋𝑋 Draw in the radii shown.𝑋𝑋 𝑃𝑃 𝐷𝐷

Triangles APQ, QBR, and PQD are equilateral, and so all angles in those triangles have measure 60°. ABQ is isosceles with an angle at Q of measure 180° 60° 60° = 60°. It follows that it△ is equilateral as well. − −

We have m XAP = 180° m PAQ m QAB=180° 60° 60° = 60°. Since XPA is isosceles, with∠ one angle of− measure∠ − 60°,∠ it follows that− it −is also equilateral. In△ particular, m XPA=60°. ∠ Thus, m XPD = 60° + 60° + 60° = 180°, showing that X, P, and D are collinear.

∠

Module 5: Circles With and Without Coordinates Date: 8/29/14 13

NYS COMMON CORE MATHEMATICS CURRICULUM Mid-Module Assessment Task M5 GEOMETRY

b. In the diagram a section is shaded. What percent of the full area of the circle with center is shaded? 𝑄𝑄 Draw the segment AD shown. We see that it divides a sector of the circle with center Q into two regions, which�� we’ve labeled I and II. DRAFT

If we can determine the area of region I, then we see that the area of the desired 1 shaded region is area full circle 4 × area I . 2 120 1 Now area(I + II) = � πr2 = πr2 where− r is the� radius of the circle. 360 3

Region II is composed of two congruent right triangles each containing a 60° angle and r each with hypotenuse r. It follows that the remaining sides of each are and 2 r 2 3r r r2 = . Thus, region II is a triangle with base 3r and height , so it has an 2 √2 2 1 r 3 1 3 area� − of � � × 3r × = r. We have area I = area(I + II) √area II = πr2 r2. 2 2 √4 3 √4 Thus, the shaded√ region in question has an area of − −

1 1 3 1 2 3 π πr2 4 πr2 r2 = πr2 πr2 = r2. 2 3 4 2 3 2 6 √ √ � − � − �� − � − � The proportion of the full area πr2 this represents is

3 π r2 2 6 3 1 √ = 10.9%. � π−r2 � 2π 6 √ − ≈

Module 5: Circles With and Without Coordinates Date: 8/29/14 14

GEOMETRY • MODULE 5

Topic C: Secants and Tangents

G-C.A.2, G-C.A.3, G-C.A.4

Focus Standard:DRAFT G-C.A.2 Identify and describe relationships among inscribed angles, radii, and chords. G-C.A.3 Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle. G-C.A.4 Use coordinates to prove simple geometric theorems algebraically. Instructional Days: 6 Lesson 11: Properties of Tangents Lesson 12: Tangent Segments Lesson 13: The Inscribe Angle Alternate a Tangent Angle Lesson 14: Secant Lines; Secant Lines That Meet Inside a Circle Lesson 15: Secant Angle Theorem, Exterior Case Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams

Topic C focuses on secant and tangent lines intersecting circles, the relationships of angles formed, and segment lengths. In Lesson 11, students study properties of tangent lines and construct tangents to a circle through a point outside the circle and through points on the circle (G-C.A.4). Students prove that at the point of tangency the tangent line and radius meet at a right angle. Lesson 12 continues the study of tangent lines proving segments tangent to a circle from a point outside the circle are congruent. Students inscribe a circle in an angle and a circle in a triangle with constructions (G-C.A.3) leading to the study of inscribed angles with one ray being part of the tangent line in Lesson 13 (G-C.A.2). Students solve a variety of missing angle problems in Lesson 13 using theorems introduced in Lessons 11–13 (MP.1). The study of secant lines begins in Lesson 14 as students study two secant lines that intersect inside a circle. Students prove that an angle whose vertex is inside a circle is equal in measure to half the sum of arcs intercepted by it and its vertical angle. Lesson 15 extends this study to secant lines that intersect outside of a circle. Students understand that an angle whose vertex is outside of a circle is equal in measure to half the difference of the degree measure of its larger and smaller intercepted arcs. This concept is extended as the secant rays rotate to form tangent rays and that relationship is developed. Topic C and the study of secant lines concludes in Lesson 16 as students discover the relationships between segment lengths of secant lines intersecting inside and outside of a circle. Students find similar triangles and use proportional sides to develop this relationship (G- SRT.A.3, G-SRT.B.5). Topic C highlights MP.1 as students persevere in solving missing angle and missing length problems and MP.6 as students extend known relationships to limiting cases.

Topic C: Secants and Tangents Date: 8/29/14 1

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

Lesson 11: Properties of Tangents

Student Outcomes . Students discover that a line is tangent to a circle at a given point if it is perpendicular to the radius drawn to that point. . Students construct tangents to a circle through a given point. . Students prove that tangent segments from the same point are equal in length.

Lesson NotesDRAFT Topic C begins our study of secant and tangent lines. Lesson 11 is the introductory lesson and requires several constructions to solidify concepts for students. The study of tangents continues in Lessons 12 and 13. During the lesson, recall the following definitions if necessary:

INTERIOR OF A CIRCLE. The interior of a circle with center and radius is the set of all points in the plane whose distance from the point is less than . 푂 푟 A point in the interior of푂 a circle is said푟 to be inside the circle. A disk is the union of the circle with its interior.

EXTERIOR OF A CIRCLE. The exterior of a circle with center and radius is the set of all points in the plane whose distance from the point is greater than . 푂 푟 A point exterior to a circle푂 is said to be outside푟 the circle.

Classwork Opening (8 minutes) . Draw a circle and a line.  Students draw a circle and a line. . Have the students tape their sketches to the board. . Let’s group together the diagrams that are alike.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

 Students should notice that some circles have lines that intersect the circle twice, others only touch the circle once, and others do not intersect Scaffolding: the circle at all. Separate them accordingly. . Post pictures of pairs of . Explain how the types of circle diagrams are different. secant lines and tangent lines on the board so  A line can intersect a circle twice, only once, or not at all. students can refer to them . Do you remember the name for a line that intersects the circle twice? when needed.  A line that intersects a circle at exactly two points is called a secant line. . Post steps for each construction on the board . Do you remember the name for a line that intersects the circle once? for easy reference.  A line that intersects a circle at exactly one point is called a tangent line. . Provide completed or . Label each group of diagrams as “secant lines,” “tangent lines,” and “don’t partially completed MP.6 intersect,” and then as a class, repeat the definitions of secant and tangent lines drawings for students with chorally. eye-hand or fine motor difficulties or a set square  SECANT LINE: A secant line to a circle is a line that intersects a circle in DRAFTto help with perpendicular exactly two points. lines and segments.  TANGENT LINE: A tangent line to a circle is a line in the same plane that . For ELL students, use a intersects the circle in one and only one point. Frayer diagram for all new  TANGENT SEGMENT: A segment is said to be a tangent segment to a vocabulary words and circle if the line it is contained in is tangent to the circle, and one of its practice with choral endpoints is the point where the line intersects the circle. repetition.

. Topic C focuses on the study of secant and tangent lines intersecting circles. . Explain to your neighbor the difference between a secant line and a tangent line.

Discussion (10 minutes) In this whole class discussion, students will need a compass, protractor, and a straight edge to complete constructions. . Draw a circle and a tangent line.  Students draw a circle and a tangent line. . Draw a point where the tangent line intersects the circle. Label it .  Students draw the point and label it . 푃 . Point is called the point of tangency. Label point as the “Point of Tangency,” and write its definition. 푃 Share your definition with your neighbor. 푃 푃  The point of intersection of the tangent line to the circle is called the point of tangency. . Draw a radius connecting the center of the circle to the point of tangency.  Students draw a radius to point P. . With your protractor, measure the angle formed by the radius and the tangent line. Write the angle measure on your diagram.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

 Students measure and write 90 . . Compare your diagram and angle measure to three people around you. What do you notice? MP.7 ⁰ &  All diagrams are different, but all angles are 90⁰. MP.8 . What can we conclude about the segment joining a radius of a circle to the point of tangency?  The radius and tangent line are perpendicular. . Let’s think about other ways we can say this. What did we learn in Module 4 about the shortest distance between a line and a point?  The shortest distance from a point to a line is the perpendicular segment from the point to the line. . So, what can we say about the center of the circle and the tangent line?  The shortest distance between the center of the circle and a tangent line Scaffolding: is at the point of tangency and is the radius. Post these steps with . We will say it one more way. This time restate what we have found relating the accompanying diagrams to tangentDRAFT line, the point of tangency, and the radius. assist/remind students.  A tangent line to a circle is perpendicular to the radius of the circle drawn Constructing a line to the point of tangency. perpendicular to a segment through a point. . State the converse of what we have just said.

 If a line through a point on a circle is perpendicular to the radius drawn . Extend the radius beyond to that point, the line is tangent to the circle. the circle with center , . Is the converse true? creating segment . . Draw point at the point퐴  Answers will vary. �� of intersection of 퐴퐵 and . Try to draw a line through a point on a circle that is perpendicular to the radius the circle, using푃 your that is not tangent to the circle. compass, measure�퐴퐵 �the��  Students will try but it will not be possible. If a student thinks he has a distance from to and drawing that works show it to the class and discuss. mark that on the extended radius. 퐴 푃 . Share with your neighbor everything that you have learned about lines tangent . Draw circle with radius to circles. .  The point where the tangent line intersects the circle is called the point of . Draw circle 퐴 with radius tangency. 퐴퐵. 퐵  A tangent line to a circle is perpendicular to the radius of the circle drawn . Mark the point of 퐵퐴 to the point of tangency. intersection of the circles points and .  A line through a point on a circle is tangent at the point if, and only if, it . Construct a line through is perpendicular to the radius drawn to the point of tangency. and . 퐶 퐷 퐶 퐷 Example 1 (12 minutes) In this example, students will construct a tangent line through a given point on a circle and a tangent line to a given circle through a given point exterior to the circle (i.e., outside the circle). This lesson may have to be modified for students with eye-hand or fine motor difficulties. It could be done as a whole class activity where the teacher models the construction for everyone. Another option is to provide these students with an already complete step-by-step construction where each drawing shows only one step of the construction at a time. Students can try the next step but then will have an accurate drawing of the construction if they need assistance. Students should refer back to Module 1

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

for help on constructions.

Have students complete constructions individually, but pair students with a buddy who can help them if they struggle. Walk around the room and use this as an informal assessment of student understanding of constructions and lines tangent to a circle. Students will need a straight edge, a protractor, and a compass.

. Draw a circle and a radius intersecting the circle at a point labeled .  Students draw a circle and a radius and label point . 푃 . Construct a line going through point and perpendicular to the radius. Write the steps that you followed. 푃  Students draw a line perpendicular to the radius through . 푃

푃

DRAFT

. Check students’ constructions. . Draw a circle and a point exterior to the circle, and label it point .  Students construct a circle and a point exterior to the circle labeled point . 퐴 푅 . Construct a line through point tangent to the circle . 퐴 푅  This construction is difficult. Give students a few minutes to try, and then follow with the 푅 퐴 instructions that are below. . Draw segment .  Students draw segment . �퐴푅�� . Construct the perpendicular bisector of to find its midpoint. Mark the midpoint . 퐴푅��  Students construct the perpendicular bisector of and mark the midpoint . 퐴푅�� 푀 . Draw an arc of radius with center intersecting the circle. Label this point of intersection as point . 퐴푅�� 푀  Students draw an arc intersecting the circle and mark the point of intersection as point . 푀퐴 푀 퐵 . Draw line and segment . 퐵  Students draw line and segment . ⃖푅퐵��⃗ �퐴퐵�� . Is ? Verify the measurement푅 with your �퐴퐵protractor.��  Students verify that the line and radius are perpendicular. ⃖푅퐵��⃗ ⊥ �퐴퐵�� . What does this mean?

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

 Line is a tangent line to circle at point . . Repeat this process, and draw another line through point tangent to circle , intersecting the circle at point ⃖푅퐵��⃗ 퐴 퐵 . 푅 퐴  Students repeat the process, and this time the tangent line intersects the other side of the circle. 퐶

DRAFT

. What is true about , , , and ?  They are all the same length. 푀퐵 푀퐴 푀푅 푀퐶 . Let’s remember that! It may be useful for us later.

Exercises 1 - 3 (7 minutes) This proof requires students to understand that tangent lines are perpendicular to the radius of a circle at the point of tangency and then to use their previous knowledge of similar right triangles to prove = . Have students work in homogeneous pairs, helping some groups if necessary. Pull the entire class together to share proofs and see different methods used. Correct any misconceptions. 푎 푏

1. and are tangent to circle at points and respectively. Use a two-column proof to prove = . �푪푫�� 푪푬�� 푨 푫 푬 풂 풃

Draw radii and and segment . �� 푨푫�� 푨푬�� 푨푪

Lesson 11: Properties of Tangents Date: 8/29/14 5

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

= , = Given

푪푫 풂 푪푬 풃 . Tangent lines are perpendicular to the radius at the point of tangency. ∠푨푫푪 풂풏풅 ∠푨푬푪 풂풓풆 풓풊품풉풕 풂풏품풍풆풔 and are right triangles. Definition of a right triangle

∆푨푫푪= ∆푨푬푪 Radii of the same circle are equal in measure.

푨푫 = 푨푬 Reflexive property

푨푪 푨푪 . HL

∆푨푫푪= ≅ ∆ 푨푬푪 CPCTC

푪푫= 푪푬 Substitution

풂 풃 2. In circle A, the radius is 9 mm and BC = 12 mm. a. Find AC. DRAFTAC = 15 mm

b. Find the area of .

2 A = 54 mm ∆푨푪푫

c. Find the perimeter of quadrilateral ABCD. P = 42 mm

3. In circle A, EF = 12 and AE = 13. AE:AC = 1:3. Find a. The radius of the circle. 5

b. BC (round to the nearest whole number) 39

c. EC 51

Closing (3 minutes) Project the picture to the right. Have students do a 30 second quick write on all they know about the diagram if:

is tangent to the circle at point .

⃖퐹퐵��⃗ is tangent to the circle at point 퐵.

퐸퐶⃖��⃗ 퐸

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

is tangent to the circle at point .

⃖퐷퐶Then��⃗ have the class as a whole share퐷 their ideas. . , , . = 퐴퐸�� ⊥ �퐶퐸�� 퐴퐵�� ⊥ �퐹퐵�� 퐴퐷�� ⊥ �퐶퐷�� . = = 퐶퐸 퐶퐷 퐴퐵 퐴퐸 퐴

Lesson Summary

THEOREMS

. A tangent line to a circle is perpendicular to the radius of the circle drawn to the DRAFT point of tangency. . A line through a point on a circle is tangent at the point if, and only if, it is perpendicular to the radius drawn to the point of tangency.

RELEVANT VOCABULARY

. INTERIOR OF A CIRCLE. The interior of a circle with center and radius is the set of all points in the plane whose distance from the point is less than . 푶 풓 . A point in the interior of a circle is said to be inside the circle. A disk is the union of 푶 풓 the circle with its interior. . EXTERIOR OF A CIRCLE. The exterior of a circle with center and radius is the set

of all points in the plane whose distance from the point is greater than . 푶 풓 . A point exterior to a circle is said to be outside the circle. 푶 풓

. TANGENT TO A CIRCLE. A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point. This point is called the point of

tangency. . TANGENT SEGMENT/RAY. A segment is a tangent segment to a circle if the line that contains it is tangent to the circle and one of the end points of the segment is a point of tangency. A ray is called a tangent ray to a circle if the line that contains it is tangent to the circle and the vertex of the ray is the point of tangency. . SECANT TO A CIRCLE. A secant line to a circle is a line that intersects a circle in exactly two points.

. POLYGON INSCRIBED IN A CIRCLE. A polygon is inscribed in a circle if all of the vertices of the polygon lie on the circle.

. CIRCLE INSCRIBED IN A POLYGON. A circle is inscribed in a polygon if each side of the polygon is tangent to the circle.

Exit Ticket (5 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

Name Date

Lesson 11: Properties of Tangents

Exit Ticket

1. If = 9, = 6, and = 15, is line tangent to circle ? Explain. 퐵퐶 퐴퐵 퐴퐶 퐵퐶⃖��⃗

퐴

DRAFT

2. Construct a line tangent to circle through point .

퐴 퐵

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

Exit Ticket Sample Solutions

1. If BC = 9, AB = 6, and AC = 15, is line tangent to circle A? Explain.

No, is not a right triangle because⃖푩푪��⃗ + . This means is not perpendicular to . ퟐ ퟐ ퟐ ∆푨푩푪 ퟗ ퟔ ≠ ퟏퟓ 푨푩�� 푩푪�� 2. Construct a line tangent to circle through point .

푨 푩

DRAFT

Problem Set Sample Solutions Problems 1 – 6 should be completed by all students. Problems 7 and 8 are more challenging and can be assigned to some students for routine work and others as a student choice challenge.

1. If = , = , and = , is tangent to circle at point ? Explain.

Yes,푨푩 ퟓ 푩푪is a rightퟏퟐ triangle푨푪 becauseퟏퟑ the⃖푩푪��⃗ Pythagorean theorem푨 holds푩 + = . Angle B is right, so ; therefore, is tangent to ∆푨푩푪 circleퟐ atퟐ pointퟐ . ퟓ ퟏퟐ ퟏퟑ 푨푩�� ⃖푩푪��⃗ 푨 푩

2. is tangent to circle at point . = and = . ⃖푩푪��⃗ 푨 푩 푫푪 ퟗ 푩푪 a. Find the radius of the circle. ퟏퟓ =

푹 ퟖ b. Find AC. =

푨푪 ퟏퟕ

Lesson 11: Properties of Tangents Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

3. A circular pond is fenced on two opposite sides ( , ) with wood and the other two sides with metal fencing. If all four sides of �� fencing are tangent to the pond, is there more wood푪푫 푭푬 or metal fencing used?

There is an equal amount of wood and metal fencing because the distance from each corner to the point of tangency is the same.

DRAFT 4. Find x if the line shown is tangent to the circle at point B.

⁰

ퟔퟕ

5. Line is tangent to the circle at point C, and CD = DE. Find

⃖퐏퐂��⃗ a. x ( )

퐦⁰퐂퐃�

ퟏퟎퟒ b. y ( )

풎∠⁰ 푪푭푬

ퟏퟎퟒ c. z ( )

퐦∠⁰ 퐏퐂퐅

ퟑퟖ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M5 T GEOMETRY

6. Construct two lines tangent to circle A through point B.

7. DRAFTFind x, the length of the common tangent line between the two circles (round to the nearest hundredth).

= .

풙 ퟏퟏ ퟖퟑ

8. is tangent to both circles and . The radius of circle is 9, and the radius of circle is 5. The circles are 2 units apart. Find the length of , (round to the nearest hundredth). �퐄퐅�� 퐀 퐂 퐀 퐂

�퐄퐅�� 퐱

= .

풙 ퟕ ퟕퟓ

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Lesson 12: Tangent Segments

Student Outcomes . Students use tangent segments and radii of circles to conjecture and prove geometric statements, especially those that rely on the congruency of tangent segments to a circle from a given point. . Students recognize and use the fact if a circle is tangent to both rays of an angle, then its center lies on the angle bisector.

Lesson Notes The common themeDRAFT of all the lesson activities is tangent segments and radii of circles can be used to conjecture and prove geometric statements. Students first conjecture and prove that if a circle is tangent to both rays of an angle, then its center lies on the angle bisector. After extrapolating that every point on an angle bisector can be the center of a circle tangent to both rays of the angle, students show that there exists a circle simultaneously tangent to two angles with a common side. Finally, students conjecture and prove that the three angle bisectors of a triangle intersect at a single point and prove that this single point is the center of a circle inscribed in the triangle.

Scaffolding: Classwork . Give students the theorem Opening Exercise (5 minutes) from Lesson 11 if they need help getting started. Students apply the theorem from Lesson 11, two segments tangent to a circle from a point . THEOREM: Two segments outside the circle are congruent. This theorem will be used in proofs of this lesson’s main tangent to a circle from an results. Once you state the theorem below and ask the question, review Exercise 1 from exterior point are Lesson 11, or have students discuss the proof to make sure students understand the congruent. theorem.

MP.7 In the diagram below, what do you think the length of z could be? How do you know? = because each successive segment is tangent to the same circle so the segments are congruent.

풛 ퟖ . Can someone say, in your own words, the theorem used to determine ?  Students explain the theorem in their own words. 푧 . How did you use this theorem to find ?  Each successive segment was tangent to the same circle, so they were 푧 congruent. . How many times did you use this theorem?  5 times

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. Summary: By applying this theorem and transitivity over and over again, you found that each of the tangent segments has equal length, so = 8.

푧 Example 1 (7 minutes)

The point of this example is to understand why the following statement holds: If a circle is tangent to both rays of an angle, then its center lies on the angle bisector. Students explore this statement by investigating the contrapositive: attempting to draw such circles when the center is not on the angle bisector and reasoning why this cannot be done.

Example 1

In each diagram, try to draw a circle with center that is tangent to both rays of the angle .

푫 ∠푩푨푪 (a) DRAFT(b) (c)

Which diagrams did it seem impossible to draw such a circle? Why did it seem impossible?

(a) Impossible; (b) possible; (c) impossible. It’s not possible to draw a circle tangent to both rays of the angle in (a) and (c) because is a different distance from the two sides of the angle. The distance is the radius of a circle tangent to that side. 푫

MP.3 What do you conjecture about circles tangent to both rays of an angle? Why do you think that? If a circle is tangent to both rays of an angle, then the center lies on the angle bisector. We saw this in all examples.

. How many people were able to draw a circle for (a)? (b)? (c)?  It was not possible to draw such a circle for (a) and (c), but it did seem possible for (b). . What is special about (b) that wasn’t true for (a) and (c)?  The point was more in the middle of the angle in (b) than in (a) or (c).  The point is on the angle bisector. 퐷 . Why did this make a difference? 퐷  You couldn’t make a circle that was tangent to both sides at the same time because the center was too far or too close to one side of the angle.

. (If students do not see that must be on the angle bisector) Suppose you were given a different angle. Create

퐷

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different angles or have students draw some samples. Which of these points would you pick as the center of a circle tangent to both rays? How about these? How about these? What do you notice about the possible centers? (Draw two points, one that could be on the angle bisector and one that is obviously not; continue giving different examples until it is apparent that the set of viable centers is likely the angle bisector.) . State what we have just discovered to your neighbor.  Conjecture: If a circle is tangent to both rays of an angle, then the center of the circle lies on the angle bisector.

Exercises 1 – 5 (25 minutes) Allow students to work in pairs or groups to complete the exercises. You may assign certain groups particular problems and call the class together to share results. Some groups may need more guidance on these exercises. Students first proveDRAFT the conjecture made in Example 1, and it becomes a theorem. This theorem allows us to resolve the mystery opened in the last lesson: Does every triangle have an inscribed circle? Exercises 2 – 5 trace the mathematical steps from the proof of the Example 1 conjecture to the construction of the circle inscribed in a given triangle. Throughout these exercises, emphasize that the definition of angle bisector is the set of points equidistant from the rays of an angle. Make sure students understand this means both that given any point on the angle bisector, the perpendiculars dropped from this point to the rays of the angle must be the same length and that if the dropped perpendiculars are the same length, then the point from which the perpendiculars are dropped must be on the angle bisector. These observations are critical for all the exercises and especially for Exercises 3 – 5. Exercise 1 notes. The point of Exercise 1 is the following theorem. THEOREM. If a circle is tangent to both rays of an angle, then its center lies on the angle bisector. To get at the proof of this theorem, you might first poll students as to whether they used congruent triangles to show the conjecture; ask which triangles. If students ask how this proof is different from the Opening Exercise, you can point out that both proofs use the same pair of congruent triangles to draw their conclusions. However, in the Opening Exercise, the application of CPCTC (corresponding parts of congruent triangles are congruent) is for showing two legs are congruent, whereas here is it for two angles. Exercise 2 notes. The point of Exercise 2 is the following construction, whose mathematical validity is a consequence of the theorem proven in Exercise 1. CONSTRUCTION. The following constructs a circle tangent to both rays of a given angle: (1) Construct the angle bisector of the given angle; (2) Select a point of the bisector to be the center; (3) Drop a perpendicular from the selected point to the angle to find the radius; (4) Construct a circle with that center and radius. A discussion could proceed: How did you construct the center of the circle? The radius? (Select a center from the points on the angle bisector; drop a perpendicular to find the radius.) We sometimes say that the angle bisector is the set of points equidistant from the two rays of the angle. How do you know that the distance from each of your centers to the two rays of the angle is the same? (If P is the center of a circle with tangents to the circle at and , then and are radii of the circle, and all radii have the same length) Why do we know that the radius is the distance from the center to the tangent line? (The radii are perpendicular to the rays.) 퐶 퐵 푃퐶 푃퐵

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Exercise 3 notes. The point of Exercise 3 is applying Exercise 2 to the condition that the desired circle must have a center lying on angle bisectors of two angles. The key points of the argument are (1) finding a potential center of the circle, (2) finding a potential radius of the circle, and (3) establishing that a circle with this center and radius is tangent to both rays of both angles. The reasoning for the key points could be the following: (1) The angle bisectors of two angles intersect in only one point, and if there is such a circle, the center of that circle must be the intersection point. (2) A potential radius is the length of the perpendicular segment from this center to the common side shared by the angles; a line intersects a circle in one point if and only if it is perpendicular to the radius at that point. (3) The circle with this radius and center is tangent to both rays of both angles, and it is the only circle. This is because the angle bisector is the set of points equidistant from the two sides, so the perpendicular segments from the potential center to the rays of the angle are all congruent (since the distance is defined as the length of the perpendicular segment). There is only one such circle because there is only one intersection point, and the distance from this point to the rays of the angle is well defined. Exercise 4 notes. The point of Exercise 4 is extending the reasoning from Exercise 3 to conclude that all three-angle bisectors of a triangleDRAFT meet at a single point, so finding the intersection of any two points suffices. Central to Exercise 4 is the definition of angle bisector. The key idea of the argument is that the intersection point of the angle bisectors of any two consecutive angles is the same distance from both rays of both angles. Exercise 5 notes. Exercise 5 applies Exercise 4.

Exercises 1–5

1. You conjectured that if a circle is tangent to both rays of a circle, then the center lies on the angle bisector. a. Rewrite this conjecture in terms of the notation suggested by the diagram.

Given:

Circle with center and tangent to and ; = ; = 푷 푨푩 푨푪 풙 Need풎∠푪 푨푷to show:풚 ∠ 푩푨푷 =

풙 풚 b. Prove your conjecture using a two column proof. = x, = y Given

풎∠푪푨푷 , 풎∠푩푨푷 Tangent is perpendicular to the radius at the point of tangency. 푷푪�� ⊥ 푨푪�� 푷푩�� ⊥ 푨푩�� are right triangles Definition of right triangle

휟푨푩푷= 풂풏풅 휟푨푪푷 Reflexive property

푨푷 = 푨푷 Radii of the same circle are equal in length.

푷푩 푷푪 HL

휟푨푩푷 ≅ =휟푨푪푷 CPCTC

풎∠x = y푩푨푷 풎∠푪푨푷 Substitution Definition of angle bisector

푨푷�� 풃풊풔풆풄풕풔 ∠푪푨푩

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2. An angle is shown below. a. Draw at least 3 different circles that are tangent to both rays of the given angle.

DRAFT Many circles are possible. The main idea is that the center must be a point on the angle bisector, and the radius is the perpendicular from the center to the rays of the angle.

b. Label the center of one of your circles with . How does the distance between and the rays of the angle compare to the radius of the circle? How do you know? 푷 푷 The distance between P and the rays of the angle is the same length as the radius of the circle because it is perpendicular to tangent segments as proved last lesson.

3. Construct as many circles as you can that are tangent to both the given angles at the same time. You can extend the rays as needed. These two angles share a side.

Explain how many circles you can draw to meet the above conditions and how you know.

There is only one circle. The center of the circle has to be on the angle bisector of each angle; the angle bisectors only intersect in one point. There is only one circle with that point as a center that is tangent to the rays. It is the one with the radius you get by dropping the perpendicular from the point to the side .

푨 4. In a triangle, let be the location where two angle bisectors meet. Must be on the third angle bisector as well? Explain your reasoning. 푷 푷 The angle bisectors of any two consecutive angles are the same distance from both rays of both angles. Let the vertices of the triangle be , , and where = , = , and = . If point is on the angle bisector of and , then it the same distance from and , and it is the same distance from and . Thus, it is the same distance푨 from푩 푪 and 풎∠푨; therefore,ퟐ풙 it풎∠푩 is on theퟐ풚 angle bisector∠푪 ퟐ풛 of . 푷 ∠푨 ∠푩 푨푩 푨푪 푩푨 푩푪 푪푨 푪푩 ∠푪

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5. Using a straightedge, draw a large triangle . a. Construct a circle inscribed in the given triangle. 푨푩푪 Construct angle bisectors between any two angles. Their intersection point will be the center of the circle. Drop a perpendicular from the intersection point to any side. This will be a radius. Draw the circle with that center and radius.

b. Explain why your construction works. The point from Exercise 4 is the same distance from all three sides of . Thus, the perpendicular segments from to each side are the same length. A line is tangent to a circle if and only if the line is perpendicular푷 to a radius where the radius meets the circle. Therefore, 휟푨푩푪a circle with center and radius the length of the perpendicular푷 segments from to the sides is inscribed in the triangle. 푷 DRAFTc. Do you know another name for the intersection푷 of the angle bisectors in relation to the triangle?

The intersection of the angle bisectors is the incenter of the triangle.

Closing (3 minutes) Have a whole-class discussion of the topics studied today. Also, go through any questions that came up during the exercises. Today we saw why any point on an angle bisector is the center of a circle that is tangent to both rays of an angle but that any point that’s not on the angle bisector cannot possibly be the center of such a circle. This mean that we can construct a circle tangent to both rays of an angle by first constructing the angle bisector, selecting a point on it, and then dropping a perpendicular to find the radius. We put this all together to solve a mystery we raised yesterday: . Do all triangles have inscribed circles?  We found the answer is yes! The theme of all our constructions today, and for the homework, is that tangent segments and radii of circles are incredibly useful for conjecturing and proving geometric statements. . What did we discover from our constructions today?  The two tangent segments to a circle from an exterior point are congruent.  If a circle is tangent to both rays of an angle, then its center lies on the angle bisector.  Every triangle contains an inscribed circle whose center is the intersection of the triangle’s angle bisectors.

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Lesson Summary

THEOREMS

. The two tangent segments to a circle from an exterior point are congruent. . If a circle is tangent to both rays of an angle, then its center lies on the angle bisector. . Every triangle contains an inscribed circle whose center is the intersection of the triangle’s angle bisectors. DRAFT

Exit Ticket (5 minutes)

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Name Date

Lesson 12: Tangent Segments

Exit Ticket

1. Draw a circleDRAFT tangent to both rays of this angle.

2. Let and be the points of tangency of your circle. Find the measures of and . Explain how you determined your answer. 퐵 퐶 ∠퐴퐵퐶 ∠퐴퐶퐵

3. Let be the center of your circle. Find the measures of the angles in .

푃 훥퐴푃퐵

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Exit Ticket Sample Solutions

1. Construct a circle tangent to both rays of this angle. Lots of circles are possible. Check that the center of the circle is on the angle bisector and the radius is from a perpendicular from the center to a ray of the angle.

2. Let and be the points of tangency of your circle. Find the measures of and . Explain how you determined your answer. 푩 푪 ∠푨푩푪 ∠푨푪푩 Triangle is isosceles because the tangent segments and are congruent, so

° ° 휟푨푩푪= = = °. 푨푩 푨푪 ퟏퟖퟎ −ퟒퟎ 풎∠푨푩푪 풎∠푨푪푩 ퟐ ퟕퟎ 3. DRAFTLet P be the center of your circle. Find the measures of the angles in . In triangle , the angle at measures °, at measures °, and휟푨푷푩 at is °.

휟푨푷푩 푨 ퟐퟎ 푷 ퟕퟎ 푩 ퟗퟎ

Problem Set Sample Solutions It is recommended to assign Problem 8. This problem is used to open Lesson 12. Problems 1 – 7 rely heavily on the fact that two tangents from a given exterior point are congruent and, hence, that if a circle is tangent to both rays of an angle, then the center of the circle lies on the angle bisector; these problems may also do arithmetic on lengths of tangent segments. Problem 8 examines angles between tangent segments and chords. Problems involving proofs may take a while, so they can be assigned as student choice.

1. On a piece of paper, draw a circle with center and a point, , outside of the circle.

a. How many tangents can you draw from푨 to the circle?푪

2 tangents. 푪

b. Draw two tangents from to the circle, and label the tangency points and . Fold your paper along the line . What do you notice about푪 the lengths of and ? About the measures of 푫the angles푬 and ? 푨푪 푪푫 푪푬 The lengths are the same; the∠푫푪푨 angles are∠푬푪푨 congruent.

c. is the ______of .

푨푪Angle bisector ∠푫푪푬

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d. and are tangent to circle A. Find AC. = by the Pythagorean theorem. �푪푫�� 푪푬��

푨푪 ퟏퟑ

4. In the figure at right, the three segments are tangent to the circle at points , , and . If = , find , , and . ퟐ Tangents to a circle푩 푭 from푮 a given풚 pointퟑ 퐱 are congruent.풙 풚 풛 So = = , GD = = , and BC = = . This allows us to set up a system of simultaneous linear equations that 푬푭can be푬푮 solved풙 for , ,푩푫 풚. 푪푭 풛

+ = , + 풙=풚 풂풏풅, +풛 =

풙x = 27,풛 y ퟒퟖ= 18,풙 z =풚 21 ퟒퟓ 풚 풛 ퟑퟗ DRAFT

5. In the figure given, the three segments are tangent to the circle at points , , and . a. Prove = + 푱 푰 푯 = and = because tangents to a circle from a 푮푭 푮푱 푯푭 given point are congruent. 푮푱 푮푰 푭푰 푭푯 = + sum of segments = + by substitution 푮푭 푮푰 푰푭

푮푭 푮푱 푯푭 b. Find the perimeter of .

= ∆푮푪푭 = , = , = tangents to a circle from a 푪푯given pointퟏퟔ are congruent. 푮푱 = 푮푰 푰푭+ ,푭푯 =푪푱 푪푯+ sum of segments Perimeter = + + 푪푱 +푮푪 +푮푱 ( 푪푯+ )푪푭 by substitution푭푯 푮푪 푪푭 푮푭 + + + 푮푪 푪푭 푮푰 푰푭 + + + by substitution 푮푪+ 푮푰 by푪푭 substitution푰푭 푮푪 =푮푱 ( 푪푭) = 푭푯 cm 푪푱 푪푯 ퟐ푪푯 ퟐ ퟏퟔ ퟑퟐ

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6. In the given figure, the three segments are tangent to the circle at point , , and . Find .

7 m 푭 푩 푮 푫푬

DRAFT

7. is tangent to circle . If points and are the intersection points of circle A and any line parallel to ⃖푬푭��⃗ 푨 푪 푫 , answer the following.

⃖푬푭�a.��⃗ Does = for any line parallel to ? Explain. 푪푮 푮푫 ⃖푬푭��⃗ Yes. No matter what, is an isosceles triangle, and is the altitude to the base of the triangle and, therefore∆푨푪푫, the angle bisector of the angle opposite푨푮 the base.

b. Suppose that coincides with . Would C, G, and D all coincide with B? 푪푫⃖��⃗ 푬푭⃖��⃗ Yes. If they approached at different times, then at some point would not be on the same line as and , but푩 is defined to be contained in the chord푮 . 푪 푫 푮 �푪푫��

c. Suppose , , and have now reached , so is tangent to the circle. What is the angle between the line and ? 푪 푮 푫 푩 ⃖푪푫��⃗ 90⁰. A line tangent to a circle is perpendicular to the radius at the point of tangency. ⃖푪푫��⃗ �푨푩��

d. Draw another line tangent to the circle from some point, , in the exterior of the circle. If the point of tangency is point , what is the measure of ? 푷 The measure of angle푻 PTA is 90 ⁰ because a line∠푷푻푨 tangent to a circle is perpendicular to a radius through the tangency point.

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8. The segments are tangent to circle A at points B and D. is a diameter of the circle. �푬푫�� a. Prove .

�푩푬=��‖푪푨�� Inscribed angle is half the central ퟏ angle that intercepts the same 풎∠푬 ퟐ 풎∠푩푨푫 arc. bisects If a circle is tangent to both rays of an angle, then its center lies on �� 푨푪 ∠푩푨푫 the angle bisector. = Substitution If two lines are cut by a 풎∠푬 풎∠푩푨푪 transversal such that the alternate �� 푩푬 ‖푪푨 interior angles are congruent, then the lines are parallel.

DRAFTb. Prove quadrilateral is a kite. = 푨푩푪푫 tangents to a circle from a given point are congruent. 푪푩 = 푪푫 radii are congruent. Quadrilateral is a kite because two pairs of adjacent sides are congruent. 푨푩 푨푫 푨푩푪푫

9. In the diagram shown, is tangent to the circle at point . What is the relationship between , the angle between the tangent ⃖��⃗ and a chord, and the arc푩푯 subtended by that chord and its inscribed푩 angle ? ∠푫푩푯 = ∠푫푪푩

풎∠푫푩푯 풎∠푫푪푩

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Lesson 13: The Inscribed Angle Alternate a Tangent Angle

Student Outcomes . Students use the inscribed angle theorem to prove other theorems in its family (different angle and arc configurations and an arc intercepted by an angle at least one of whose rays is tangent). . Students solve a variety of missing angle problems using the inscribed angle theorem.

Lesson Notes The Opening Exercise reviews and solidifies the concept of inscribed angles and their intercepted arcs. Students then extend that knowledgeDRAFT in the remaining examples to the limiting case of inscribed angles, one ray of the angle is tangent. Example 1 looks at a tangent and secant intersecting on the circle. Example 2 uses rotations to show the connection between the angle formed by the tangent and a chord intersecting on the circle and the inscribed angle of the second arc. Students then use all of the angle theorems studied in this topic to solve missing angle problems. Scaffolding: . Post diagrams showing key theorems for students to Classwork refer to. . Use scaffolded questions Opening Exercise (5 minutes) with a targeted small This exercise solidifies the relationship between inscribed angles and their intercepted group such as, “What do arcs. Have students complete this exercise individually and then share with a neighbor. we know about the Pull the class together to answer questions and discuss part (g). measure of the intercepted arc of an inscribed angle?” In circle A, = , and is a diameter. Find the listed measure, and explain your answer. ퟎ a. 풎푩푫� ퟓퟔ �푩푪��

풎∠°푩푫푪, angles inscribed in a diameter b. ퟗퟎ ⁰

풎∠⁰푩푪푫, inscribed angle is half measure of intercepted arc c. ퟐퟖ

풎∠⁰푫푩푪, sum of angles of a triangle is 180⁰ d. ퟔퟐ

풎∠⁰푩푭푮, inscribed angle is half measure of intercepted arc e. ퟐퟖ

f. 풎 푩푪� ⁰, semicircle ퟏퟖퟎ⁰, intercepted arc풎 푫푪is� double inscribed angle

g. Isퟏퟐퟒ the = ? Explain.

No, the풎∠ central푩푮푫 angleퟓퟔ ⁰of arc would be ⁰. is not a 푩푫� ퟓퟔ ∠푩푮푫

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central angle because its vertex is not the center of the circle. h. How do you think we could determine the measure of ?

Answers will vary. This leads to today’s lesson. ∠퐁퐆퐃 Example 1 (15 minutes)

In the Lesson 12 homework, students were asked to find a relationship between the measure of an arc and an angle. The point of Example 1 is to establish the following conjecture for the class community and prove the conjecture.

CONJECTURE: Let be a point on a circle, let be a tangent ray to the circle, and let be a point on the circle such that = = is a secant to the circle. If and�� ⃗ is the angle measure of the arc intercepted by , then . 퐴 퐴퐵 퐶 1 The⃖퐴퐶��⃗ opening exercise establishes푎 empirical푚∠퐵퐴퐶 evidence푏 toward the conjecture and helps students determine∠퐵퐴퐶 whether푎 2 푏 their reasoning on the homework may have had flaws; it can be used to see how well students understand the diagram and to review how to DRAFTmeasure arcs. Students will need a protractor and a ruler.

Example 1

Diagram 1 Diagram 2

Examine the diagrams shown. Develop a conjecture about the relationship between and .

= 풂 풃 ퟏ 풂 풃 ퟐ Test your conjecture by using a protractor to measure and .

풂 풃

풂 풃 Diagram 1

Diagram 2

Do your measurements confirm the relationship you found in your homework?

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If needed, revise your conjecture about the relationship between and :

풂 풃 Now test your conjecture further using the circle below.

DRAFT

푎 푏

. What did you find about the relationship between and ?

 = . An angle inscribed between a tangent푎 line푏 and secant line is equal to half of the angle 1 measure of its intercepted arc. 푎 2 푏 . How did you test your conjecture about this relationship?  Look for evidence that students recognized that the angle should be formed by a secant intersecting a tangent at the point of tangency and that they knew to measure the arc by taking its central angle. . What conjecture did you come up with? Share with a neighbor.  Let students discuss, and then state a version of the conjecture publically.

Now, we will prove your conjecture, which is stated below as a theorem.

THE TANGENT-SECANT THEOREM: Let be a point on a circle, let be a tangent ��⃗ 푨 푨푩

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ray to the circle, and let be a point on the circle such that is a secant to the circle. If = and is the angle = measure of the arc intercepted by , then . ⃖��⃗ 푪 ퟏ 푨푪 풂 풎∠푩푨푪 풃 Given circle A with tangent , prove∠푩푨푪 what we 풂haveퟐ just풃 discovered using what you know about the properties of a circle and tangent and secant lines. 푩푮⃖��⃗ a. Draw triangle ABC. What is the measure of ? Explain.

The central angle is equal to the degree measure∠푩푨푪 of the arc it intercepts.

b. 풃⁰What is the measure of ? Explain.

The radius is perpendicular∠푨푩푮 to the tangent line at the point of tangency. ퟎ c. Expressퟗퟎ the measure of the remaining two angles of triangle ABC in terms of “ ” and explain.

The angles are congruent because the triangle is isosceles. Each angle has a measure풂 of ( – )° since + = . ퟗퟎ 풂 풎∠푨푩푪 d. DRAFT풎∠What푪푩푮 is the ퟗퟎmeasure⁰ of BAC in terms of “ ”? Show how you got the answer. The sum of the angles of∠ a triangle is °,풂 so – – + = °. Therefore, = = . ퟏ e. Explain to your neighbor what we haveퟏퟖퟎ just proveퟗퟎ n.풂 ퟗퟎ 풂 풃 ퟏퟖퟎ 풃 ퟐ풂 풐풓 풂 ퟐ 풃 An inscribed angle formed by a secant and tangent line is half of the angle measure of the arc it intercepts.

Example 2 (5 minutes)

We have shown that the inscribed angle theorem can be extended to the case when one of the angle’s rays is a tangent segment and the vertex is the point of tangency. Example 2 develops another theorem in the inscribed angle theorem’s family: the angle formed by the intersection of the tangent line and a chord of the circle on the circle and the inscribed angle of the same arc. This example is best modeled with dynamic Geometry software. Alternatively, the teacher may ask students to create a series of sketches that show point E moving towards point A. THEOREM. Suppose is a chord of circle , and is a tangent segment to the circle at point . If is any point other than or in the arc of on the opposite side of from , then = . �퐴퐵�� 퐶 �퐴퐷�� 퐴 퐸 퐴 .퐵 Draw a circle퐶 and label it C. �퐴퐵�� 퐷 푚∠퐵퐸퐴 푚∠퐵퐴퐷  Students draw circle C.

. Draw a chord .

 Students�퐀퐁�� draw chord . . Construct a segment tangent to퐀퐁 the circle through point , and label it .

 푨 Students construct�푨푫�� tangent segment . . Now draw and label point that is between �푨푫� and�� but on the other side of chord from . 푬 푨 푩  Students draw point �푨푩.� �� 푫 . Rotate point on the circle toward푬 s point . What happens to ? 푬 푨 푬푩��

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M5 T GEOMETRY

 moves closer and closer to lying on top of as E gets closer and closer to A.

. What happens푬푩�� to ? �푨푩��  moves�푬푨� �closer� and closer to lying on top of . . What happens�푬푨�� to ? �푨푫��  moves∠푩푬푨 closer and closer to lying on top of . . Does the∠ 푩푬푨 change as it rotates? ∠푩푨푫  No,풎∠ it푩푬푨 remains the same because the intercepted arc length does not change. . Explain how these facts show that = ?

 The measure of angle 풎∠ does푩푬푨 not change.풎∠푩푨푫 The segments are just rotated, but the angle measure is conserved. ∠푩푬푨 DRAFT Exercises (12 minutes) Students should work on the exercises individually and then compare answers with a neighbor. Walk around the room, and use this as a quick informal assessment.

Exercises

Find x, y, a, b, and/or c.

1. 2.

MP.1

a = 34⁰, b = 56⁰, c = 52⁰ a = 16⁰, b = 148⁰

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M5 T GEOMETRY

3. 4.

DRAFTb

a = 86⁰, b = 43⁰ x = 5, y = 2.5

MP.1

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 13 M5 T GEOMETRY

a = 60⁰

Closing (3 minutes) Have students do a 30-second quick write of everything that we have studied in Topic C on tangent lines to circles and their segment and angle relationships. Bring the class back together and share, allowing students to add to their list. . What have we learned about tangent lines to circles and their segment and angle relationships?  A tangent line intersects a circle at exactly one point (and is in the same plane).  The point where the tangent line intersects a circle is called a point of tangency. The tangent line is perpendicular to a radius whose end point is the point of tangency.  The two tangent segments to a circle from an exterior point are congruent.

 The measure of an angle formed by a tangent segment and a chord is the angle measure of DRAFT1 its intercepted arc. 2  If an inscribed angle intercepts the same arc as an angle formed by a tangent segment and a chord, then the two angles are congruent.

Lesson Summary

THEOREMS

. CONJECTURE: Let A be a point on a circle, let AB be a tangent ray to the circle, and C AC a = let be a point on the circle such that is a secant��⃗ to the circle. If and b a = b is the angle measure of the arc intercepted⃖��⃗ by , then . 1 m∠BAC . THE TANGENT-SECANT THEOREM: Let be a point∠BAC on a circle, let2 be a tangent ray to the circle, and let be a point on the circle such that is a 퐀 �퐀퐁��⃗ secant to the circle. If = and is the angle measure of the arc 퐂 ⃖퐀퐂��⃗ intercepted by , then = . 퐚 퐦∠퐁퐀퐂ퟏ 퐛 . Suppose AB is a∠ chord퐁퐀퐂 of circle퐚 Cퟐ, and퐛 AD is a tangent segment to the circle at point A. If E is any point other than A or B in the arc of C on the opposite side of AB from ��

��

Exit Ticket (5 minutes)

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Name Date

Lesson 13: The Inscribed Angle Alternate a Tangent Angle

Exit Ticket

1. Find , , and .

풂 풃 풄 DRAFT

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Exit Ticket Sample Solutions

Find a, b, and c.

a = 56⁰, b = 63⁰, c = 61⁰

Problem Set Sample Solutions

The first 6 problems are easy entry problems and are meant to help students struggling with the concepts of this lesson. They show the same problems with varying degrees of difficulty. Problems 7 – 11 are more challenging. Assign problems based on student ability. In problems 1 – 9, solve for a, b, and/or c. DRAFT 1. 2. 3.

a = 67⁰ a = 67⁰ a = 67⁰

4. 5. 6.

a = 116⁰ a = 116⁰ a = 116⁰, b = 64⁰, c = 26⁰

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7. 8. 9.

DRAFT a = 45⁰, b = 45⁰ a = 47⁰, b = 47⁰ a = 57⁰

10. is tangent to circle A. is a diameter. Find

⃖푩푯��⃗ �푫푭�� a.

풎∠푩푪푫⁰ b. ퟓퟎ

풎∠푩푨푭⁰ c. ퟖퟎ

풎∠40⁰푩푫푨 d.

풎∠40⁰푭푩푯 e.

풎∠98⁰푩푮푭

11. is tangent to circle . is a diameter. Prove: (i) = and (ii) =

(i)�푩푮�� = 푨 tangent�푩푬�� perpendicular to radius풇 풂 풅 풄 = sum of angles 풎∠푬푩푮 = ퟗퟎ⁰ angle inscribed in semi-circle 풇 ퟗퟎ −, 풆 풎∠+푬푪푩+ ퟗퟎ= ⁰ sum of angles of a triangle 푰풏=∆푬푪푩 풃 = ퟗퟎ 풆 ퟏퟖퟎ angles inscribed in same arc congruent 풃 = ퟗퟎ − 풆 substitution (ii) 풂 + 풃= inscribed in opposite arcs 풂 = 풇 inscribed in same arc 풂 + 풄 = ퟏퟖퟎ linear pairs form supplementary angles 풂+ 풇= + substitution 풇 = 풅 ퟏퟖퟎ 풄 풇 풇 풅 풄 풅

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 M5 GEOMETRY

Lesson 14: Secant Lines; Secant Lines That Meet Inside a Circle

Student Outcomes . Students understand that an angle whose vertex lies in the interior of a circle intersects the circle in two points and that the edges of the angles are contained within two secant lines of the circle. . Students discover that the measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle.

Lesson NotesDRAFT Lesson 14 begins the study of secant lines. The study actually began in Lessons 4 – 6 with inscribed angles, but we did not call the lines secant then. Therefore, students have already studied the first case, lines that intersect on the circle. In this lesson, students study the second case, secants intersecting inside the circle. The third case, secants intersecting outside the circle, will be introduced in Lesson 15.

Classwork Opening Exercise (5 minutes) This exercise reviews the relationship between tangent lines and inscribed angles, preparing students for our work in Lesson 14. Have students work on this exercise individually and then compare answers with a neighbor. Finish with a class discussion.

Opening Exercise

⃡ is tangent to the circle as shown.

a. Find the values of a and b.

a = 13, b = 80

b. Is ̅ ̅̅ ̅ a diameter of the circle? Explain.

No, if ̅ ̅̅ ̅ was a diameter, then would be .

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Discussion (10 minutes)

In this discussion, we remind students of the definitions of tangent and secant lines and Scaffolding: then have students draw circles and lines to see the different possibilities of where . Post the theorem tangent and secant lines can intersect with respect to a circle. Every student should draw definitions from previous the sketches called for and then, as a class, classify the sketches and talk about why the lessons in this module on classifications were chosen. the board so that students . Draw a circle and a line that intersects the circle. can easily review them if necessary. Add definitions  Students draw a circle and a line. and theorems as they are . Have the students tape their sketches to the board. studied. MP.7 . Let’s group together the diagrams that are alike.  Students should notice that some circles have lines that intersect the circle twice and others only touch the circle once, and students should separate them accordingly. DRAFT

. Explain how the groups are different.  A line and a circle in the same plane that intersect can intersect in one or two points. . Does anyone know what we call each of these lines?  A line that intersects a circle at exactly two points is called a secant line.  A line in the same plane that intersects a circle at exactly one point is called a tangent line. . Label each group of diagrams as “secant lines” and “tangent lines.” Then, as a class, have students write their own definition of each. . Secant Line: A secant line to a circle is a line that intersects a circle in exactly two points. . Tangent Line: A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point. . This lesson focuses on secant lines. We studied tangent lines in Lessons 11 – 13. . Starting with a new piece of paper, draw a circle and draw two secant lines. (Check to make sure that students are drawing two lines that each intersect the circle twice. This is an informal assessment of their understanding of the definition of a secant line.)  Students draw a circle and two secant lines. . Again, have students tape their sketches to the board. . Let’s group together the diagrams that are alike.

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Students should notice that some lines intersect outside of the circle, others inside the circle, others on the circle, and others are parallel and don’t intersect. Teachers may want to have a case of each prepared ahead of time in case all are not created by the students.

. We have four groups. Explain the differences between the groups.  DRAFTSome lines intersect outside of the circle, others inside the circle, others on the circle, and others are parallel and don’t intersect. . Label each group as “intersect outside the circle,” “intersect inside the circle,” “intersect on the circle,” “intersect on the circle,” and “parallel.” . Show students that the angles formed by intersecting secant line have edges that are contained in the secant lines. . Today we will talk about three of the cases of secant lines of a circle and the angles that are formed at the point of intersection.

Exercises 1 – 2 (5 minutes) Exercises 1 – 2 deal with secant lines that are parallel and secant lines that intersect on the circle (Lessons 4 – 6). When exercises are presented, students should realize that we already know how to determine the angles in these cases.

1. In circle P, ̅̅̅ ̅ is a radius, and ̂ . Find and explain how you know.

Since ̅̅̅ ̅ is a radius and extends to a diameter, the measure of the arc intercepted by the diameter is 180⁰. ̂ , so the arc intercepted by is 180⁰ - 142⁰ = 38⁰. is inscribed in this arc, so its measure is half

the degree measure of the arc or ( ) .

2. In the circle shown, ̂ . Find and ̂ . Explain your answer.

̂ ̂ ̂ ̂ because arcs between parallel lines are congruent.

By substitution, ̂ .

m ̂ ( ) because it is inscribed in a 55⁰ arc.

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Example 1 (12 minutes)

In this example, students are introduced for the first time to secant lines that intersect inside a circle.

Example 1:

a. Find . Justify your answer.

80⁰. If you draw and because they are half of the measures of their inscribed arcs. That means because the sums of the angles of a triangle total 180⁰. are supplementary, so .

. What do you think the measure of is?  DRAFTResponses will vary and many will just guess.  This is not an inscribed angle or a central angle and the chords are not congruent, so students won’t actually know the answer. That is what we want them to realize – they don’t know. . Is there an auxiliary segment you could draw that would help determine the measure of BGE? MP.7  Draw chord ̅ ̅̅ ̅. . Can you determine any of the angle measures in ∆BDG? Explain.  Yes, all of them. because it is half of the degree measure of the intercepted arc, which is 40⁰. because it is half of the degree measure of the intercepted arc, which is 120⁰. because the sum of the angles of a triangle are 180⁰. . Does this help us determine ?  Yes, DGB and BDE are supplementary, so their sum is 180⁰. That means . . The angle in part (a) above is often called a secant angle because it sides are contained in two secants of the circle such that each side intersects the circle in at least one point other than the angle’s vertex. . Is the vertical angle also a secant angle?  Yes, rays and intersect the circle at points and respectively. Let’s try another problem. Have students work in groups to go through the same process to determine . Scaffolding:

. Advanced students should b. Find x. determine x and the 31.5 general result independently.

. Use scaffolded questions with the whole class or a targeted small group.

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. Can we determine a general result? . What equation would represent the result we are looking to prove?



. Draw ̅ ̅̅ ̅  Students draw chord BD. . What are the measures of the angles in ∆BDG?



. WhatDRAFT is x?

 – ( )

. Simplify that.



. What have we just determined? Explain this to your neighbor.  The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle. . Does this formula also apply to secant lines that intersect on the circle (an inscribed angle) as in Exercise 1? . Have students look at Exercise 1 again. . What are the angle measures of the two intercepted arcs?  There is only one intercepted arc and its measure is 38⁰. . The vertical angle doesn’t intercept an arc since its vertex lies on the circle. Suppose for a minute, however, that the “arc” is that vertex point. What would be the angle measure of that “arc”?  It would have a measure of 0⁰. . Does our general formula still work using for the measure or the “arc” given by the vertical angle?

 . It does work.

. Explain this to your neighbor.  The measure of an inscribed angle is a special case of the general formula when suitably interpreted.

We can state the results of part (b) of this example as the following theorem:

SECANT ANGLE THEOREM: INTERIOR CASE. The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle.

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Exercises 3 – 7 (5 minutes) The first three exercises are straight forward, and all students should be able to use the formula found in this lesson to solve. The final problem is a little more challenging. Assign some students only Exercises 3 - 5 and others 5 - 7. Have students complete these individually and then compare with a neighbor. Walk around the room, and use this as an informal assessment of student understanding.

In Exercises 3 – 5, find x and y.

3. . 4.

DRAFT

x = 115, y = 65 x = 59, y = 76

x = 34, y = 146

6. In circle, ̅ ̅̅ ̅ is a diameter. Find x and y.

x = 24, y = 53

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7. In the circle shown, ̅̅̅ ̅ is a diameter. DC: BE = 2:1. Prove using a two-column proof.

̅̅̅ ̅ is a diameter of circle A Given Given ̂ Arc is double angle measure of inscribed angle

̂ DC:BE = 2:1 ̂ ̂ ̂ Semi-circle measures 180⁰ ̂ Arc addition ̂ Arc addition

( )

Measure of angle whose vertex lies in a circle is half the angle measures of arcs intercepted by it and its vertical angles.

Substitution and simplification.

DRAFT

Closing (3 minutes) Project the circles below on the board, and have a class discussion with the following questions.

. What types of lines are drawn through the three circles? o Secant lines . Explain the relationship between the angles formed by the secant lines and the intercepted arcs in the first two circles. o The first circle has angles with a vertex inside the circle. The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle. o The second circle has an angle on the vertex, an inscribed angle. Its measure is half the angle measure of its intercepted arc. . How is the third circle different? o The lines are parallel, and no angles are formed. The arcs are congruent between the lines.

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Lesson Summary

THEOREMS

. SECANT ANGLE THEOREM: INTERIOR CASE. The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle. RELEVANT VOCABULARY

. TANGENT TO A CIRCLE. A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point. This point is called the point of tangency.

. TANGENT SEGMENT/RAY. A segment is a tangent segment to a circle if the line that DRAFTcontains it is tangent to the circle and one of the end points of the segment is a point of tangency. A ray is called a tangent ray to a circle if the line that contains it is tangent to the circle and the vertex of the ray is the point of tangency.

. SECANT TO A CIRCLE. A secant line to a circle is a line that intersects a circle in exactly two points.

Exit Ticket (5 minutes)

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Name Date

Lesson 14: Secant Lines; Secant Lines That Meet Inside a Circle

Exit Ticket

1. Lowell says that ( ) because it is half of

the intercepted arc. Sandra says that you can’t determine the measure of because you don’t have enough information. Who is correct and why? DRAFT

2. If , find and explain how you determined your answer.

b. ̂

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Exit Ticket Sample Solutions

1. Lowell says that ( ) because it is half of the intercepted arc. Sandra says that you can’t

determine the measure of because you don’t have enough information. Who is correct and why?

Sandra is correct. We would need more information to determine the answer. Lowell is incorrect because DFC is not a central angle.

2. If , find and explain how you determined your answer.

81⁰, (supplementary angles), so – .

b. ̂

39⁰, ( ) using formula for an angle with vertex inside a circle. DRAFT

Problem Set Sample Solutions

Problems 1 – 4 are more straight forward problems. The other problems are more challenging and could be given as a student choice or specific problems assigned to different students.

In Problems 1 – 4, find x.

1. 2.

x = 85 x = 65

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3. 4.

DRAFT

x = 7 x = 9

5. Find x ( ̂ ) and y ( ̂ ).

x = 70, y = 100

6. Find the ratio of ̂ ̂ .

3:4

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7. ̅̅̅ ̅ is a diameter of circle A.

DRAFT 8. Show that the general formula we discovered in Example 1 also works for central angles. (Hint: Extend the radii to form 2 diameters, and use relationships between central angles and arc measure.)

Extend the radii to form two diameters.

Let the measure of the central angle = x⁰.

The measure ̂ because the angle measure of the arc intercepted by a central angle is equal to the measure of the central angle.

The measure of the vertical angle is also x because vertical angles are congruent.

The angle of the arc intercepted by the vertical angle is also x⁰.

The measure of the central angle is half the sum of angle measures of the arcs intercepted by the central angle and its vertical angle

( ( )).

This formula also works for central angles.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Lesson 15: Secant Angle Theorem, Exterior Case

Student Outcomes . Students find the measures of angle/arcs and chords in figures that include two secant lines meeting outside a circle, where the measures must be inferred from other data.

Lesson Notes The Opening Exercise reviews and solidifies the concept of secants intersecting inside of the circle and the relationships between the angles and the subtended arcs. Students then extend that knowledge in the remaining examples. Example 1 looks at a tangentDRAFT and secant intersecting on the circle. Example 2 moves the point of intersection of two secant lines outside of the circle and continues to allow students to explore the angle/arc relationships.

Classwork Opening Exercise (10 minutes) This Opening Exercise reviews Lesson 14, secant lines that intersect inside circles. Students must have a firm understanding of this concept to extend this knowledge to secants intersecting outside the circle. Students need a protractor for this exercise. Have students initially work individually and then compare answers and work with a partner. Use this as a way to informally assess student understanding.

1. Shown below are circles with two intersecting secant chords. Scaffolding: . Post pictures of the different types of angle and arc relationships that we have studied so far with the associated formulas to help students.

a. Measure , , and in the two diagrams. Make a conjecture about the relationship between them.

CONJECTURE about the relationship between , , and :

. The measure is the average of and .

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

2. We will prove the following. SECANT ANGLE THEOREM: INTERIOR CASE. The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle. We can interpret this statement in terms of the diagram below. Let and be the angle measures of the arcs

intercepted by the angles and . Then measure is the average of and ; that is, .

DRAFT

a. Find as many pairs of congruent angles as you can in the diagram below. Express the measures of the angles in terms of and whenever possible.

b. Which triangles in the diagram are similar? Explain how you know.

. All angles in each pair have the same measure.

c. See if you can use one of the triangles to prove the Secant Angle Theorem: Interior Case. (Hint: Use the MP.3 Exterior Angle Theorem.) – MP.3

By the exterior angle theorem, . We can conclude .

Turn to your neighbor and summarize what we've learned so far in this exercise.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Example 1 (10 minutes) Scaffolding: We have shown that the Inscribed Angle Theorem can be extended to the case when one . For advanced learners, this of the angle’s rays is a tangent segment and the vertex is the point of tangency. Example 1 example could be given as develops another theorem in the Inscribed Angle Theorem’s family, the Secant Angle individual or pair work Theorem: Exterior Case. without leading questions. . Use scaffolded questions THEOREM (Secant Angle Theorem: Exterior Case). The measure of an angle whose vertex with a targeted small lies in the exterior of the circle, and each of whose sides intersect the circle in two points, is group. equal to half the difference of the angle measures of its larger and smaller intercepted . For example: Look at the arcs. table that you created. Do Example 1 you see a pattern between the sum of b and c and the value of a? ShownDRAFT below are two circles with two secant chords intersecting outside the circle.

For example: Look at the table that you created. Do you see a pattern between

the sum of b and c and the

value of a?and the value of

Measure , , and . Make a conjecture about the relationship between them.

CONJECTURE about the relationship between , , and :

Test your conjecture with another diagram.

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Example 2 (7 minutes)

In this example, we will rotate the secant lines one at a time until one and then both are tangent to the circle. This should be easy for students to see but can be shown with dynamic geometry software.

. Let’s go back to our circle with two secant lines intersecting in the exterior of the circle (show circle at right). . Remind me how I would find the measure of angle C.

 Half the difference between the longer intercepted arc and the shorter intercepted arc.

 ̂ ̂ DRAFT . Rotate one of the secant segments so that it becomes tangent to the circle (show circle at right).

. Can we apply the same formula?

 Answers will vary, but the answer is yes.

. What is the longer intercepted arc? The shorter intercepted arc?

 The longer arc is ̂ . The shorter arc is ̂ .

. So do you think we can apply the formula? Write the formula.

 Yes. ̂ ̂

. Why is it not identical to the first formula?

 Point D is an endpoint that separates the two arcs.

. Now rotate the other secant line so that it is tangent to the circle. (Show circle at right).

. Does our formula still apply?

 Answers will vary, but the answer is yes.

. What is the longer intercepted arc? The shorter intercepted arc?

 The longer arc is ̂ . The shorter arc is ̂ .

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

. How can they be the same?

 They aren’t. We need to add a point in between so that we can show they are two different arcs.

. So what is the longer intercepted arc? The shorter intercepted arc?

 The longer arc is ̂ . The shorter arc is ̂ .

. So do you think we can apply the formula? Write the formula.

 Yes. ̂ ̂ . DRAFT

. Why is this formula different from the first two?

 Points D and E are the endpoints that separate the two arcs. Turn to your neighbor and summarize what you have learned in this exercise.

Exercises (8 minutes) Have students work on the exercises individually and check their answers with a neighbor. Use this as an informal assessment and clear up any misconceptions. Have students present problems to the class as a wrap-up.

Exercises

Find x, y, and/or z.

1. 2.

x = 28 x = 72

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

3. 4.

DRAFTx = 35 x = 66, y = 57, z = 57

Closing (5 minutes) Have students complete the summary table, and then share as a class to make sure students understand concepts.

Lesson Summary:

We have just developed proofs for an entire family of theorems. Each theorem in this family deals with two shapes and how they overlap. The two shapes are two intersecting lines and a circle.

In this exercise, you’ll summarize the different cases.

The Inscribed Angle Theorem and its Family of Theorems Relationship How the two shapes Diagram between , overlap and

Intersection is on circle.

(Inscribed Angle Theorem)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Intersection is on the

circle.

(SecantDRAFT – Tangent)

Intersection is in the

interior of the circle.

(Secant Angle Theorem: Interior)

Intersection is exterior to

the circle.

(Secant Angle Theorem: Exterior)

Lesson 15: Secant Angle Theorem, Exterior Case Date: 8/29/14 7

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Intersection of two

tangent lines to a circle.

(Two Tangent Lines)

DRAFT

Lesson Summary THEOREMS . SECANT ANGLE THEOREM: INTERIOR CASE. The measure of an angle whose vertex lies in the interior of a circle is equal to half the sum of the angle measures of the arcs intercepted by it and its vertical angle.

. SECANT ANGLE THEOREM: EXTERIOR CASE . The measure of an angle whose vertex lies in the exterior of the circle, and each of whose sides intersect the circle in two points, is equal to half the difference of the angle measures of its larger and smaller intercepted arcs.

RELEVANT VOCABULARY

. SECANT TO A CIRCLE. A secant line to a circle is a line that intersects a circle in exactly two points.

Exit Ticket (5 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Name Date

Lesson 15: Secant Angle Theorem, Exterior Case

Exit Ticket

1. Find x. Explain your answer.

DRAFT

2. Use the diagram to show that ̂ and ̂ . Justify your work.

Lesson 15: Secant Angle Theorem, Exterior Case Date: 8/29/14 9

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 M5 GEOMETRY

Exit Ticket Sample Solutions

1. Find x. Explain your answer.

x = 40. Major arc ̂ .

2. Use the diagram to show that ̂ and ̂ . Justify your work.

( ̂ ̂ ) ̂ ̂ . Angle whose vertex lies exterior of circle is equal to half the

difference of the angle measures of its larger and smaller intercepted arcs.

( ̂ ̂ ) ̂ ̂ . Angle whose vertex lies in a circle is equal to half the sum of the

arcs intercepted by the angle and its vertical angle. DRAFTAdding the two equations gives ̂ ̂ . Subtracting the two equations gives ̂ ̂ .

Problem Set Sample Solutions

1. Find x. 2. Find and .

x= 32

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3. Find . 4. Find .

DRAFT

5. Find x and y. 6. The radius of circle A is 4. ̅ ̅̅ ̅ ̅̅̅ ̅ are tangent to the circle with DC = 12. Find ̂ and the area of quadrilateral DAEC rounded to the nearest hundredth.

x = 64⁰, y = 30⁰ ̂ square units

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7. Find ̂ ̂ ̂ . 8. Find x and y.

DRAFT

̂ ̂ ̂ x = 105, y = 40

9. The radius of a circle is 6. a. If the angle formed between two tangent lines to the circle is 60⁰, how long are the segments between the point of intersection of the tangent lines and the circle?

b. If the angle formed between the two tangent lines is 120⁰, how long are the segments between the point of intersection of the tangent lines and the circle? Round to the nearest hundredth.

6.93

10. ̅̅̅ ̅ and ̅̅̅ ̅ are tangent to circle A. Prove BD = BE.

Join AD, AE, BD, and BE.

radii of same circle

reflexive property

radii perpendicular to tangent lines at point of tangency

CPCTC

̂ ̂ congruent angles intercept congruent arcs

congruent arcs intercept chords of equal measure

Lesson 15: Secant Angle Theorem, Exterior Case Date: 8/29/14 12

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

Lesson 16: Similar Triangles in Circle-Secant (or Circle- Secant-Tangent) Diagrams

Student Outcomes . Students find “missing lengths” in circle-secant or circle-secant-tangent diagrams.

Lesson Notes The Opening ExerciseDRAFT reviews Lesson 15, secant lines that intersect outside of circles. In this lesson, students continue the study of secant lines and circles, but the focus changes from angles formed to segment lengths and their relationships to each other. Examples 1 and 2 allow students to measure the segments formed by intersecting secant lines and develop their own formulas. Example 3 has students prove the formulas that they developed in the first two examples. This lesson will focus heavily on MP.8, as students work to articulate relationships among segment lengths by noticing patterns in repeated measurements and calculations.

Classwork Opening Exercise (5 minutes) We have just studied several relationships between angles and arcs of a circle. This exercise should be completed individually and asks students to state the type of angle and the angle/arc relationship, and then find the measure of an arc. Use this as an informal assessment to monitor student understanding.

Identify the type of angle and the angle/arc relationship, and then find the measure of x.

1. 2.

; inscribed angle is equal to half intercepted ; angle with vertex inside arc. circle is half sum of arcs intercepted by angle and its vertical angle.

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 1

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

3. 4.

DRAFT ; Angle with vertex outside circle has measure , central angle has measure of half the difference of larger intercepted arc of intercepted arc

Example 1 (10 minutes) Scaffolding: In Example 1, we study the relationships of segments of secant lines intersecting inside of . Model the process of circles. Students will measure and then find a formula. Allow students to work in pairs measuring and recording and have them construct more circles with secants crossing at exterior points until they values. see the relationship. Students will need a ruler. . Ask advanced students to If chords of a circle intersect, the product of the lengths of the segments of one chord is generate an additional equal to the product of the lengths of the segments of the other chord. diagram that illustrates the pattern shown and explain it. Example 1

Measure the lengths of the chords in centimeters and record them in the table.

1. 2.

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 2

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

3. 4.

CircleDRAFT # a (cm) b (cm) c (cm) d (cm) Do you notice a relationship? 1

MP.8 . What relationship did you discover?  . Say that to your neighbor in words.  If chords of a circle intersect, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

Example 2 (10 minutes)

In the second example, the point of intersection is outside of the circle and students try to develop an equation that works. Students should continue this work in groups.

Example 2

Measure the lengths of the chords in centimeters and record them in the table.

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 3

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

1. 2.

3. DRAFT 4.

Circle # a (cm) b (cm) c (cm) d (cm) Do you notice a relationship?

. Does the same relationship hold?  No. MP.8 . Did you discover a different relationship?  Yes, ( ) ( ).

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 4

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

. Explain the two relationships that you just discovered to your neighbor and when to use each formula.  When secant lines intersect inside a circle, use .  When secant lines intersect outside of a circle, use ( ) ( ).

Example 3 (12 minutes)

Students have just discovered relationships between the segments of secant and tangent line and circles. In Example 3, they will prove why the formulas work mathematically. Display the diagram at right on the board. . We are going to prove mathematically why the formulas we found in Examples 1 and 2 are valid using similar triangles. . Draw DRAFT̅ ̅̅ ̅ ̅̅̅ ̅. . Take a few minutes with a partner and prove that ∆BFD is similar to ∆EFC. . Allow students time to work while you circulate around the room. Help groups that are struggling. Bring the class back together and have students share their proofs.  Vertical angles  Inscribed in same arc  Inscribe in same arc  AA . What is true about similar triangles?  Corresponding sides are proportional. . Write a proportion involving sides ̅ ̅̅ ̅ ̅̅̅ ̅ ̅̅̅ ̅ ̅̅̅ ̅.



. Can you rearrange this to prove the formula discovered in Example 1?  (BF)(FC) = (DF)(FE) . Display the next diagram on the board. . Display the diagram at right on the board. . Not let’s try to prove the formula we found in Example 2. . Name two triangles that could be similar.  ∆CFB and ∆CED

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 5

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

. Take a few minutes with a partner and prove that ∆CFB is similar to ∆CED. . Allow students time to work while you circulate around the room. Help groups that are struggling. Bring the class back together and have students share their proofs.  Common angle  Inscribed in same arc  AAA . Write a proportion that will be true.



. Can you rearrange this to prove the formula discovered in Example 2?  DRAFT(CE)(CB) = (CF)(CD) . What if one of the lines is tangent and the other is secant? Show diagram.  Students should be able to reason that ( )  ( )

 a = √ ( )

Closing (3 minutes) We have just concluded our study of secant lines, tangent lines, and circles. In lesson 15, you completed a table about angle relationships. This summary completes the table adding segment relationships. Complete the table below and compare your answers with your neighbor. Bring class back together to discuss answers to ensure students have the correct formulas in their tables.

Lesson Summary: The Inscribed Angle Theorem and Its Family

Diagram How the two shapes overlap Relationship between , and

Intersection is in the interior of the

circle.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

Intersection is on the exterior of ( ) ( ) the circle.

Tangent and secant are ( ) intersecting.

DRAFT

Lesson Summary

THEOREMS

. When secant lines intersect inside a circle, use .

. When secant lines intersect outside of a circle, use ( ) ( ).

RELEVANT VOCABULARY . SECANT TO A CIRCLE. A secant line to a circle is a line that intersects a circle in exactly two points.

Exit Ticket (5 minutes)

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 7

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

Name Date

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant- Tangent) Diagrams

Exit Ticket

1. In the circle below, ̂ = 30⁰, ̂ = 120⁰, CG = 6, GH = 2, FH = 3, CF = 4, HE = 9, and FE = 12. DRAFT

a. Find a ( ).

b. Find b ( ) and explain your answer.

c. Find x (HD) and explain your answer.

d. Find y (DG).

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 8

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

Exit Ticket Sample Solutions

1. In the circle below, ̂ = 30⁰, ̂ = 120⁰, CG = 6, GH = 2, FH = 3, CF = 4, HE = 9, and FE = 12. a. Find a ( ).

a = 75⁰

b. Find b ( ).

b = 45⁰; b is an angle with vertex outside of the circle, so it has measure half the difference between its larger and smaller intercepted arcs.

c. Find x (HD). DRAFTx = 6; x is part of a secant line inside the circle, so

d. Find y (DG).

Problem Set Sample Solutions

1. Find x. 2. Find x.

x = 8 x = 3

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 9

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

3. Prove DF = 3 4. CE = 6, CB = 9, CD = 18. Show CF = 3.

7⋅6 = 42, so DF⋅FE must equal 42. IF DF < FE, 6⋅9 = 54 and 18⋅CF = 54. This means CF = 3. DRAFTDF could equal 1, 3, or 6. DF ≠1 and DF < FB, so DF must equal 3.

4. Find x. 6. Find x.

x = √ x = 11.25

7. 8.

x = 32 x = 9

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 16 M5 GEOMETRY

9. In the circle shown, DE = 11, BC = 10, DF = 8. Find FE, BF, FC.

FE = 3, BF = 4, FC = 6

DRAFT 10. In the circle shown, ̂ ̂

a. Find .

60⁰

b. Prove .

Angles have same measure.

Vertical angles are congruent.

c. Set up a proportion using sides ̅ ̅̅ ̅ and ̅̅̅ ̅.

d. Set up an equation with ̅ ̅̅ ̅ and ̅̅̅ ̅ using a theorem for segment lengths from this section.

( )

e. Solve for CE and GE.

CE = 8, GE = 4

Lesson 16: Similar Triangles in Circle-Secant (or Circle-Secant-Tangent) Diagrams Date: 8/29/14 11

GEOMETRY • MODULE 5

Topic D: Equations for Circles and Their Tangents

G-GPE.A.1, G-GPE.A.4

Focus Standard:DRAFT G-GPE.A.1 Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation. G-GPE.A.4 Use coordinates to prove simple geometric theorems algebraically. Instructional Days: 3 Lesson 17: Writing the Equation for a Circle (P) Lesson 18: Recognizing Equations of Circles (P) Lesson 19: Equations for Tangent Lines to Circles (P)

Topic D consists of three lessons focusing on MP.7. Students see the structure in the different forms of equations of a circle and lines tangent to circles. In Lesson 17, students deduce the equation for a circle in center-radius form using what they know about the Pythagorean theorem and the distance between two points on the coordinate plane (G-GPE.A.1). Students first understand that a circle whose center is at the origin of the coordinate plane is given by 2 + 2 = 2, where is the radius. Using their knowledge of translation, students derive the general formula for a circle as ( )2 + ( )2 = 2, where is the radius and ( , ) is the center of the circle.𝑥𝑥 In Lesson𝑦𝑦 𝑟𝑟 18, students𝑟𝑟 use their algebraic skills of factoring and completing the square to transform equations into center-radius.𝑥𝑥 Students −𝑎𝑎 𝑦𝑦prove −𝑏𝑏 that 𝑟𝑟2 + 2 +𝑟𝑟 + + = 0 is the𝑎𝑎 equation𝑏𝑏 of a circle and find the formula for the center and radius of this circle (G-GPE.4). Students know how to recognize the equation of a circle once the equation format is in𝑥𝑥 center𝑦𝑦 -radius.𝐴𝐴𝐴𝐴 In𝐵𝐵 𝐵𝐵 Lesson𝐶𝐶 19, students again use their algebraic skills to write the equations of lines, specifically lines tangent to a circle, using information about slope and/or points on the line. Bringing together students’ understanding of tangent from Lesson 11 and the equations of circles from Lessons 17 and 18, students determine the equation of tangent lines to a circle from points outside of the circle.

Topic D: Equations for Circles and Their Tangents Date: 8/29/14 1

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

Lesson 17: Writing the Equation for a Circle

Student Outcomes . Students write the equation for a circle in center-radius form, ( ) ( ) using the Pythagorean theorem or the distance formula. . Students write the equation of a circle given the center and radius. Students identify the center and radius of a circle given the equation.

Lesson Notes In this lesson, studentsDRAFT deduce the equation for a circle in center-radius form, ( ) ( ) using what they already know about the Pythagorean theorem and the distance formula: the distance between two points, ( ) and ( ) is √( ) ( ) . Exercise 11 foreshadows the work of the next lesson where students will need to complete the square in order to determine the equation of a circle.

Classwork Opening Exercise (4 minutes)

1. What is the length of the segment shown on the coordinate plane below?

Students may use the Pythagorean theorem or the distance formula to determine the length of the segment to be units.

2. Use the distance formula to determine the distance between points ( ) and ( )

√( ) ( ) √

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

Example 1 (10 minutes)

. If we graph all of the points whose distance from the origin is equal to , what shape will be formed?  By definition, the set of all points in the plane whose distance from the origin is 5 units is called a circle. . Our goal now is to find the coordinates of eight of those points that comprise the circle. Four are very easy to find. What are they? Provide time for students to think and discuss how to find the coordinates of the four “easy” points. Have students explain how they got their coordinates.  The four points are ( ) ( ) ( ) and ( ) To find these points, we went right, left, up, and down units from the origin. . Now we need to locate more points on the circle. We need the distance from the origin, i.e., the center of the circle, to be Graphically, we are looking for the coordinates ( ) that are exactly units from the center of the circle: DRAFT

. Identify one other point on the circle that appears to be units from the center of the circle.  Students may identify ( ) ( ) ( ) ( ) ( ) ( ) ( ) or ( ) . What can we do to be sure that the distance between the center of the circle and the identified point is in fact

Provide time for students to discuss the answer to this question. Some students may say they could use the Pythagorean theorem and others may say they could use the distance formula. Since the distance formula is derived from the Pythagorean theorem, both answers are correct. Encourage students to explain their use of either strategy. For example, using the Pythagorean theorem and point ( ) we have the following:

Lesson 17: Writing the Equation for a Circle Date: 8/29/14 2

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

DRAFT Scaffolding: Using the coordinates, we know that one leg of the right triangle formed above has length . Using a chart to organize and the other has length . We must check that the hypotenuse is equal to To that calculations could be end, is true, and the point ( ) is units from the center of the circle. This helpful. See example process can be repeated to check the three other points, but it is not necessary. below.

The distance formula will bring students to the same conclusion. To find the distance between the origin and the point ( ), we must calculate √( ) ( ) . We must show that the distance between the two points is . Make clear to students that using the distance formula in this case, where the center is at the origin, is no different from the strategy of using the Pythagorean theorem because √( ) ( ) √ . Based on our work using the Pythagorean theorem, we can say that any point ( ) on this circle whose center is at the origin ( ) and whose radius is must satisfy the equation In other words, all solutions to the equation are the points of the circle.

Example 2 (10 minutes)

Let’s look at another circle’ one whose center is not at the origin. Shown below is a circle with center ( ) with radius

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

DRAFT

. Again, there are four points that are easy to locate and others that can be verified using the Pythagorean theorem or distance formula. What are the differences between this circle and the one we just looked at in Example 1? Provide students time to discuss the answer to this question.  Both of the circles have a radius of , but their centers are different, which makes the points that comprise the circles different. MP.3 . Are the circles congruent? Is there a sequence of basic rigid motions that would take this circle to the origin? Explain.  Yes, the circles are congruent because both have a radius equal to We could map one circle onto the other using a translation. For example, we could translate the circle with center at ( ) to the origin by translating along a vector from point Scaffolding: ( ) ( ) to point . Give students specific . What effect does the translation have on all of the points from the circle above? points to compare. For example, compare the Show the circles side by side. Provide time for students to discuss this with partners. point ( ) from the circle  Each -coordinate is decreased by and each -coordinate is decreased above to the point ( ) by on the circle whose center is at the origin. . The effect that translation has on the points can be expressed as the following. Let ( ) be any point on the circle with center ( ) Then, the coordinates of all of the points ( ) after the translation are: (( ) ( )) . Since the radius is equal to , we can locate any point ( ) on the circle using the Pythagorean theorem as we did before. ( ) ( )

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

The solutions to this equation are all the points of a circle whose radius is and center is at ( ) . What do the numbers and represent in the equation above?  The and represent the location of the center ( ) and the is the radius. . Assume we have a circle with radius whose center is at ( ) What is an equation whose graph is that circle? Provide time for students to discuss this in pairs.  The circle with radius and center at ( ) is given by the graph of the equation ( ) ( ) . . Assume we have a circle with radius whose center is at ( ) What is an equation whose graph is the that circle? Provide time for students to discuss this in pairs.  DRAFTThe circle with radius and center at ( ) is given by the graph of the equation ( ) ( ) . . The last equation, ( ) ( ) is the general equation for any circle with radius and center ( )

Exercises 3–11 (12 minutes) Students should be able to complete Exercises 3–5 independently. Check that the answers to Exercises 3–5 are correct before assigning the remaining exercises in the set.

Exercises 3–11 (same heading used in SF)

3. Write an equation for the circle whose center is at ( ) and has radius

( )

4. Write an equation whose graph is the circle below.

( ) ( )

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

5. What is the radius and center of the circle given by the equation ( ) ( )

The radius is and the center is ( )

6. Petra is given the equation ( ) ( ) and identifies its graph as a circle whose center is ( ) and radius is Has Petra made a mistake? Explain.

Petra did not identify the correct center. The general form for the equation of a circle is given by ( ) ( ) where ( ) is the center and is the radius. Petra noted the value of as when it is really and the value of as when it is really Therefore, Petra should have identified the center as ( ). The radius was identified correctly.

7. a. What is the radius of the circle with center ( ) that passes through ( )

( ) ( ) ( ) ( )

DRAFT√

b. What is the equation of this circle?

( ) ( )

8. A circle with center ( ) is tangent to the -axis. a. What is the radius of the circle?

b. What is the equation of the circle?

9. Two points in the plane, ( ) and ( ), represent the endpoints of the diameter of a circle. a. What is the center of the circle? Explain. ( ); the midpoint of the diameter.

b. What is the radius of the circle? Explain. 10, the distance from one endpoint to the center

c. Write the equation of the circle. ( ) ( )

10. Consider the circles with equations:

( ) ( ) a. What are the radii of the circles?

The radii are and 10.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

b. What is the distance between the centers of the circles?

√( ) ( )

c. Make a rough sketch of the two circles to explain why the circles must be tangent to one another.

DRAFT

The circles must be tangent because there is just one point that is common to both graphs (or there is only one solution that satisfies both equations), i.e., ( )

11. A circle is given by the equation ( ) ( ) Scaffolding: a. What is the center of the circle? . If students are struggling, The center is at ( ) have them practice factoring perfect square b. What is the radius of the circle? trinomials. . a2 + 2ab + b2 = (a + b)2 The radius is . a2 - 2ab + b2 = (a - b)2

. Factor c. Describe what you had to do in order to determine the center and the radius of the 2 circle. o x + 4x + 4 2 I had to factor each of the trinomials to get the equation in the proper form in order o x – 6x + 9 to identify the center of the circle. To get the radius, I had to take the square root of

Closing (4 minutes) Have students summarize the main points of the lesson in writing, by talking to a partner, or as a whole class discussion. Use the questions below, if necessary. . Which fundamental theorem was critical for allowing us to write the equation of a circle? . The equation of a circle can always be rewritten into what form? . What parts of the equation give information about the center of the circle? The radius?

Lesson 17: Writing the Equation for a Circle Date: 8/29/14 7

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

Lesson Summary

. ( ) ( ) is the general equation for any circle with radius and center ( )

Exit Ticket (5 minutes)

DRAFT

Lesson 17: Writing the Equation for a Circle Date: 8/29/14 8

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

Name Date

Lesson 17: Writing the Equation for a Circle

Exit Ticket

1. Describe the circle given by the equation ( ) ( )

DRAFT 2. Write the equation for a circle with center ( ) and radius

3. Write the equation for the circle shown below.

4. A circle has a diameter with endpoints at ( ) and ( ) Write the equation for the circle.

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Exit Ticket Sample Solutions

1. Describe the circle given by the equation ( ) ( )

The circle has a center at ( ) and a radius of

2. Write the equation for a circle with center ( ) and radius

( )

3. DRAFTWrite the equation for the circle shown below.

4. A circle has a diameter with endpoints at ( ) and ( ) Write the equation for the circle.

( ) ( )

Problem Set Sample Solutions

1. Write the equation for a circle with center ( ) and radius √

( ) ( )

2. What is the center and radius of the circle given by the equation ( )

The center is located at ( ) and the radius is

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3. A circle is given by the equation Which of the following points are on the circle? a. ( ) This point is on the circle.

b. ( ) This point is on the circle.

c. ( ) This point is not on the circle.

d. ( ) DRAFTThis point is not on the circle.

e. ( ) This point is on the circle.

4. Determine the center and radius of each circle: a. The center is at ( ) and the radius is

b. ( ) ( )

The center is at ( ) and the radius is √

c. ( ) ( ) The center is at ( ) and the radius is

5. A circle has center ( ) and passes through the point ( ) a. What is the radius of the circle?

( ) ( ) ( ) ( )

b. Write the equation of the circle.

( ) ( )

6. Two points in the plane, ( ) and ( ) represent the endpoints of the diameter of a circle. a. What is the center of the circle?

( )

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 17 M5 GEOMETRY

b. What is the radius of the circle?

c. Write the equation of the circle.

( ) ( )

7. DRAFTWrite the equation of the circle shown below.

( ) ( )

8. Write the equation of the circle shown below.

( ) ( )

9. Consider the circles with equations:

( ) ( ) a. What are the radii of the two circles?

The radii are √ and √

b. What is the distance between their centers?

√ √ √ √ √

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c. Make a rough sketch of the two circles to explain why the circles must be tangent to one another.

The circles must be tangent because there is just one point that is common to both graphs (or there is only one solution that satisfies both equations), i.e., ( )

DRAFT

Lesson 17: Writing the Equation for a Circle Date: 8/29/14 13

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

Lesson 18: Recognizing Equations of Circles

Student Outcomes . Students complete the square in order to write the equation of a circle in center-radius form. . Students recognize when a quadratic in and is the equation for a circle.

푥 푦 Lesson Notes This lesson builds from the understanding of equations of circles in Lesson 17. The goal is for students to recognize when a quadratic equation in and is the equation for a circle. The Opening Exercise reminds students of algebraic skills that are neededDRAFT in this lesson, specifically multiplying binomials, factoring trinomials, and completing the square. Throughout the lesson, students푥 will푦 need to complete the square (A-SSE.3b) in order to determine the center and radius of the circle. Completing the square was taught in Grade 9, Module 4, Lessons 11 and 12.

Classwork Opening Exercise (6 minutes) It is important that students can factor trinomial squares and complete the square in order to recognize equations of circles. Use the following exercises to determine which students may need remediation of these skills.

Opening Exercise

x 3 x x2 3x

40 3 3x ? =

x 3 x x2 3x

49 3 3x 9 =

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a. Express this as a trinomial: ( ) . ퟐ Scaffolding: ( ) = + 풙 − ퟓ . ퟐ ퟐ For English language learners and 풙 − ퟓ 풙 − ퟏퟎ풙 ퟐퟓ students that may be below grade level, b. Express this as a trinomial: ( + ) . a visual approach may help students ퟐ understand completing the square. ( + ) = + + 풙 ퟒ . For example, students may view an ퟐ ퟐ 2 풙 ퟒ 풙 ퟖ풙 ퟏퟔ equation such as x + 6x = 40 first by c. Factor the trinomial : + + . developing a visual that supports the ퟐ left side of the equation. + + = ( + ) 풙 ퟏퟐ풙 ퟑퟔ ퟐ ퟐ 풙 ퟏퟐ풙 ퟑퟔ 풙 ퟔ x 3 2 d. Complete the square to solve the following equation: + = . x x 3x ퟐ + = 풙 ퟔ풙 ퟒퟎ + ퟐ + = + DRAFTퟐ (풙 + ퟔ풙) = ퟒퟎ 3 3x 풙 ퟔ풙 + ퟗퟐ = ퟒퟎ ퟗ 풙 ퟑ = ퟒퟗ 풙 ퟑ ퟕ 풙 ퟒ

Ask students to share answers and discuss parts that they may have found challenging. . Students then view solving the equation as “finding the missing piece” (literally “completing the square”). Example 1 (4 minutes)

. The following is the equation of a circle with radius 5 and center (1,2). Do you see why? 2 + 1 + 4 + 4 = 25 2 2 Provide time for students to think about this in pairs푥 − or푥 small groups.푦 − If푦 necessary, guide their thinking by telling students that parts (a)–(c) in the Opening Exercise are related to the work they will be doing in this example. Allow individual students or groups of students to share their reasoning as to how they make the connection between the MP.1 information given about a circle and the equation. Once students have shared their thinking, continue with the reasoning below. . We know that the equation 2 + 1 + 4 + 4 = 25 is a circle with Scaffolding: radius 5 and center (1, 2) because2 when we multiply2 out the equation . Ask students to identify a ( 1) + ( 2) = 5 , we푥 get− 푥 2 +푦 1− + 푦 4 + 4 = 25. point (perhaps by 2 2 2 2 2 sketching the graph) that Provide students time to verify that these equations are equal. 푥 − 푦 − 푥 − 푥 푦 − 푦 should be on the circle, . Recall the equation for a circle with center ( , ) and radius from the previous and verify that it satisfies lesson. the equation given. 푎 푏 푟  ( ) + ( ) = . Multiply out each2 of the binomials2 2 to write an equivalent equation. 푥 − 푎 푦 − 푏 푟  2 + + 2 + = . Sometimes2 equations 2of circles2 are presented2 2 in this “simplified” form. To easily identify the center and the 푥 − 푎푥 푎 푦 − 푏푦 푏 푟 radius of the graph of the circle, we sometimes need to factor and/or complete the square in order to rewrite the equation back in its standard form ( ) + ( ) = . 2 2 2 푥 − 푎 푦 − 푏 푟

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Exercise 1 (3 minutes) Use the exercise below to assess students’ understanding of the content in Example 1.

Exercise 1 Scaffolding: 1. Rewrite the following equations in the form ( ) + ( ) = . a. + + + + = ퟐ ퟐ ퟐ . Students can draw squares as in the 풙 − 풂 풚 − 풃 풓 ퟐ ퟐ Opening Exercise to complete the 풙( + ퟒ풙) +ퟒ( 풚 )− ퟔ풙= ퟗ ퟑퟔ ퟐ ퟐ square. 풙 ퟐ 풚 − ퟑ ퟑퟔ . For advanced learners, offer b. + + + + = equations with coefficients on one ퟐ ퟐ or both of the squared terms. For 풙( − ퟏퟎ풙) + (ퟐퟓ+ 풚) = ퟏퟒ풚. ퟒퟗ ퟒ ퟐ ퟐ example, 풙 − ퟓ 풚 ퟕ ퟒ 3 + 12 + 4 48 = 197. DRAFT 2 2 Example 2 (5 minutes) 푥 푥 푦 − 푦 . What is the center and radius of the following circle? + 4 + 12 = 41 2 2 Provide time for students to think about this in pairs푥 or small푥 groups.푦 − If푦 necessary, guide their thinking by telling students that part (d) in the Opening Exercise is related to the work they will be doing in this example. Allow individual students or groups of students to share their reasoning as to how they determine the radius and center of the circle. . We can complete the square, twice, in order to rewrite the equation in the necessary form. First, + 4 : MP.1 2 + 4 + 4 = ( + 2) . Second,푥 푥 12 : 2 2 2 푥 12푥 + 36 =푥( 6) . Then, 푦 − 푦 2 2 + 4 + 푦 −12 푦 = 41 푦 − ( 2 + 4 + 42) + ( 12 + 36) = 41 + 4 + 36 (푥 2+ 2)푥 +푦( − 6)푦2 = 81. 푥 2푥 푦2 − 푦 Therefore, the center is at (푥 2, 6), and 푦the − radius is 9.

− Exercises 2–4 (6 minutes) Use the exercise below to assess students’ understanding of the content in Example 2.

Exercises 2–4

2. Identify the center and radius for each of the following circles. a. + + = ퟐ ퟐ 풙( − ퟐퟎ풙) +풚( +ퟔ풚) =ퟑퟓ ퟐ ퟐ The풙 − centerퟏퟎ is (풚 , ퟑ ), andퟏퟒퟒ the radius is . ퟏퟎ −ퟑ ퟏퟐ

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b. + = ퟐ ퟐ ퟏퟗ 풙 − ퟑ풙 풚 − ퟓ풚 ퟐ ퟐ + ퟐ = ퟑ ퟓ �풙 − � �풚 − � ퟏퟖ The centerퟐ is , ퟐ , and the radius is . Scaffolding: ퟑ ퟓ � � ퟏퟖ ퟐ ퟐ √ . Allow accelerated groups 3. Could the circle with equation + = have a radius of 4? Why or why not? of students time to ퟐ ퟐ complete the square on (풙 − ퟔ풙) + 풚 =− ퟕ ퟎ ퟐ ퟐ their own. Yes, the radius is 4. 풙 − ퟑ 풚 ퟏퟔ . For students who are struggling, ask questions 4. Stella says the equation + + = has a center of (4, -2) and a radius of . such as the following: Is she correct? Why or whyퟐ not? ퟐ . 풙 − ퟖ퐱 풚 ퟐ퐲 ퟓ √ퟐퟐ “What is the final form we ( ) + ( + ) = are trying to put this DRAFTퟐ ퟐ equation in?” The center is (4, -2), but the 풙radius − ퟒ is 5. 풚She ퟐdid notퟐퟓ add the values to the left that she added to the right when completing the square. . “What would we need to add to x2 + Ax to create a

perfect square trinomial?” Example 3 (8 minutes)

. Could + + + + = 0 represent a circle?  2Let students2 think about this. Answers will vary. 푥 푦 퐴푥 퐵푦 퐶 The goal is to be able to recognize when the graph of an equation is, in fact, a circle. Suppose we look at ( ) + ( ) = another way: 2 푥 − 푎 2 2 2 + + 2 + = . 푦 − 푏 푟 This equation can be rewritten using the commutative2 and a2ssociative2 properties2 of2 addition. We will also use the 푥 − 푎푥 푎 푦 − 푏푦 푏 푟 subtraction property of equality to set the equation equal to zero: + 2 2 + ( + ) = 0. . What is the equation of a circle?2 2 2 2 2 푥 푦 − 푎푥 − 푏푦 푎 푏 − 푟  ( ) + ( ) = . Expand the equation.2 2 2 푥 − 푎 푦 − 푏 푟  2 + + 2 + = . Now rearrange2 terms 2using 2the commutative2 2and associative properties of addition to match + + + 푥 − 푎푥 푎 푦 − 푏푦 푏 푟 + = 0. 2 2 푥 푦 퐴푥  + 2 2 + ( + ) = 0 퐵푦 퐶 . What exp2ressions2 are equivalent to2 , ,2 and 2 ? 푥 푦 − 푎푥 − 푏푦 푎 푏 − 푟  = 2 , = 2 , and = + , where , , and are constants 퐴 퐵 퐶 . So, could x + y + Ax + By + C = 0 represent2 2 a 2circle? 퐴 − 푎 퐵 − 푏 퐶 푎 푏 − 푟 퐴 퐵 퐶  Yes2. 2 . The graph of the equation + + + + = 0 is either 1) a circle, 2) a point, or 3) an empty set. We want to be able to determine2 which2 of the three possibilities is true for a given equation. 푥 푦 퐴푥 퐵푦 퐶

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

. Let’s begin by completing the square: + + + + = 0. 2 2 푥 푦 퐴푥 퐵푦 퐶 + + + + = 0 2 + 2 + + + = 0 + +푥 2 푦+ 퐴푥+2 퐵푦+ 퐶 = 2 푥 퐴푥 2 푦 퐵푦 퐶 푥 + 퐴푥+ + 푦 + 퐵푦+ = −퐶 + + 2 2 2 2 2 2 2 2 2 퐴 2 퐵 퐴 퐵 푥 퐴푥 � � +푦 +퐵푦 +� � = −퐶 + � � + � � 2 2 2 2 2 2 2 2 퐴 퐵 4 퐴 퐵 �푥 + � + �푦 + � = −퐶 +� � + � � 2 2 2 2 4 42 42 퐴 퐵 +퐶 퐴 4 퐵 �푥 + � + �푦 + � = − 2 2 2 2 2 42 퐴 퐵 퐴 퐵 − 퐶 DRAFT�푥 � �푦 � . Just as the discriminant is used to determine how many times a graph will intersect the -axis, we use the right side of the equation above to determine what the graph of the equation will look like. 푥

. If the fraction on the right is positive, then we can identify the center of the circle as , and the radius 퐴 퐵 as = = + 4 . �− 2 − 2� 2 2 � 2 2 퐴 +퐵 −4퐶 퐴 +퐵 −4퐶 1 2 2 . What� do4 you think it means2 if 2the√퐴 fraction퐵 −on the퐶 right is zero?  It means we could locate a center, , , but the radius is 0; therefore, the graph of the equation 퐴 퐵 is a point, not a circle. �− 2 − 2� . What do you think it means if the fraction on the right is negative?  It means that we have a negative length of a radius, which is impossible. . When the fraction on the right is negative, we have an empty set. That is, there are no possible solutions to this equation because the sum of two squared numbers cannot be negative. If there are no solutions, then there are no ordered points to graph. . We can apply this knowledge to equations whose graphs we cannot easily identify. For example, write the equation + + 2 + 4 + 5 = 0 in the form of ( ) + ( ) = .  x +2 y +2 2x + 4y + 5 = 0 2 2 2 푥 푦 푥 푦 푥 − 푎 푦 − 푏 푟 2 2 ( + 2 + 1) + ( + 4 + 4) + 5 = 5 ( + 1) + ( + 2) = 0 푥 푥 푦 푦 Since the right side of the equation is zero, then2 the graph of2 the equation is a point, ( 1, 2). 푥 푦

− −

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Exercise 5 (6 minutes) Use the exercise below to assess students’ understanding of the content in Example 3.

Exercise 5

5. Identify the graphs of the following equations as a circle, a point, or an empty set. a. + + = ퟐ ퟐ 풙 풚 ퟒ풙 ퟎ + + + = ퟐ ( + ) + ퟐ = 풙 ퟒ풙 ퟒퟐ 풚ퟐ ퟒ The right side of the equation is positive, so the풙 graphퟐ of풚 the ퟒequation is a circle.

b. + + + = ퟐ ퟐ 풙 풚 ퟔ풙 − ퟒ풚 ퟏퟓ ퟎ( + + ) + ( + ) + = DRAFTퟐ ( ퟐ ) + ( ) = 풙 ퟔ풙 ퟗ 풚 − ퟒ풚ퟐ ퟒ ퟏퟓퟐ ퟏퟑ The right side of the equation is negative, so the풙 graph − ퟑ of this풚 − equation ퟐ −ퟐ cannot be a circle; it is an empty set.

c. + + + = ퟐ ퟐ ퟏퟑ ퟐ풙 ퟐ풚 − ퟓ풙 풚 ퟒ ퟎ + + + = ퟏ ퟐ ퟐ ퟏퟑ �ퟐ풙 ퟐ풚 +− ퟓ풙+ 풚 + � = ퟎ ퟐ ퟒ ퟐ ퟓ ퟐ ퟏ ퟏퟑ 풙 − 풙 풚 풚 ퟎ + ퟐ + ퟐ+ + ퟐ ퟐ + ퟖ = ퟐ + ퟐ ퟐ ퟓ ퟓ ퟐ ퟏ ퟏ ퟏퟑ ퟓ ퟏ 풙 − 풙 � � 풚 풚 � � � � � � ퟐ ퟒ ퟐ ퟐ +ퟒ + ퟖퟐ = ퟒ + ퟒ ퟐ + ퟐ ퟓ ퟏ ퟏퟑ ퟓ ퟏ �풙 − � �풚 � − � � � � ퟒ ퟐ + + ퟒ ퟐ = ퟖ + ퟒ ퟒ ퟓ ퟏ ퟏퟑ ퟐퟔ �풙 − � �풚 � − ퟒ ퟐ + + ퟒ ퟐ = ퟖ ퟏퟔ ퟓ ퟏ �풙 − � �풚 � ퟎ The right side of the equation is equal to zeroퟒ, so the graphퟒ of the equation is a point.

Closing (2 minutes) Have students summarize the main points of the lesson in writing, by talking to a partner, or as a whole class discussion. Use the questions below, if necessary. . Describe the algebraic skills that were necessary to work with equations of circles that were not of the form ( ) + ( ) = . . Given the2 graph of a2 quadratic2 equation in and , how can we recognize it as a circle or something else? 푥 − 푎 푦 − 푏 푟 푥 푦 Exit Ticket (5 minutes)

Lesson 18: Recognizing Equations of Circles Date: 8/29/14 6

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

Name Date

Lesson 18: Recognizing Equations of Circles

Exit Ticket

1. The graph of the equation below is a circle. Identify the center and radius of the circle. + 10 + 8 8 = 0 2 2 푥 푥 푦 − 푦 −

DRAFT

2. Describe the graph of each equation. Explain how you know what the graph will look like. a. + 2 + = 1 2 2 푥 푥 푦 −

b. + = 3 2 2 푥 푦 −

c. + + 6 + 6 = 7 2 2 푥 푦 푥 푦

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

Exit Ticket Sample Solutions

1. The graph of the equation below is a circle. Identify the center and radius of the circle. + + = ퟐ ퟐ (풙 + ퟏퟎ풙) + ( 풚 −)ퟖ풚=− ퟖ ퟎ ퟐ ퟐ The center is ( , ), and the radius is . 풙 ퟓ 풚 − ퟒ ퟒퟗ −ퟓ ퟒ ퟕ 2. Describe the graph of each equation. Explain how you know what the graph will look like. a. + + = ퟐ ퟐ 풙 ퟐ풙 풚 −ퟏ ( + ) + = ퟐ ퟐ The graph of this equation is a point. I can풙 placeퟏ a center풚 ퟎ at ( , ), but the radius is equal to zero. Therefore, the graph is just a single point. −ퟏ ퟎ DRAFT b. + = ퟐ ퟐ The풙 right풚 side−ퟑ of the equation is negative, so its graph cannot be a circle; it is an empty set.

c. + + + = ퟐ ퟐ 풙 풚 ퟔ풙 ퟔ풚 ퟕ ( + ) + ( + ) = ퟐ ퟐ The graph of this equation is a circle with풙 ퟑ center 풚( ퟑ, ) andퟐퟓ radius . Since the equation can be put in the form ( ) + ( ) = , I know the graph of it is a circle. ퟐ ퟐ ퟐ −ퟑ −ퟑ ퟓ 풙 − 풂 풚 − 풃 풓 Problem Set Sample Solutions

1. Identify the center and radii of the following circles. a. ( + ) + = ퟐ The풙 centerퟐퟓ is풚 ( ퟏ, ), and the radius is . −ퟐퟓ ퟎ ퟏ b. + + = ퟐ ퟐ 풙( + ퟐ풙) +풚( − ퟖ풚) =ퟖ ퟐ ퟐ The풙 centerퟏ is 풚( − ퟒ, ), andퟐퟓ the radius is . −ퟏ ퟒ ퟓ c. + + = ퟐ ퟐ (풙 − ퟐퟎ풙) +풚( − ퟏퟎ풚) = ퟐퟓ ퟎ ퟐ ퟐ The풙 − centerퟏퟎ is (풚 −, ퟓ), andퟏퟎퟎ the radius is . ퟏퟎ ퟓ ퟏퟎ d. + = ퟐ ퟐ The풙 center풚 ퟏퟗis ( , ), and the radius is .

ퟎ ퟎ √ퟏퟗ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

e. + + + = ퟐ ퟐ ퟏ 풙 풙 풚 풚 − ퟒ + ퟐ + + ퟐ = ퟏ ퟏ ퟏ �풙 � �풚 � The centerퟐ is ퟐ, ,ퟒ and the radius is . ퟏ ퟏ ퟏ �− − � ퟐ ퟐ ퟐ 2. Sketch a graph of the following equations. a. + + + = ퟐ ퟐ (풙 + 풚) +ퟏퟎ풙( − ퟒ풚) = ퟑퟑ ퟎ ퟐ ퟐ The풙 rightퟓ side풚 of − the ퟐ equation−ퟒ is negative, so its graph cannot be a circle; it is an empty set, and since there are no solutions, there is nothing to graph.

b. + + + = DRAFTퟐ ퟐ (풙 + 풚) +ퟏퟒ풙( − ퟏퟔ풚) = ퟏퟎퟒ ퟎ ퟐ ퟐ The풙 graphퟕ of 풚the − ퟖequationퟗ is a circle with center ( , ) and radius . −ퟕ ퟖ ퟑ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M5 GEOMETRY

c. + + + = ퟐ ퟐ (풙 + 풚) +ퟒ풙( − ퟏퟎ풚) = ퟐퟗ ퟎ ퟐ ퟐ The풙 graphퟐ of 풚this − ퟓequationퟎ is a point at ( , ). −ퟐ ퟓ

DRAFT

3. Chante claims that two circles given by ( + ) + ( ) = and + + + = are externally tangent. She is right. Show that she is. ퟐ ퟐ ퟐ ퟐ 풙 ퟐ 풚 − ퟒ ퟒퟗ 풙 풚 − ퟔ풙 ퟏퟔ풚 ퟑퟕ ퟎ ( + ) + ( ) = ퟐ ퟐ This graph of this equation is a circle with center풙 ퟐ ( , 풚) −and ퟒ radiusퟒퟗ . ( ) −ퟐ+ (ퟒ + ) = ퟕ ퟐ ퟐ This graph of this equation is a circle with center풙 − ퟑ ( , 풚) andퟖ radiusퟑퟔ . By the distance formula, the distance between theퟑ two−ퟖ centers is ,ퟔ which is precisely the sum of the radii of the two circles. Therefore, these circles will touch only at one point, i.e., they are externally tangent. ퟏퟑ

4. Draw a circle. Randomly select a point in the interior of the circle; label the point . Construct the greatest radius circle with center that lies within the circular region defined by the original circle. Hint: Draw a line through the center, the circle, and point . 푨 푨 Constructions will vary. Ensure푨 that students have drawn two circles such that the circle with center is in the interior and tangent to the original circle. 푨

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 19 M5 T

GEOMETRY

Lesson 19: Equations for Tangent Lines to Circles

Student Outcomes . Given a circle, students find the equations of two lines tangent to the circle with specified slopes. . Given a circle and a point outside the circle, students find the equation of the line tangent to the circle from that point.

Lesson Notes This lesson builds on the understanding of equations of circles in Lessons 17 and 18 and on the understanding of tangent lines developedDRAFT in Lesson 11. Further, the work in this lesson relies on knowledge from Module 4 related to G-GPE.B.4 and G-GPE.B.5. Specifically, students must be able to show that a particular point lies on a circle, compute the slope of a line, and derive the equation of a line that is perpendicular to a radius. The goal is for students to understand how to find equations for both lines with the same slope tangent to a given circle and to find the equation for a line tangent to a given circle from a point outside the circle.

Classwork Opening Exercise (5 minutes) Students are guided to determine of the equation of a line perpendicular to a chord of a given circle.

Opening Exercise

A circle of radius passes through points ( , ) and ( , ). Scaffolding: a. Whatퟓ is the special name for segment푨 −ퟑ ퟑ ? 푩 ퟑ ퟏ . Provide struggling Segment is called a chord. 푨푩 students with a picture of 푨푩 the two circles containing b. How many circles can be drawn that meet the given criteria? Explain how you know. the indicated points. Two circles of radius pass through points and . Two distinct circles, at most, can . Alternatively, struggling have two points in common. students may benefit from ퟓ 푨 푩 simply graphing the two points. c. What is the slope of ? �� The slope of is 푨푩. ퟏ 푨푩�� − ퟑ d. Find the midpoint of .

The midpoint of is푨푩�� (�� , ).

푨푩�� ퟎ ퟐ

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e. Find the equations of the line containing a diameter of the given circle perpendicular to .

= ( ) �푨푩��

풚 − ퟐ ퟑ 풙 − ퟎ f. Is there more than one answer possible for question ?

Although two circles may be drawn that meet the givenퟓ criteria, the diameters of both lie on the line perpendicular to . That line is the perpendicular bisector of . 푨푩�� 푨푩��

Example 1 (10 minutes)

Example 1

Consider the circle with equation ( ) + ( ) = . Find the equations of two tangent lines to the circle that ퟐ ퟐ eachDRAFT have slope . 풙 − ퟑ 풚 − ퟓ ퟐퟎ ퟏ − ퟐ Provide time for students to think about this in pairs or small groups. If necessary, guide their thinking by reminding students of their work in Lesson 11 on tangent lines and their work in Lessons 17–18 on equations of circles. Allow MP.1 individual students or groups of students to share their reasoning as to how to determine the needed equations. Once students have shared their thinking, continue with the reasoning below. . What is the center of the circle? The radius of the circle?  The center is (3,5), and the radius is 20.

. If the tangent lines are to have a slope of √, what must be the slope of the radii to those tangent lines? Why? 1  2 The slope of each radius must be −. 2 A tangent is perpendicular to the radius at the point of tangency. Since the tangent lines must have slopes of , the radii must have slopes of 2, the negative reciprocal 1 of . − 2 1 . ( , ) 2 Label the center− 2 . We need to find a point on the circle with a slope of . We have 푂 = 2, and ( 3) + ( 퐴 푥5)푦 = 20. 푦−5 2 2 Since푥−3 5 =푥 2( − 3), 푦 − then ( 5) = 4( 3) . 푦 − 푥 − Substituting into2 the equation2 for the circle results in ( 3) + 4( 3) = 20. 푦 − 푥 − . At this point, ask students to solve the equation for and call on volunteers to2 share their results.2 푥 − 푥 − Ask if students noticed that using the distributive property made solving the equation easier. 푥 ( 3) (1 + 4) = 20 ( 3)2 = 4 3 = 2 or 3 = 2 = 5,1 푥 − . As expected,2 there are two possible values for , 1 5. Why are two values expected? 푥 − 푥 − 푥 − − 푥  There should be two lines tangent to a circle for any given slope. 푥 표푟

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. What are the coordinates of the points of tangency? How can you determine the y-coordinates?  (1,1) and (5,9); it is easiest to find the -coordinate by plugging each -coordinate into the slope formula above ( = 2). 푦−5 푦 푥 . How can we verify that these푥−3 two points lie on the circle?  Substituting the coordinates into the equation given for the circle proves that they do, in fact, lie on the circle. . How can we find the equations of these tangent lines? What are the equations?  We know both the slope and a point on each of the two tangent lines, so we can use the point-slope formula. The two equations are: 1 = ( 1) and 9 = ( 5). 1 1 Provide students time to discuss the process for finding푦 − the equations− 2 푥 − of the 푦tangent − lines− 2 ;푥 then − , ask students how the solution would have changed if we were looking for two tangent lines whose slopes were 4 instead of . Students 1 , should respondDRAFT that the slope of the radii to the tangent lines would change to which would have impacted2 all other − calculations related to slope and finding the equations. 1 − 4

Exercise 1 (5 minutes, optional) The exercise below can be used to check for understanding of the process used to find the equations of tangent lines to circles. This exercise should be assigned to groups of students who struggled to respond to the last question from the previous discussion. Consider asking the question again—how would the solution have changed if the slopes of the tangent lines were instead of 2—after students finish work on Exercise 1. If this exercise is not used, Exercises 3–4 can be assigned at the end1 of the lesson. 3

Exercise 1

Consider the circle with equation ( ) + ( ) = . Find the equations of two tangent lines to the circle that each have slope . ퟐ ퟐ 풙 − ퟒ 풚 − ퟓ ퟐퟎ = ( ퟐ) and = ( ) 풚 − ퟕ ퟐ 풙 − ퟎ 풚 − ퟑ ퟐ 풙 − ퟖ Example 2 (10 minutes) Scaffolding: Example 2 . Consider labeling point

as ( , ) for struggling Refer to the diagram below. 푄 students.1 √2 Ask them to 2 2 show that the point does Let > . What is the equation of the tangent line to the lie on the circle. Then, ask circle + = through the point ( , ) on the -axis them to find the slope of 풑 ퟐퟏ ퟐ with a point of tangency in the upper half-plane? the tangent line at that 풙 풚 ퟏ 풑 ퟎ 풙 point.

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. Use ( , )as the point of tangency, as shown in the diagram provided. Label the center as (0,0). What do we know about segments and ? 푄 푥 푦 푂  They are perpendicular. 푂푃 푂푄 . Write an equation that takes this into account.  = , giving ( ) + = 0, or + = 0 푦 푥−푝 2 2 2 . Combine푥 the푦−표 two equations푥 푥 −to 푝find an푦 expression푥 for− 푥푝 . 푦 2 2  Since + = 1, we get 1 = 0, or =푥 . 1 . Use the expression푥 푦for to find an expression− 푥푝 for 푥. 푝

 = 1 , which푥 simplifies to = 푦 1 1 1 2 2 . What are푦 the �coordinates− 푝 of the point (the푦 point푝 �푝 of− tangency)?  DRAFT( , 1) 푄 1 1 2 . What is the푝 푝slope�푝 of− in terms of ?  1 푂푄�� 푝 . What is the 2slope of in terms of ? �푝 −  = �푄푃�� 푝 2 1 �푝 −1 2 2 . What is the− � 푝equation−1 1−푝 of line ?

 = ( ) 푄푃 2 �푝 −1 2 푦 1−푝 푥 − 푝 Exercise 2 (3 minutes) The following exercise continues the thinking that began in Example 2. Allow students to work on the exercise in pairs or small groups if necessary.

Exercise 2

Use the same diagram from Example 2 above, but label the point of tangency in the lower half-plane as .

a. What are the coordinates of ? 푸′ ′ ( , ) 푸 ퟏ ퟏ ퟐ 풑 − 풑 �풑 − ퟏ

b. What is the slope of ?

푶푸��′ ퟐ − �풑 − ퟏ

c. What is the slope of ?

�푸′푷�� ퟐ 풑 − ퟏ � ퟐ 풑 − ퟏ

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d. Find the equation of the second tangent line to the circle through ( , ).

풑 ퟎ = ( ) ퟐ �풑 − ퟏ 풚 ퟐ 풙 − 풑 풑 − ퟏ Discussion (4 minutes)

. As the point ( , 0) on the -axis slides to the right, that is, as we choose larger and larger values of , to what coordinate pair does the point of tangency ( ) of the first tangent line (in Example 2) converge? Hint: It might 푝 푥 푝 be helpful to rewrite the coordinates of as푄 ( , 1 ). 1 1 2 푄 푝 � − 푝  (0,1) . WhatDRAFT is the equation of the tangent line in this limit case?  = 1 . Suppose instead that we let the value of be a value very close to = 1. What can you say about the point of 푦 tangency and the tangent line to the circle in this case? 푝 푝  The point of tangency would approach (1,0), and the tangent line would have the equation = 1. . For the case of = 2, what angle does the tangent line make with the -axis? 푥  30°; a right triangle is formed with base 1 and hypotenuse 2. 푝 푥 . What value of gives a tangent line intersecting the -axis at a 45° angle?  There is no solution that gives a 45° angle. If = 1 (which results in the required number of degrees), 푝 푥 then the tangent line is = 1, which is perpendicular to the -axis and, therefore, not at a 45° angle. 푝 . What is the length of ? 푥 푥  1 (Pythagorean�푄푃�� theorem) 2 �푝 − Exercises 3–4 (5 minutes) The following two exercises can be completed in class, if time, or assigned as part of the problem set. Consider posing this follow up question to Exercise 4: How can we change the given equations so that they would represent lines tangent to the circle. Students should respond that the slope of the first equation should be , and the slope of the 4 . second equation should be − 3 3 − 4

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Exercises 3–4

3. Show that a circle with equation ( ) + ( + ) = has two tangent lines with equations + = ( ퟐ ퟐ ퟏ ) and = ( + ). 풙 − ퟐ 풚 ퟑ ퟏퟔퟎ 풚 ퟏퟓ ퟑ 풙 − ퟏ ퟔ 풚 − ퟗ 풙 ퟐ Assume that theퟑ circle has the tangent lines given by the equations above. Then, the tangent lines have slope , and the slope of the radius to those lines must be . If we can show that the points ( , ) and ( , ) are on theퟏ circle, and that the slope of the radius to the tangent lines is , then we will have shown that the given circle ퟑhas the two tangent lines given. −ퟑ ퟔ −ퟏퟓ −ퟐ ퟗ −ퟑ MP.3 ( ) + ( + ) = + ( ) = ퟐ ퟐ ퟐ ퟐ ퟔ − ퟐ −ퟏퟓ ퟑ ퟒ −ퟏퟐ ( ퟏퟔퟎ) + ( + ) = ( ) + = ퟐ ퟐ ퟐ ퟐ −ퟐ − ퟐ ퟗ ퟑ −ퟒ ퟏퟐ Since both points satisfy the equation, thenퟏퟔퟎ the points ( , ) and ( , ) are on the circle. ( ퟔ)−ퟏퟓ −ퟐ ퟗ DRAFT= = = −ퟏퟓ − −ퟑ ퟏퟐ 풎 ( ) − −ퟑ = ퟔ − ퟐ = ퟒ= ퟗ − −ퟑ ퟏퟐ 풎 − −ퟑ The slope of the radius is . −ퟐ − ퟐ ퟒ −ퟑ 4. Could a circle given by the equation ( ) + ( ) = have tangent lines given by the equations = ( ) = ( )? ퟐ ퟐ and 풙 − ퟓ Explain풚 −how ퟏ youퟐퟓ know. ퟒ ퟑ Though풚 − ퟒ theퟑ 풙points − ퟏ ( , 풚) and − ퟓ ( ,ퟒ )풙 are − ퟖon the circle, the given equations cannot represent tangent lines. For the equation = ( ), the slope of the tangent line is . To be tangent, the slope of the radius must be – , but ퟏ ퟒ ퟖ ퟓ ퟒ ퟒ ퟑ ퟑ the slope −ퟒof the ퟑradius풙 − is ퟏ ; therefore, the equation does not represent a tangent line. Similarly for the secondퟒ ퟑ equation, the slope is ; ퟒto be tangent to the circle, the radius must have slope – , but the slope of the radius is . Neither of the given equationsퟑ represents lines that are tangent to the circle. ퟒ ퟒ ퟒ ퟑ ퟑ = ퟏퟔퟎ Closing (3 minutes) Have students summarize the main points of the lesson in writing, by talking to a partner, or as a whole class discussion. Use the questions below, if necessary. . Describe how to find the equations of lines that are tangent to a given circle. . Describe how to find the equation of a tangent line given a circle and a point outside of the circle.

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Lesson Summary

Theorems

A tangent line to a circle is perpendicular to the radius of the circle drawn to the point of tangency.

Relevant Vocabulary

TANGENT TO A CIRCLE. A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point. This point is called the point of tangency.

Exit Ticket (5DRAFT minutes)

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Name Date

Lesson 19: Equations for Tangent Lines to Circles

Exit Ticket

Consider the circle ( + 2) + ( 3) = 9. There are two lines tangent to this circle having a slope of 1. 2 2 1. Find the coordinates푥 of the 푦two − points of tangency. −

DRAFT

2. Find the equations of the two tangent lines.

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Exit Ticket Sample Solutions

Consider the circle ( + ) + ( ) = . There are two lines tangent to this circle having a slope of . ퟐ ퟐ 1. Find the coordinates풙 ퟐ of the풚 two − ퟑ pointsퟗ of tangency. −ퟏ

+ , and , or ~( . , . ) and ( . , . ) � � ퟑ ퟐ−ퟒ ퟑ ퟐ ퟔ −ퟑ√ퟐ−ퟒ −ퟑ√ퟐ+ퟔ � � ퟎ ퟏퟐ ퟓ ퟏퟐ −ퟒ ퟏퟐ ퟎ ퟖퟖ ퟐ ퟐ ퟐ ퟐ � �

2. Find the equations of the two tangent lines. + + = and = or . = ( . ) and . = � � � � ( ퟑ+ ퟐ. ퟔ ) ퟑ ퟐ−ퟒ −ퟑ ퟐ ퟔ −ퟑ ퟐ−ퟒ 풚 − ퟐ − �풙 − ퟐ � 풚 − ퟐ − �풙 − ퟐ � 풚 − ퟓ ퟏퟐ − 풙 − ퟎ ퟏퟐ 풚 − ퟎ ퟖퟖ DRAFT− 풙 ퟒ ퟏퟐ Problem Set Sample Solutions

1. Consider the circle ( ) + ( ) = . There are two lines tangent to this circle having a slope of . a. Find the coordinatesퟐ of the pointsퟐ of tangency. 풙 − ퟏ 풚 − ퟐ ퟏퟔ ퟎ ( , ) and ( , )

ퟏ ퟔ ퟏ −ퟐ b. Find the equations of the two tangent lines. = and =

풚 ퟔ 풚 −ퟐ 2. Consider the circle + + + = . There are two lines tangent to this circle having a slope of . ퟐ ퟐ ퟐ a. Find the coordinates풙 − ퟒ풙 of 풚the twoퟏퟎ풚 pointsퟏퟑ of tangency.ퟎ ퟑ + , and , or ~( . , . ) and ( . , . ) � � −ퟖ ퟏퟑ ퟐퟔ ퟏퟐ ퟏퟑ−ퟔퟓ ퟖ√ퟏퟑ+ퟐퟔ −ퟏퟐ√ퟏퟑ−ퟔퟓ � � −ퟎ ퟐ −ퟏ ퟕ ퟒ ퟐ −ퟖ ퟑ ퟏퟑ ퟏퟑ � ퟏퟑ ퟏퟑ � b. Find the equations of the two tangent lines. + + = + and + = � � � � ퟏퟐ ퟏퟑ−ퟔퟓ ퟐ ퟖ ퟏퟑ ퟐퟔ ퟏퟐ ퟏퟑ−ퟔퟓ ퟐ ퟖ ퟏퟑ ퟐퟔ 풚 − �풙 � 풚 �풙 − � or + .ퟏퟑ= ( ퟑ+ . ) andퟏퟑ + . = ( ퟏퟑ. ) ퟑ ퟏퟑ ퟐ ퟐ 풚 ퟏ ퟕ ퟑ 풙 ퟎ ퟐ 풚 ퟖ ퟑ ퟑ 풙 − ퟒ ퟐ 3. What are the coordinates of the points of tangency of the two tangent lines through the point ( , ) each tangent to the circle + = ? ퟐ ퟐ ퟏ ퟏ ( , ) and풙 ( , 풚) ퟏ

ퟎ ퟏ ퟏ ퟎ 4. What are the coordinates of the points of tangency of the two tangent lines through the point ( , ) each tangent to the circle + = ? ퟐ ퟐ −ퟏ −ퟏ ( , ) and ( , ) 풙 풚 ퟏ

ퟎ −ퟏ −ퟏ ퟎ

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5. What is the equation of the tangent line to the circle + = through the point ( , )? ퟐ ퟐ = ( ) 풙 풚 ퟏ ퟔ ퟎ � ퟑퟓ 풚 − ퟑퟓ 풙 − ퟔ 6. D’Andre said that a circle with equation ( ) + ( ) = has a tangent line represented by the equation = ( + ). ퟐ ퟐ Is he correct? Explain.풙 − ퟐ 풚 − ퟕ ퟏퟑ ퟑ Yes,풚 − ퟓD’Andre− ퟐ is풙 correct.ퟏ The point ( , ) is on the circle, and the slopes are negative reciprocals. −ퟏ ퟓ

7. Kamal gives the following proof that = ( + ) is the equation of a line that is tangent to a circle given by ( + ) + ( ) = . ퟖ 풚 − ퟏ ퟗ 풙 ퟏퟎ The circleퟐ has centerퟐ ( , ) and radius . The point ( , ) is on the circle because 풙 ퟏ 풚 − ퟗ ퟏퟒퟓ ( + ) + ( ) = ( ) + ( ) = . −ퟏ ퟗ ퟏퟐ −ퟏퟎ ퟏ ퟐ ퟐ ퟐ ퟐ MP.3 The slope of the radius is −=ퟏퟎ; thereforeퟏ ퟏ, the−ퟗ equation−ퟗ of the tangent−ퟖ lineퟏퟒퟓ is = ( + ). DRAFT— ퟗ−ퟏ ퟖ ퟖ 풚 − ퟏ 풙 ퟏퟎ −ퟏ ퟏퟎ ퟗ ퟗ a. Kerry, said that Kamal has made an error. What was Kamal’s error? Explain what he did wrong. Kamal used the slope of the radius as the slope of the tangent line. To be tangent, the slopes must be negative reciprocals of one another, not the same.

b. What should the equation for the tangent line be?

= ( + ) ퟗ 풚 − ퟏ − ퟖ 풙 ퟏퟎ 8. Describe a similarity transformation that maps a circle given by + + = to a circle of radius , that is tangent to both axes in the first quadrant. ퟐ ퟐ 풙 ퟔ풙 풚 − ퟐ풚 ퟕퟏ ퟑ The given circle has center ( , ) and radius . A circle that is tangent to both axes in the first quadrant with radius must have a center at ( , ). Then, a translation of + + = along a vector from ( , ) to −ퟑ ퟏ ퟗ ퟐ ퟐ ( , ), ퟑand a dilation by a scale factorퟑ ퟑ of from the center ( 풙, ) willퟔ풙 map풚 circle− ퟐ풚 +ퟕퟏ + = onto−ퟑ theퟏ . ퟏ ퟐ ퟐ circleퟑ ퟑ in the first quadrant with radius ퟑ ퟑ 풙 ퟔ풙 풚 − ퟐ풚 ퟕퟏ ퟑ ퟑ

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New York State Common Core

Mathematics Curriculum

GEOMETRY • MODULE 5

Topic E: Cyclic Quadrilaterals and Ptolemy’s Theorem G-C.A.3 DRAFT Focus Standard: G-C.A.3 Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle. Instructional Days: 2 Lesson 20: Cyclic Quadrilaterals (P) Lesson 21: Ptolemy’s Theorem (E)

Topic E is a two lesson topic pulling together several concepts from the year, e.g., Pythagorean theorem, similarity, and trigonometry, as well as concepts from Module 5 related to arcs and angles. In Lesson 20 students are introduced to the term cCyclic quadrilaterals and define the term informally as a quadrilateral whose vertices lie on a circle. Students then prove that a quadrilateral is cyclic if and only if the opposite angles of the quadrilateral are supplementary. They use this reasoning and the properties of quadrilaterals inscribed in circles (G-C.A.3) to develop the area formula for a cyclic quadrilateral in terms of side length. Lesson 21 continues the study of cyclic quadrilaterals as students prove Ptolemy’s theorem and understand that the area of a cyclic quadrilateral is a function of its side lengths and an acute angle formed by its diagonals (G-SRT.9(+)). Students must identify features within complex diagrams to inform their thinking, highlighting MP.7. For example, students use the structure of an inscribed triangle in a half-plane separated by the diagonal of a cyclic quadrilateral to conclude that a reflection of the triangle along the diagonal produces a different cyclic quadrilateral with an area equal to the original cyclic quadrilateral. Students use this reasoning to make sense of Ptolemy’s theorem and its origin.

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Lesson 20: Cyclic Quadrilaterals

Student Outcomes . Students show that a quadrilateral is cyclic if and only if its opposite angles are supplementary. . Students derive and apply the area of cyclic quadrilateral as sin( ) where is the measure 1 of the acute angle formed by diagonals and . 퐴퐵퐶퐷 2 퐴퐶 ⋅ 퐵퐷 ⋅ 푤 푤 퐴퐶 퐵퐷 Lesson Notes In Lessons 20 andDRAFT 21, students experience a culmination of the skills they learned in the previous lessons and modules to reveal and understand powerful relationships that exist among the angles, chord lengths, and areas of cyclic quadrilaterals. Students will apply reasoning with angle relationships, similarity, trigonometric ratios and related formulas, and relationships of segments intersecting circles. They begin exploring the nature of cyclic quadrilaterals and use the lengths of the diagonals of cyclic quadrilaterals to determine their area. Next, students construct the circumscribed circle on three vertices of a quadrilateral (a triangle) and use angle relationships to prove that the fourth vertex must also lie on the circle (G-C.A.3). They then use these relationships and their knowledge of similar triangles and trigonometry to prove Ptolemy’s theorem, which states that the product of the lengths of the diagonals of a cyclic quadrilateral is equal to the sum of the products of the lengths of the opposite sides of the cyclic quadrilateral.

Classwork Opening (5 minutes) Students first encountered a cyclic quadrilateral in Lesson 5 Exercise 1, part (a), though it was referred to simply as an inscribed polygon. Begin the lesson by discussing the meaning of a cyclic quadrilateral. . Quadrilateral shown in the opening exercise is an example of a cyclic quadrilateral. What do you believe the term cyclic means in this case? 퐴퐵퐶퐷  The vertices , , , and lie on a circle. . Discuss the following question with a shoulder partner and then share out: What is the relationship of and 퐴 퐵 퐶 퐷 in the diagram? 푥 푦  and must be supplementary since they are inscribed in two adjacent arcs that form a complete circle. 푥 푦 . Make a clear statement to students that a cyclic quadrilateral is a quadrilateral that is inscribed in a circle.

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Opening Exercise (5 minutes)

Opening Exercise MP.7 Given cyclic quadrilateral shown in the diagram, prove that + = °.

푨푩푪푫 풙 풚 ퟏퟖퟎ Statements Reasons/Explanations 1. + = ° 1. and are non- overlapping arcs that form a complete � � � � �풎푩푪푫� �풎푫푨푩� ퟑퟔퟎ circle.푨풓풄 푩푪푫 풂풓풄 푫푨푩

2. + = ( °) 2. Multiplicative property of equality ퟏ ퟏ ퟐ � � ퟐ 3. ��풎푩푪푫�+ �(풎푫푨푩�)� = ퟑퟔퟎ° 3. Distributive property ퟏ ퟏ ퟐ � ퟐ � 4. �=풎푩푪푫( � ) and풎푫푨푩 = ( ퟏퟖퟎ ) 4. Inscribed angle theorem DRAFTퟏ ퟏ 5. 풙 + ퟐ =풎푩푪푫�° 풚 ퟐ 풎푫푨푩� 5. Substitution

풙 풚 ퟏퟖퟎ

Example 1 (5 minutes) Scaffolding: . Remind students that a Pose the question below to students before starting Example 1, and ask them to triangle can be inscribed in hypothesize their answers. a circle, or a circle . The Opening Exercise shows that if a quadrilateral is cyclic, then its opposite circumscribed about a angles are supplementary. Let’s explore the converse relationship. If a triangle. This allows us to quadrilateral has supplementary opposite angles, is the quadrilateral necessarily draw a circle on three of a cyclic quadrilateral? the four vertices of the quadrilateral. It is our job  Yes. to show that the fourth . How can we show that your hypothesis is valid? vertex also lies on the  Student answers will vary. circle. . Do concrete work by having students create Example 1 cyclic quadrilaterals and Given quadrilateral with measuring angles to see + = °, prove that patterns. quadrilateral 푨푩푪푫 is cyclic; i.e., prove 풎∠푨that points풎∠푪 , , ퟏퟖퟎ, and lie on the . Explain proof by same circle. 푨푩푪푫 contradiction by 푨 푩 푪 푫 presenting a simple proof such as 2 points define a line, and have students try to prove this is not true.

. First, we are given that angles and are supplementary. What does this mean about angles and , and MP.7 why? 퐴 퐶 퐵 퐷  The angle sum of a quadrilateral is 360°, and since it is given that angles and are supplementary, Angles and must then have a sum of 180°. 퐴 퐶 퐵 퐷

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. We, of course, can draw a circle through point , and we can further draw a circle through points and (infinitely many circles, actually). Can we draw a circle through points , , and ? 퐴 퐴 퐵  Three non-collinear points can determine a circle, and since the points were given to be vertices of a 퐴 퐵 퐶 quadrilateral, the points are non-collinear; so, yes! . Can we draw a circle through points , , , and ?  Not all quadrilaterals are cyclic (for example a non-rectangular parallelogram), so we cannot assume 퐴 퐵 퐶 퐷 that a circle can be drawn through vertices , , , and . . Where could point lie in relation to the circle? 퐴 퐵 퐶 퐷  could lie on the circle, in the interior of the circle, or on the 퐷 exterior of the circle. 퐷 . To show that lies on the circle with , , and , we need to consider the cases where it is not, and show that those cases are impossible. First, let’s considerDRAFT 퐷the case where is outside퐴 퐵 the circle.퐶 On the diagram, use a red pencil to locate and label point such that it is outside the circle; then, draw segments and퐷 . 퐷′ . What do you notice about sides and if vertex is outside the 퐶퐷′ 퐴퐷′ circle? 퐴퐷′ 퐶퐷′ 퐷′  The sides intersect the circle and are, therefore, secants.

Statements Reasons/Explanations 1. + = ° 1. Given

2. 풎Assume∠푨 풎 point∠푪 ퟏퟖퟎ lies outside the circle determined by 2. Stated assumption for case 1. points , , and . 푫′ 3. Segments푨 푩 and푪 intersect the circle at distinct 3. If the segments intersect the circle at the same points and , thus > °. point, then lies on the circle, and the stated 푪푫′ 푨푫′ assumption (Statement 2) is false. 푷 푸 풎푷푸� ퟎ 푫′ 4. = 4. Secant angle theorem: exterior case ′ ퟏ 풎∠푫 �풎푨푩푪� − 풎푷푸� � 5. = ퟐ( ) 5. Inscribed angle theorem ퟏ � 6. 풎∠푩 + ퟐ 풎푨푷푪+ + = ° 6. The angle sum of a quadrilateral is °. 7. 풎∠푨 + 풎∠푩 =풎∠푪° 풎∠푫′ ퟑퟔퟎ 7. Substitution (Statements 1 and 6) ퟑퟔퟎ

풎∠푩 풎∠푫′ ퟏퟖퟎ 8. Substitution (Statements 4, 5, and 7) 8. ( ) + ( ) = ° ퟏ ퟏ 풎푨푷푪 풎푨푩푪 − 풎푷푸 ퟏퟖퟎ 9. ퟐ ° � ퟐ= � � 9. Arcs and are non-overlapping arcs that form a complete circle with a sum of °. ퟑퟔퟎ − 풎푨푷푪� 풎푨푩푪� 푨푷푪� 푨푩푪� 10. Substitution (Statements 8 and 9) ퟑퟔퟎ 10. ( ) + ( ° ) = ° ퟏ ퟏ 풎푨푷푪 ퟑퟔퟎ − 풎푨푷푪 − 풎푷푸 ퟏퟖퟎ 11. Distributive property 11. ퟐ ( � ) + ퟐ � ° ( �) �=� ° ퟏ ퟏ 풎푨푷푪 ퟏퟖퟎ − 풎푨푷푪 − 풎푷푸 ퟏퟖퟎ 12. ퟐ =� ° ퟐ � � 12. Subtraction property of equality 13. 풎푷푸 �cannotퟎ lie outside the circle. 13. Statement 12 contradicts our stated assumption that and are distinct with > ° 푫′ (Statement 3). 푷 푸 풎푷푸� ퟎ

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Exercise 1 (5 minutes) Students use a similar strategy to show that vertex cannot lie inside the circle.

퐷 Exercises

1. Assume that vertex lies inside the circle as shown in the diagram. Use a similar argument to Example 1 to show that vertex cannot lie inside the circle. 푫′′ 푫′′

DRAFTStatements Reasons/Explanations 1. + = ° 1. Given

2. 풎Assume∠푨 풎 point∠푪 ퟏퟖퟎ lies inside the circle determined by points 2. Stated assumption for case 2. , , and . 푫′ 3. 푨 푩 and 푪 intersect the circle at points and 3. If the segments intersect the respectively, thus > . circle at the same point, then 푪푫⃖��′′⃗ 푨푫⃖��′′⃗ 푷 푸 lies on the circle, and the 풎푷푸� ퟎ stated assumption (Statement 푫′2) is false.

4. = + 4. Secant angle theorem: ′′ interior case ퟏ � � 풎∠푫 ퟐ �풎푷푸 풎푨푩푪� 5. = 5. Inscribed angle theorem ퟏ � 6. 풎∠푩 + ퟐ �풎푨푷푪+ � + = ° 6. The angle sum of a ′′ quadrilateral is °. 풎∠푨 풎∠푩 풎∠푪 풎∠푫 ퟑퟔퟎ 7. + = ° 7. Substitution (Statementsퟑퟔퟎ 1 ′′ and 6) 풎∠푩 풎∠푫 ퟏퟖퟎ 8. Substitution (Statements 4, 5, 8. ( ) + ( + ) = ° and 7) ퟏ ퟏ 풎푨푷푪 풎푷푸 풎푨푩푪 ퟏퟖퟎ 9. ퟐ �= °ퟐ � � 9. Arcs and are non- overlapping arcs that form a � � � � 풎푨푩푪 ퟑퟔퟎ − 풎푨푷푪 complete푨푷푪 circle푨푩푪 with a sum of °.

10. ퟑퟔퟎSubstitution (Statements 8 10. ( ) + + ( ° ) = ° and 9) ퟏ ퟏ 풎푨푷푪 풎푷푸 ퟑퟔퟎ − 풎푨푷푪 ퟏퟖퟎ ퟐ ퟐ 11. Distributive property 11. ( � ) + (� �) + ° (� �) = ° ퟏ ퟏ ퟏ 풎푨푷푪 풎푷푸 ퟏퟖퟎ − 풎푨푷푪 ퟏퟖퟎ 12. ퟐ =� ° ퟐ � ퟐ � 12. Subtraction property of equality 풎푷푸� ퟎ 13. cannot lie inside the circle. 13. Statement 12 contradicts our stated assumption that and 푫′′ are distinct with > ° (Statement 3). 푷 푸 풎푷푸� ퟎ

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. In Example 1 and Exercise 1, we showed that the fourth vertex cannot lie outside the circle or inside the circle. What conclusion does this leave us with? 퐷  The fourth vertex must then lie on the circle with points , , and . . In the Opening Exercise, you showed that the opposite angles in a cyclic quadrilateral are supplementary. In 퐴 퐵 퐶 Example 1 and Exercise 1, we showed that if a quadrilateral has supplementary opposite angles, then the vertices must lie on a circle. This confirms the following theorem:

Theorem. A quadrilateral is cyclic if and only if its opposite angles are supplementary.

. Take a moment to discuss with a shoulder partner what this theorem means and how we can use it.  Answers will vary. DRAFT Exercises 2–3 (5 minutes) Students now apply the theorem about cyclic quadrilaterals.

2. Quadrilateral is a cyclic quadrilateral. Explain why ~ .

푷푸푹푺 If is a cyclic quadrilateral,△ 푷푸푻 draw△ 푺푹푻 circle on points , , , and . Then and are angles inscribed푷푸푹푺 in the same arc and are therefore 푷equal푸 푹 in measure.푺 Also,∠푷푸푺 and∠푷푹푺 are � both inscribed in the same푺푷 arc and are therefore the same measure.∠푸푷푹 Therefore∠푸푺푹 by � criterion for similar triangles,푸푹 ~ . 푨푨 △ 푷푸푻 △ 푺푹푻(Students may also use vertical angles relationship at .)

푻 3. A cyclic quadrilateral has perpendicular diagonals. What is the area of the quadrilateral in terms of , , , and as shown? 풂 풃 풄 풅 Using the area formula for a triangle = , the area of the quadrilateral is the sum of the areas of the four right triangular regions. 푨풓풆풂 풃풂풔풆 ⋅ 풉풆풊품풉풕 = + + + ퟏ ퟏ ퟏ ퟏ 푨풓풆풂 OR 풂풄 풄풃 풃풅 풅풂 ퟐ ퟐ ퟐ ퟐ = ( + + + ) ퟏ 푨풓풆풂 ퟐ 풂풄 풄풃 풃풅 풅풂

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Discussion (5 minutes) Redraw the cyclic quadrilateral from Exercise 3 as shown in diagram to the right. . How does this diagram relate to the area(s) that you found in Exercise 3 in terms of , , , and ?  Each right triangular region in the cyclic quadrilateral is half of a 푎 푏 푐 푑 rectangular region. The area of the quadrilateral is the sum of the areas of the triangles, and also half the area of the sum of the four smaller rectangular regions. . What are the lengths of the sides of the large rectangle?  The lengths of the sides of the large rectangle are + and + . . Using the lengths of the large rectangle, what is its area? 푎 푏 푐 푑  = ( + )( + ) . DRAFT How is the area of the given cyclic quadrilateral related to the area of the large rectangle? 퐴푟푒푎 푎 푏 푐 푑  The area of the cyclic quadrilateral is [( + )( + )]. 1 . What does this say about the area of a cyclic quadrilateral2 푎 푏 푐 with푑 perpendicular diagonals?  The area of a cyclic quadrilateral with perpendicular diagonals is equal to one-half the product of the lengths of its diagonals. . Can we extend this to other cyclic quadrilaterals? (i.e., cyclic quadrilaterals whose diagonals intersect to form an acute angle °.) Discuss this question with a shoulder partner before beginning Example 2.

푤 Exercises 4–5 (5 minutes) (Optional) These exercises may be necessary for review of the area of a non-right triangle using one acute angle. You may assign these problems as additional homework problems in the previous lesson because the skills have been taught previously. If students demonstrate confidence with the content, go directly to Exercise 6.

4. Show that the triangle in the diagram has area ( ). ퟏ 풂풃 퐬퐢퐧 풘 Draw altitude toퟐ side with length as shown in the diagram to form adjacent right triangles. Using right triangle trigonometry, the sine of the acute angle with degree measure is: 풃

= 풘 where is the length of the altitude to side . 풉 It풔풊풏 then풘 follows풂 that풉 = . 풃

Using the area formula풉 for풂 풔풊풏 a triangle풘 = × × : ퟏ = ( ( )) 푨풓풆풂 ퟐ 풃풂풔풆 풉풆풊품풉풕 ퟏ 푨풓풆풂 풃 풂 풔풊풏 풘 = ퟐ ( ) ퟏ 푨풓풆풂 ퟐ 풂풃 풔풊풏 풘

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5. Show that the triangle with obtuse angle ( )° has area ( ). ퟏ ퟏퟖퟎ − 풘 풂풃 퐬퐢퐧 풘 Draw altitudeퟐ to the line that includes the side of the triangle with length . Using right triangle trigonometry:

( ) = where 풃 is the length of the altitude drawn. 풉 It풔풊풏 follows풘 then풂 that 풉= ( ). Using the area formula풉 for풂 풔풊풏 a triangle풘

= × × : ퟏ 푨풓풆풂 = ퟐ (풃풂풔풆( 풉풆풊품풉풕)) ퟏ 푨풓풆풂 풃 풂 풔풊풏 풘 = ퟐ ( ) ퟏ DRAFT 푨풓풆풂 ퟐ 풂풃 풔풊풏 풘

Exercise 6 (5 minutes)

Students in pairs apply the area formula = sin( ), first encountered in Module 2, Lesson 31, to show that the 1 area of a cyclic quadrilateral is one-half the product of the lengths of its diagonals and the sine of the acute angle formed 퐴푟푒푎 2 푎푏 푤 by their intersection.

Exercise 6

6. Show that the area of the cyclic quadrilateral shown in the diagram is = ( + )( + ) ( ). ퟏ 푨풓풆풂 풂 풃 풄 풅 퐬퐢퐧 풘 Using the area formula for a triangle = ( ): ퟐ ퟏ = ( ) 푨풓풆풂 ퟐ 풂풃 ⋅ 풔풊풏 풘 ퟏ 푨풓풆풂ퟏ 풂풅 ⋅ 풔풊풏 풘 = ퟐ ( )

ퟐ ퟏ 푨풓풆풂 = ퟐ 풂풄 ⋅ 풔풊풏(풘)

ퟑ ퟏ 푨풓풆풂 = ퟐ 풃풄 ⋅ 풔풊풏(풘 )

ퟒ ퟏ 푨풓풆풂 ퟐ 풃풅 ⋅ 풔풊풏 풘 = + + +

풕풐풕풂풍 ퟏ ퟐ ퟑ ퟒ 푨풓풆풂 = 푨풓풆풂 푨풓풆( )풂+ 푨풓풆풂 ( 푨풓풆) +풂 ( )

풕풐풕풂풍 ퟏ ퟏ ퟏ +푨풓풆풂 ( 풂풅) ⋅ 풔풊풏 풘 풂풄 ⋅ 풔풊풏 풘 풃풄 ⋅ 풔풊풏 풘 ퟐ ퟐ ퟐ ퟏ 풃풅 ⋅ 풔풊풏 풘 ퟐ = ( ) ( + + + ) ퟏ 푨풓풆풂풕풐풕풂풍 � 풔풊풏 풘 � 풂풅 풂풄 풃풄 풃풅 = ퟐ ( ) [( + )( + )] ퟏ 푨풓풆풂풕풐풕풂풍 � 풔풊풏 풘 � 풂 풃 풄 풅 = ퟐ( + )( + ) ( ). ퟏ 푨풓풆풂풕풐풕풂풍 풂 풃 풄 풅 풔풊풏 풘 ퟐ

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Closing (3 minutes) Ask students to verbally provide answers to the following closing questions based on the lesson. . What angle relationship exists in any cyclic quadrilateral?  Both pairs of opposite angles are supplementary. . If a quadrilateral has one pair of opposite angles supplementary, does it mean that the quadrilateral is cyclic? Why?  Yes. We proved that if the opposite angles of a quadrilateral are supplementary, then the fourth vertex of the quadrilateral must lie on the circle through the other three vertices. . Describe how to find the area of a cyclic quadrilateral using its diagonals.  The area of a cyclic quadrilateral is one-half the product of the lengths of the diagonals and the sine of the acute angle formed at their intersection. DRAFT Lesson Summary

Theorems Given a convex quadrilateral, the quadrilateral is cyclic if and only if one pair of opposite angles are supplementary. The area of a triangle with side lengths and and acute included angle with degree measure : = ( ). 풂 풃 풘 ퟏ 푨풓풆풂 ퟐ 풂풃 ⋅ 퐬퐢퐧 풘 The area of a cyclic quadrilateral whose diagonals and intersect to form an acute or right angle with ( ) = ( ). degree measure : 푨푩푪푫 푨푪�� 푩푫�� ퟏ 풘 푨풓풆풂 푨푩푪푫 ퟐ ⋅ 푨푪 ⋅ 푩푫 ⋅ 퐬퐢퐧 풘

Relevant Vocabulary

CYCLIC QUADRILATERAL: A quadrilateral inscribed in a circle is called a cyclic quadrilateral.

Exit Ticket (5 minutes)

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Name Date

Lesson 20: Cyclic Quadrilaterals

Exit Ticket

1. What value of guarantees that the quadrilateral shown in the diagram below is cyclic? Explain.

푥

DRAFT

2. Given quadrilateral , + = 180°, = 60°, = 4, = 48, = 8, and = 24, find the area of quadrilateral . Justify your answer. 퐺퐾퐻퐽 푚∠퐾퐺퐽 푚∠퐾퐻퐽 푚∠퐻푁퐽 퐾푁 푁퐽 퐺푁 푁퐻

퐺퐾퐻퐽

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Exit Ticket Sample Solutions

1. What value of guarantees that the quadrilateral shown in the diagram below is cyclic? Explain.

풙 + =

ퟒ풙 ퟓ풙=− ퟗ ퟏퟖퟎ ퟗ풙 −= ퟗ ퟏퟖퟎ ퟗ풙= ퟏퟖퟗ If풙 =ퟐퟏ , then the opposite angles shown are supplementary, and any quadrilateral with supplementary opposite angles is cyclic. 풙 ퟐퟏ

2. Given quadrilateral , + = °, = °, = , = , = , and = , find the area of quadrilateral . Justify your answer. 푮푲푯푱 풎∠푲푮푱 풎∠푲푯푱 ퟏퟖퟎ 풎∠푯푵푱 ퟔퟎ 푲푵 ퟒ 푵푱 ퟒퟖ 푮푵 ퟖ 푵푯 ퟐퟒ DRAFTOpposite angles and are supplementary, so 푮푲푯푱 quadrilateral is cyclic. 푲푮푱 푲푯푱 The area of a 푮푲푯푱cyclic quadrilateral:

= ( + )( + ) ( ) ퟏ 푨풓풆풂 ퟐ ퟒ ퟒퟖ ퟖ ퟐퟒ ⋅ 풔풊풏 ퟔퟎ = ( )( ) ퟏ √ퟑ 푨풓풆풂 ퟐ ퟓퟐ ퟑퟐ ⋅ ퟐ = √ퟑ 푨풓풆풂 = ퟒ ⋅ ퟏퟔퟔퟒ.

The area of quadrilateral is square units. 푨풓풆풂 ퟒퟏퟔ√ퟑ 푮푲푯푱 ퟒퟏퟔ√ퟑ

Problem Set Sample Solutions The problems in this Problem Set get progressively more difficult and require use of recent and prior skills. The length of the Problem Set may be too time consuming for students to do the entire set. Problems 10–12 are the most difficult and may be passed over, especially for struggling students.

1. Quadrilateral is cyclic, is the center of the circle, and = °. Find .

푩푫푪푬 푶 By the inscribed angle theorem,풎∠퐁퐎퐂 ퟏퟑퟎ= 풎∠퐁퐄퐂, so = °. ퟏ Opposite angles of cyclic quadrilaterals풎∠푩푫푪 are supplementary,ퟐ 풎∠푩푶푪 풎∠ so 푩푫푪 ퟔퟓ + ° = °, and thus = °.

풎∠ 푩푬푪 ퟔퟓ ퟏퟖퟎ 풎∠푩푬푪 ퟏퟏퟓ

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2. Quadrilateral is cyclic, = , = , = , and = °. Find the area of quadrilateral . 푭푨푬푫 푨푿 ퟖ 푭푿 ퟔ 푿푫 ퟑ 풎∠푨푿푬 ퟏퟑퟎ 푭푨푬푫 Using the two-chord power rule, ( )( ) = ( )( ).

( ) = ( ) , thus = . 푨푿 푿푫 푭푿 푿푬

ퟖTheퟑ areaퟔ of푿푬 a cyclic quadrilateral푿푬 ퟒ is equal to the product of the lengths of the diagonals and the sine of the acute angle formed by them. The acute angle formed by the diagonals is °.

= ( + )( + ) ퟓퟎ( )

푨풓풆풂 = (ퟖ )(ퟑ ퟔ) ퟒ (⋅ 풔풊풏) ퟓퟎ 푨풓풆풂 ퟏퟏ. ퟏퟎ ⋅ 풔풊풏 ퟓퟎ 푨풓풆풂 ≈ ퟖퟒ ퟑ 3. In the diagram below, , and = °. Find the value of and . DRAFT�푩푬�� ∥ 푪푫�� 풎∠Quadrilateral푩푬푫 ퟕퟐ is cyclic, so opposite풔 풕 angles are supplementary. + ° = 푩푪푫푬°

풔 = ퟕퟐ ° ퟏퟖퟎ Parallel풔 ퟏퟎퟖ chords and intercept congruent arcs and . By angle addition, = , so it follows by the inscribed angle theorem that � � = = °. 푩푬 푪푫 푪푩 푬푫 풎푪푩푬� 풎푩푬푫� 풔 풕 ퟏퟎퟖ

4. In the diagram below, is a diameter, = °, and . Find .

�푩푪�� Triangle풎∠푩푪푫 isퟐퟓ inscribed�푪푬 �in�� a≅ semi�푫푬��-circle,풎∠ so by푪푬푫 Thales’ theorem, is a right angle. By the angle sum of a triangle, = °. Quadrilateral is 푩푪푫cyclic, so opposite angles are supplementary. ∠푩푫푪 풎∠푫푩푪 ퟔퟓ 푩푪푬푫° + = °

ퟔퟓ 풎∠=푪푬푫 ° ퟏퟖퟎ Triangle풎∠푪푬푫 ퟏퟏퟓ is isosceles since , and by base ’s, . ( 푪푬푫) = ° °, by�푪푬� �the� ≅ angle�푫푬�� sum of a triangle.∠ ∠ 푬푫푪 ≅ ∠푬푪푫

ퟐ 풎∠푬푫푪= .ퟏퟖퟎ° − ퟏퟏퟓ

풎∠푬푫푪 ퟑퟐ ퟓ 5. In circle A, = °. Find .

풎∠푨푩푫 ퟏퟓ 풎∠Draw푩푪푫 diameter such that is on the circle, then draw . Triangle is inscribed in a semi-circle, therefore angle is a right angle. By the angle �� sum of a triangle,푩푨푿 = 푿 °. 푫푿 푩푫푿 푩푫푿 Quadrilateral 풎∠ is푩푿푫 a cyclicퟕퟓ quadrilateral, so its opposite angles are supplementary. 푩푪푫푿 + = °

풎∠푩푪푫 + 풎∠°푩푿푫 = ° ퟏퟖퟎ 풎∠푩푪푫 ퟕퟓ ퟏퟖퟎ= °

풎∠푩푪푫 ퟏퟎퟓ

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6. Given the diagram below, is the center of the circle. If = °, find .

푶 Draw point풎∠ 푵푶푷 on majorퟏퟏퟐ arc , and풎∠ draw푷푸푬 chords and to form cyclic quadrilateral . By the inscribed angle theorem, 푿 푵푷� �푿푵�� 푿푷�� = = ° 푸푵푿푷, so . ퟏ 풎∠An exterior푵푿푷 angleퟐ 풎∠ 푵푶푷 of cyclic풎∠푵푿푷 quadrilateralퟓퟔ is equal in measure to the angle opposite from vertex , which is . Therefore, 푷푸푬= °. 푸푵푿푷 푸 ∠푵푿푷 풎∠푷푸푬 ퟓퟔ

7. Given the angle measures as indicated in the diagram below, prove that vertices , , , and lie on a circle.

Using the angle sum of a triangle, = 푪°.푩 Angles푬 푫 and are vertical angles and, therefore, have the same measure. Angles and are DRAFTalso vertical angles and have the same풎∠푭푩푬 measure.ퟒퟑ Angles at푮푩푪 a point sum푭푩푬 to °, so = = °. Since + = , angles 푮푩푭 and 푬푩푪 of quadrilateral are supplementary. If a quadrilateral has opposite anglesퟑퟔퟎ that 풎∠are 푪푩푬supplementary,풎∠푭푩푮 thenퟏퟑퟕ the quadrilateralퟏퟑퟕ isퟒퟑ cyclicퟏퟖퟎ, which means푬푩푪 that vertices푬푫푪 , , , and 푩푪푫푬 lie on a circle.

푪 푩 푬 푫

8. In the diagram below, quadrilateral is cyclic. Find the value of .

푱푲푳푴 Angles and form풏 a linear pair and are supplementary, so angle has measure of °. 푯푲푱 푳푲푱 푳푲푱Opposite angles of a ퟗퟖcyclic quadrilateral are supplementary and thus = = °.

풎∠Angles푱푴푳 풎∠ and푱푲푯 formퟖퟐ a linear pair and are supplementary, so angle has measure of °. Therefore, = . 푱푴푳 푰푴푳 푰푴푳 ퟗퟖ 풏 ퟗퟖ

9. Do all four perpendicular bisectors of the sides of a cyclic quadrilateral pass through a common point? Explain. Yes. A cyclic quadrilateral has vertices that lie on a circle, which means that the vertices are equidistant from the center of the circle. The perpendicular bisector of a segment is the set of points equidistant from the segment’s endpoints, and since the center of the circle is equidistant from all of the vertices (endpoints of the segments that make up the sides of the cyclic quadrilateral), the center lies on all four perpendicular bisectors of the quadrilateral.

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10. The circles in the diagram below intersect at points and . If = ° and = °, find and . 푨 푩 풎∠푭푯푮 ퟏퟎퟎ 풎∠푯푮푬 ퟕퟎ 풎∠푮푬푭 풎∠푬푭푯 Quadrilaterals and are both cyclic since their vertices lie on circles. Opposite angles in cyclic quadrilaterals are supplementary. 푮푯푩푨 푬푭푩푨 + = °

풎∠푮푨푩 + 풎∠푭푯푮° = ퟏퟖퟎ° 풎∠푮푨푩 = ퟏퟎퟎ° ퟏퟖퟎ 풎∠푮푨푩 and ퟖퟎ are supplementary since they form a linear pair, so ∠푬푨푩 =∠푮푨푩°.

풎∠푬푨푩 and ퟏퟎퟎ are supplementary also since they are opposite angles in a cyclic quadrilateral, so = °. ∠푬푨푩 ∠푬푭푩 Using a similar argument,풎∠ 푬푭푩 ퟖퟎ DRAFT+ = + = + = ° 풎∠푯푮푬 + 풎∠푯푩푨 = 풎∠푯푩푨 + 풎∠푭푩푨 = 풎∠푭푩푨 + 풎∠푮푬푭 = ퟏퟖퟎ°. 풎∠푯푮푬 풎∠+푯푩푨 풎∠=푯푩푨 ° 풎∠푭푩푨 풎∠푭푩푨 풎∠푮푬푭 ퟏퟖퟎ 풎∠푯푮푬° + 풎∠푮푬푭= ퟏퟖퟎ° ퟕퟎ 풎∠푮푬푭= ퟏퟖퟎ° 풎∠푮푬푭 ퟏퟏퟎ 11. A quadrilateral is called bicentric if it is both cyclic and possesses an inscribed circle. (See diagram on the right.) a. What can be concluded about the opposite angles of a bicentric quadrilateral? Explain.

Since a bicentric quadrilateral must be also cyclic, its opposite angles must be supplementary.

b. Each side of the quadrilateral is tangent to the inscribed circle. What does this tell us about the segments contained in the sides of the quadrilateral?

Two tangents to a circle from an exterior point form congruent segments. The distances from a vertex of the quadrilateral to the tangent points where it meets the inscribed circle are equal.

c. Based on the relationships highlighted in part (b), there are four pairs of congruent segments in the diagram. Label segments of equal length with , , , and .

See diagram풂 풃 풄 on the풅 right.

d. What do you notice about the opposite sides of the bicentric quadrilateral?

The sum of the lengths of one pair of opposite sides of the bicentric quadrilateral is equal to the sum of the lengths of the other pair of opposite sides:

( + ) + ( + ) = ( + ) + ( + ).

풂 풃 풄 풅 풅 풂 풃 풄

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12. Quadrilateral is cyclic such that is the diameter of the circle. If , prove that is a right angle, and show that , , , and lie on a circle. 푷푺푹푸 �푷푸�� ∠푸푹푻 ≅ ∠푸푺푹 ∠푷푻푹 푺 푿 푻 Angles푷 and are congruent since they are the same angle. Since it was given that , it follows that ~ by similarity criterion.푹푸푿 Corresponding푺푸푹 angles in similar figures are congruent, so , and∠ by푸푹푻 ≅. ∠푸푺푹, . △ 푸푿푹 △ 푸푹푺 푨푨 ∠푸푹푺 ≅ ∠Quadrilateral푸푿푹 풗풆풓풕 is∠′ cyclic,풔 ∠푸푿푹 so its≅ opposite ∠푻푿푺 angles and are supplementary. Since , it follows that angles and are also supplementary.푷푺푹푸 If a quadrilateral has a pair of 푸푷푺opposite angles푸푹푺 that are supplementary, then the∠푸푿푹 quadrilateral≅ ∠푻푿푺 is cyclic. Thus, quadrilateral푻푿푺 푸푷푺 is cyclic. 푺푿푻푷 Angle is a right angle since it is inscribed in a semi-circle. If a quadrilateral is cyclic, then its opposite angles are supplementary; thus, angle must be supplementary푷푺푸 to angle , and, therefore, is also a right angle. 푷푻푹 DRAFT푷푺푸

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Lesson 21: Ptolemy’s Theorem

Student Outcomes . Students determine the area of a cyclic quadrilateral as a function of its side lengths and the acute angle formed by its diagonals. . Students prove Ptolemy’s theorem, which states that for a cyclic quadrilateral , = + . They explore applications of the result. 퐴퐵퐶퐷 퐴퐶 ⋅ 퐵퐷 퐴퐵 ⋅ 퐶퐷 퐵퐶 ⋅ 퐴퐷 Lesson Notes In this lesson, studentsDRAFT work to understand Ptolemy’s theorem which says that for a cyclic quadrilateral , = + . As such, this lesson focuses on the properties of quadrilaterals inscribed in circles. Ptolemy’s single result, and the proof for it, codify many geometric facts, the Pythagorean theorem (G-GPE.A.1, G-GPE.B.4퐴퐶), area⋅ 퐵퐷 퐴퐵 ⋅ formulas,퐶퐷 퐵퐶 ⋅and퐴퐷 trigonometry results, for instance, and serves as a capstone experience to our year-long study of geometry. Students will use the area formulas they established in the previous lesson to prove the theorem. A set square, patty paper, compass, and straight edge are needed to complete the Exploratory Challenge.

Classwork Opening (2 minutes) The Pythagorean theorem, credited to the Greek mathematician Pythagoras of Samos (ca. 570 – ca. 495 BCE), describes a universal relationship among the sides of a right triangle. Every right triangle (in fact every triangle) can be circumscribed by a circle. Six centuries later, Greek mathematician Claudius Ptolemy (ca. 90 – ca. 168 CE) discovered a relationship between the side-lengths and the diagonals of any quadrilateral inscribed in a circle. As we shall see, Ptolemy’s result can be seen as an extension of the Pythagorean theorem.

Opening Exercise (5 minutes) Students are given the statement of Ptolemy’s theorem and are asked to test the theorem by measuring lengths on specific cyclic quadrilaterals they are asked to draw. Students conduct this work in pairs and then gather to discuss their ideas afterwards in class as a whole.

Opening Exercise

Ptolemy’s theorem says that for a cyclic quadrilateral , = + .

With ruler and a compass, draw an example of a cyclic quadrilateral.푨푪 ⋅ 푩푫 푨푩 Label⋅ 푪푫 its푩푪 vertices⋅ 푨푫 A, B, C, and D. MP.7 Draw the two diagonals and .

�푨푪�� 푩푫��

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With a ruler, test whether or not the claim that = + seems to hold true.

MP.7 Repeat for a second example of a cyclic quadrilateral.푨푪 ⋅ 푩푫 푨푩 ⋅ 푪푫 푩푪 ⋅ 푨푫 CHALLENGE: Draw a cyclic quadrilateral with one side of length zero. What shape is the this “cyclic quadrilateral”? Does Ptolemy’s claim hold true for it?

Students will see, within measuring error, that the relationship = + seems to hold. For a “quadrilateral” with one side of length zero, the figure is a triangle inscribed in a circle. If the length AB=0, then the points A and B coincide, and Ptolemy’s theorem states = 푨푪+⋅ 푩푫 푨푩, which⋅ 푪푫 is 푩푪true.⋅ 푨푫

푨푪 ⋅ 푨푫 ퟎ ⋅ 푪푫 푨푪 ⋅ 푨푫 Exploratory Challenge: A Journey to Ptolemy’s Theorem (30 minutes) This Exploratory Challenge will lead students to a proof of Ptolemy’s theorem. Students should work in pairs. The teacher will guide as necessary. DRAFT Exploratory Challenge: A Journey to Ptolemy’s Theorem

The diagram shows cyclic quadrilateral with diagonals and intersecting to form an acute angle with degree measure . �� = , = , = , and = 푨푩푪푫. 푨푪 �푩푫�� 풘 푨푩 a.풂 푩푪From풃 last푪푫 lesson,풄 what푫푨 is the풅 area of quadrilateral in terms of the lengths of its diagonals and the angle w? Remember this formula for later on! 푨푩푪푫

( ) = ퟏ 푨풓풆풂 푨푩푪푫 ퟐ 푨푪 ⋅ 푩푫 ⋅ 풔풊풏 풘 b. Explain why one of the angles, or , has a measure less than or equal to °. ∠푩푪푫 ∠푩푨푫 Opposite angles of a cyclic quadrilateralퟗퟎ are supplementary. These two angles can’t both have measures greater than °.

ퟗퟎ

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c. Let’s assume that in our diagram is the angle with a measure less than or equal to °. Call its measure degrees. What is the area of triangle in terms Scaffolding: of , , and ? What∠푩푪푫 is the area of triangle in terms of , , and ? What is the . For part (c) of the area of quadrilateralퟗퟎ in 풗terms of , , , , and ? 푩푪푫 풃 풄 풗 푩푨푫 풂 풅 풗 Exploratory Challenge, 푨푩푪푫 풂 풃 풃If 풅 represents풗 the degree review Exercises 4 and 5 measure of an acute angle, from Lesson 20 which 풗 then would be the show that the area degree measure of angle formula for a triangle, ퟏퟖퟎ since− opposite 풗 angles of a cyclic quadrilateral are ( ) = sin( ), 푩푨푫 supplementary. The area can be used where1 of triangle could then 퐴푟푒푎 퐴퐵퐶 2 푎푏 푐 represents an obtuse be calculated using 푩푪푫 angle with a targeted푐 small ( ), and the area group. ퟏ of풃풄 triangle⋅ 풔풊풏 풗 could be ퟐ . Allow advanced learners calculated by푨푩푫 to work through and DRAFTퟏ 풂풅 ⋅ struggle with the ( ) , or ( ). So, the area of theퟐ quadrilateral is ퟏ exploration on their own. the풔풊풏 sumퟏퟖퟎ of− theퟏퟖퟎ areas− of 풗 triangles풂풅 ⋅ 풔풊풏 and풗 , which provides the following푨푩푪푫: � � ퟐ ( ) = + 푨푩푫 푩푪푫. ퟏ ퟏ 푨풓풆풂 푨푩푪푫 ퟐ 풂풅 ⋅ 풔풊풏 풗 ퟐ 풃풄 ⋅ 풔풊풏 풗 d. We now have two different expressions representing the area of the same cyclic quadrilateral . Does it seem to you that we are close to a proof of Ptolemy’s claim? 푨푩푪푫 Equating the two expressions gives as a relationship that does, admittedly, use the four side lengths of the quadrilateral and the two diagonal lengths, but we also have terms that involve ( ) and ( ). These terms are not part of Ptolemy’s equation. 풔풊풏 풘 풔풊풏 풗

In order to reach Ptolemy’s conclusion, students, in Exploratory Challenge, parts (e)–(j), will use rigid motions to convert the cyclic quadrilateral ABCD to a new cyclic quadrilateral of the same area with the same side- lengths (but in an alternative order), and with its matching angle “ ” congruent to angle w in the original diagram. Equating the areas of these two cyclic quadrilaterals will yield the result we need. Again, have 푣 students complete this work in homogeneous pairs or small groups. Offer to help students as needed.

e. Trace the circle and points , , , and onto a sheet of patty paper. Reflect triangle about the perpendicular bisector of diagonal . Let푨 A’,푩 B’,푪 C’ be푫 the images of the points A, B, C, respectively. 푨푩푪 �푨푪�� i. What does the reflection do with points and ? Because the reflection was done about the 푨 푪 perpendicular bisector of the chord , the endpoints of the chord are images of each other; i.e., = and �� = . 푨푪 푨 푪′

푪 푨′

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ii. Is it correct to draw as on the circle? Explain why or why not. Scaffolding: Reflections preserve angle measure. Thus, . Also, 푩′ is supplementary to . Thus, is, too. This means . The argument provided in part that AB’CD is a quadrilateral with one pair (and∠푨푩 hence′푪 ≅ ∠both푨푩푪 pairs) of (e),(ii) follows the previous lesson. ∠opposite푨푩푪 angles supplementary.∠푪푫푨 It is, thus∠, cyclic.푨푩′푪 This means that An alternative argument: the lies on the circle that passes through A, C, and D. And, this is the perpendicular bisector of a chord of original circle. 푩′ a circle passes through the center of the circle. Reflecting a circle, or iii. Explain why quadrilateral has the same area as quadrilateral points on a circle, about the . 푨푩′푪푫 perpendicular bisector of the chord Triangle is congruent to triangle by a congruence 푨푩푪푫 is, therefore, a symmetry of the transformation, and so these triangles have the same area. It now follows that푨푩 ′푪the two quadrilaterals have푨푩푪 the same area. circle; thus, must go to a point on the same circle. 퐵 퐵′ f. The diagram shown to the right shows angles having degree measures , , , , and . Find and label any other angles having degree measures , , , , or , and justify your answers. DRAFT풖 풘 풙 풚 풛 See diagram. Justifications include the inscribed angle theorem and vertical angles. 풖 풘 풙 풚 풛

g. Explain why = + in your diagram from part (f).

The angle with풘 degree풖 풛 measure is an exterior angle to a triangle with two remote interior angles and . It follows that = + . 풘 풖 풛 풘 풖 풛

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h. Identify angles of measures , , , , and in your diagram of the cyclic quadrilateral AB’CD from part (e). See diagram below. 풖 풙 풚 풛 풘 DRAFT

i. Write a formula for the area of triangle in terms of , , and . Write a formula for the area of triangle in terms of , , and . ′ ′ 푩 푨푫 풃 풅 풘 푩 푪푫( ) = 풂 풄 ( 풘) ′ 푨풓풆풂and 푩 푨푫 풃풅 ⋅ 풔풊풏 풘 ( ) = ( ) ′ 푨풓풆풂 푩 푪푫 풂풄 ⋅ 풔풊풏 풘 j. Based on the results of part (i), write a formula for the area of cyclic quadrilateral In terms of a, , , , and . 푨푩푪푫 풃 풄 풅 풘 ( ) = ( ) = ( ) + ( ) ′ ퟏ ퟏ 푨풓풆풂 푨푩푪푫 푨풓풆풂 푨푩 푪푫 ퟐ 풃풅 ⋅ 풔풊풏 풘 ퟐ 풂풄 ⋅ 풔풊풏 풘 k. Going back to part (a), now establish Ptolemy’s theorem.

( ) = ( ) + ( ) The two formulas represent the same area. ퟏ ퟏ ퟏ 푨푪 ⋅ 푩푫 ⋅ 풔풊풏 풘 풃풅 ⋅ 풔풊풏 풘 풂풄 ⋅ 풔풊풏 풘 ퟐ ( ) ( )ퟐ= ( ) (ퟐ + ) Distributive property ퟏ ퟏ ⋅ 풔풊풏 풘= ⋅ 푨푪+ ⋅ 푩푫 ⋅ 풔풊풏 풘 ⋅ 풃풅 풂풄 Multiplicative property of equality ퟐ ퟐ or 푨푪 ⋅ 푩푫 풃풅 풂풄 = ( ) + ( ) Substitution 푨푪 ⋅ 푩푫 푩푪 ⋅ 푨푫 푨푩 ⋅ 푪푫 Closing (3 minutes) Gather the class together and ask the following questions: . What was most challenging in your work today?  Answers will vary. Students might say that it was challenging to do the algebra involved or to keep track of congruent angles, for example. . Are you convinced that this theorem holds for all cyclic quadrilaterals?  Answers will vary, but students should say “yes.”

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. Will Ptolemy’s theorem hold for all quadrilaterals? Explain.  At present, we don’t know! The proof seemed very specific to cyclic quadrilaterals, so we might suspect it holds only for these types of quadrilaterals. (If there is time, students can draw an example of non- cyclic quadrilateral and check that the result does not hold for it.)

Lesson Summary

Theorems

PTOLEMY’S THEOREM: For a cyclic quadrilateral , = + .

푨푩푪푫 푨푪 ⋅ 푩푫 푨푩 ⋅ 푪푫 푩푪 ⋅ 푨푫 Relevant Vocabluary CYCLICDRAFT QUADRILATERAL: A quadrilateral with all vertices lying on a circle is known as a cyclic quadrilateral.

Exit Ticket

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Name Date

Lesson 21: Ptolemy’s Theorem

Exit Ticket

1. What is the length of the chord ? Explain your answer. DRAFT퐴퐶��

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Exit Ticket Sample Solutions

What is the length of the chord ? Explain your answer.

Chord is 푨푪�a�� diameter of the circle, and = + = . By Ptolemy’s theorem: = + giving = ퟐ . ퟐ �푩푫�� 푩푫 �ퟐ ퟖ √ퟔퟖ ퟑퟐ 푨푪 ∙ ퟔퟖ ퟐ ∙ ퟖ ퟐ ∙ ퟖ 푨푪 � √ ퟔퟖ

Problem Set Sample Solutions

1. An equilateral triangle is inscribed in a circle. If is a point on the circle, what does Ptolemy’s theorem have to say about the distances from this point to the three vertices of the triangle? 푷 DRAFT It says that the sum of the two shorter distances is equal to the longer distance.

2. Kite is inscribed in a circle. The kite has an area of square in., and the ratio of the lengths of the non-congruent adjacent sides is : . What is the perimeter푨푩푪푫 of the kite? ퟏퟎퟖ ퟑ ퟏ = ퟏ = + = 푨푪 ⋅ 푩푫 ퟏퟎퟖ ퟐ Since푨푪 ⋅ 푩푫= 푨푩 and⋅ 푪푫 =푩푪 ⋅ 푨푫, then ퟐퟏퟔ 푨푩 =푩푪 푨푫+ 푪푫 = = = . 푨푪 ⋅ 푩푫 푨푩 ⋅ 푪푫 푨푩 ⋅ 푪푫 ퟐ푨푩 ⋅ 푪푫 ퟐퟏퟔ Let푨푩 be⋅ 푪푫 the lengthퟏퟎퟖ of , then 풙 = 푪푫 = 풙 ⋅ ퟐퟑ풙= ퟏퟎퟖ ퟑ풙ퟐ= . ퟏퟎퟖ 풙 ퟑퟔ The풙 lengthퟔ = = , and the length = = . Therefore, the perimeter of kite is in.

퐂퐃 퐀퐃 ퟔ 퐀퐁 퐁퐂 ퟏퟖ 퐀퐁퐂퐃 ퟒퟖ

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3. Draw a right triangle with leg lengths and , and hypotenuse length . Draw a rotated copy of the triangle such that the figures form a rectangle. What does Ptolemy have to say about this rectangle? 풂 풃 풄

We get + = , the Pythagorean theorem! ퟐ ퟐ ퟐ 풂 풃 풄

4. Draw a regular pentagon of side length 1 in a circle. Let b be the length of its diagonals. What does Ptolemy’s theorem say about the quadrilateral formed by four of the vertices of the pentagon?

+ = + , = . (This is the famous golden ratio!) � ퟐ ퟓ ퟏ DRAFT풃 풃 ퟏ 풔풐 풃 ퟐ

5. The area of the inscribed quadrilateral is . Determine the circumference of the circle. ퟐ √ퟑퟎퟎ 퐦퐦 Since is a rectangle, then = , and the diagonals are diameters of the circle. 푨푩푪푫 푨푫 ⋅ 푨푩 √ퟑퟎퟎ The length of = , and the length of = , so = , and 푨푩 = ( 푨푩). ퟐ풓 풔풊풏 ퟔퟎ 푩푪 ퟐ풓 풔풊풏 ퟑퟎ ퟑ 푨푫 √ 푨푩 √ퟑ 푨푫( ) = = 푨푫ퟐ⋅=√ퟑ 푨푫 √ퟑퟎퟎ 푨푫 = ퟏퟎퟎ 푨푫 ퟏퟎ 푨푩 ퟏퟎ+√ퟑ = + = 푨푩 ⋅+푫푪 푨푫= ⋅ 푩푪 푨푪 ⋅ 푩푫 ퟏퟎ√ퟑ=⋅ ퟏퟎ√ ퟑ ퟏퟎퟐ ⋅ ퟏퟎ 푨푪 ⋅ 푨푪 ퟑퟎퟎ= ퟏퟎퟎ ퟐ 푨푪 ퟒퟎퟎ 푨푪 ퟐퟎSince 푨푪 = , the radius of the circle is , and the circumference of the circle is mm. 푨푪 ퟐퟎ ퟏퟎ ퟐퟎ흅

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 21 M5 GEOMETRY

6. EXTENSION: Suppose x and y are two acute angles, and the circle has a diameter of 1 unit. Find , , , and in terms of and . Apply Ptolemy’s theorem, and determine the exact value of ( ). 풂 풃 풄 풅 풙 풚 ퟎ 퐬퐢퐧 ퟕퟓ Use scaffolded questions below as needed.

a. Explain why equals the diameter of the circle. ( ) 풂 If the diameter풔풊풏 is 풙1, this is a right triangle because it is inscribed in a semicircle, so ( ) = , or the = . Since the diameter is 1, the diameter is ( ) 풂 풂 equal풔풊풏 to풙 . ퟏ (ퟏ) 풔풊풏 풙 풂 풔풊풏 풙 b. If the circle has a diameter of 1, a = ? DRAFT= ( ) 풂 풔풊풏 풙 c. Use Thales’ theorem to write the side lengths in the original diagram in terms of x and y.

Since both are right triangles, the side lengths are = ( ), = ( ), = ( ), = ( ). 풂 풔풊풏 풙 풃 풄풐풙 풙 풄 풄풐풔 풚 풂풏풅 풅 풔풊풏 풚 d. If one diagonal of the cyclic quadrilateral is 1, what is the other? ( + )

풔풊풏 풙 풚 e. What does Ptolemy’s theorem give? ( + ) = ( ) ( ) + ( ) ( )

ퟏ ∙ 풔풊풏 풙 풚 풔풊풏 풙 풄풐풔 풚 풄풐풔 풙 풔풊풏 풚 f. Using the result from part (e), determine the exact value of ( ). ퟎ + ( ) = ( + ) = ( ) ( ) + ( )풔풊풏( ퟕퟓ) = + = � � � � � ퟏ ퟐ ퟑ ퟐ ퟐ ퟔ 풔풊풏 ퟕퟓ 풔풊풏 ퟑퟎ ퟒퟓ 풔풊풏 ퟑퟎ 풄풐풔 ퟒퟓ 풄풐풔 ퟑퟎ 풔풊풏 ퟒퟓ ퟐ ∙ ퟐ ퟐ ∙ ퟐ ퟒ

Lesson 21: Ptolemy’s Theorem Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

Name Date

1. Let be the circle in the coordinate plane that passes though the points (0, 0), (0, 6), and (8, 0).

a. 𝐶𝐶What are the coordinates of the center of the circle?

DRAFT

b. What is the area of the portion of the interior of the circle that lies in the first quadrant? (Give an exact answer in terms of .)

𝜋𝜋

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

c. What is the area of the portion of the interior of the circle that lies in the second quadrant? (Give an approximate answer correct to one decimal place.)

DRAFT

d. What is the length of the arc of the circle that lies in the first quadrant with endpoints on the axes? (Give an exact answer in terms of .)

𝜋𝜋

e. What is the length of the arc of the circle that lies in the second quadrant with endpoints on the axes? (Give an approximate answer correct to one decimal place.)

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

f. A line of slope 1 is tangent to the circle with point of contact in the first quadrant. What are the coordinates of that point of contact? −

DRAFT

g. Describe a sequence of transformations that show circle is similar to a circle with radius one centered at the origin. 𝐶𝐶

h. If the same sequence of transformations is applied to the tangent line described in part (f), will the image of that line also be a line tangent to the circle of radius one centered about the origin? If so, what are the coordinates of the point of contact of this image line and this circle?

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

2. In the figure below the circle with center circumscribes .

Points , , and are collinear and the line𝑂𝑂 through and△ 𝐴𝐴𝐴𝐴 is 𝐴𝐴tangent to the circle at . The center of the circle lies inside . 𝐴𝐴 𝐵𝐵 𝑃𝑃 𝑃𝑃 𝐶𝐶 𝐶𝐶 △ 𝐴𝐴𝐴𝐴𝐴𝐴 DRAFT

a. Find two angles in the diagram that are congruent, and explain why they are congruent.

b. If is the midpoint of and = 7, what is ?

𝐵𝐵 𝐴𝐴��𝐴𝐴� 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

c. If = 50°, and the measure of the arc is 130°, what is ?

𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝑚𝑚∠𝑃𝑃

DRAFT

3. The circumscribing circle and the inscribed circle of a triangle have the same center.

a. By drawing three radii of the circumscribing circle, explain why the triangle must be equiangular, and hence equilateral.

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

b. Prove again that the triangle must be equilateral, but this time by drawing three radii of the inscribed circle.

DRAFT

c. Describe a sequence of straightedge and compass constructions that allow you to draw a circle inscribed in a given equilateral triangle.

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

4. a. Show that

( 2)( 6) + ( 5)( + 11) = 0

is the equation of a circle. What𝑥𝑥 −is the𝑥𝑥 center − of𝑦𝑦 this − circle?𝑦𝑦 What is the radius of this circle?

DRAFT

b. A circle has diameter with endpoints ( , ) and ( , ). Show that the equation of this circle can be written as 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑑𝑑 ( )( ) + ( )( ) = 0.

𝑥𝑥 −𝑎𝑎 𝑥𝑥 −𝑏𝑏 𝑦𝑦 −𝑐𝑐 𝑦𝑦 −𝑑𝑑

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

5. Prove that opposite angles of a cyclic quadrilateral are supplementary. DRAFT

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

A Progression Toward Mastery STEP 1 STEP 2 STEP 3 STEP 4 Assessment Missing or incorrect Missing or incorrect A correct answer A correct answer Task Item answer and little answer but with some evidence supported by evidence of evidence of some of reasoning or substantial reasoning or reasoning or application of evidence of solid application of application of mathematics to reasoning or mathematics to mathematics to solve the problem, application of solve the problem. solve the problem. or an incorrect mathematics to answer with solve the problem. substantial evidence of solid reasoning or DRAFTapplication of mathematics to solve the problem. 1 G-C.1. Students struggle Students complete only Students complete the Students carry out all the G-C.2 beginning the majority a proportion of the tasks majority of tasks tasks with precision and of tasks at hand, provide required and provide correctly and provide care, articulate complete G-C.5 little evidence of first only partial explanations. clear justifications to the justifications, and G-GPE.1 steps (drawing a majority of the items provide correct final G-GPE.4 diagram), and don’t asked. solutions. articulate clearly rationale and reasoning. 2 G-C.2 Student does not Student identifies similar Student correctly Student correctly identify similar triangles triangles without identifies similar identifies similar and makes little progress justification and makes triangles, with triangles, with with this question. some progress towards justification. The student justification, finds identifying matching may show some correct matching ratios, ratios and determining difficulty in identifying and can successfully angle measures. correct matching ratios computes all angle or angle relations. measures needed in the diagram. 3 G-C.3 Student does not Student recognizes just Student recognizes the Student recognizes the recognize angle bisectors the angle bisectors or angle bisectors and angle bisectors and or perpendicular just the perpendicular perpendicular bisectors perpendicular bisectors bisectors and makes bisectors needed to and makes progress and provides two little progress on either complete these proofs towards completing complete, well- of the proofs. Student and completes, or is each of the two proofs. articulated proofs. does not describe a close to completing, one Student makes good Student also gives a suitable construction. proof. Student makes progress towards thorough description of some progress for describing a complete a construction process. describing a suitable construction process. construction process.

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

4 G-GPE.1 Student does not Student confuses the Student completes part Student provides complete the square signs of the coordinates (a) correctly and makes complete and correct correctly and does not of the center or fails to progress on part (b). solutions to both interpret the center and give the square root of problems. radius from the equation the quantity for the obtained. Little or no radius after conducting progress is made on part the correct algebraic (b). work for (a). Makes some progress on part (b). 5 G-C.3 Student does not set up Student describes a The student makes some Student provides a a suitable scenario for potentially suitable good progress for thorough and elegant constructing the proof. setup for constructing a establishing a proof with proof. proof but does not only minor complete the proof. inconsistencies in DRAFTreasoning or explanation.

Module 5: Circles With and Without Coordinates Date: 8/29/14 10

NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

Name Date

1. Let be the circle in the coordinate plane that passes though the points (0, 0), (0, 6), and (8, 0).

a. 𝐶𝐶What are the coordinates of the center of the circle?

Since the angle formed by the points (0, 6), (0, 0), and (8, 0) is a right angle, the line segment connecting (0, 6) to (8, 0) must be the DRAFTdiameter of the circle. Therefore, the center of the circle is (4, 3), the midpoint of this diameter.

b. What is the area of the portion of the interior of the circle that lies in the first quadrant? (Give an exact answer in terms of .)

𝜋𝜋 The distance between (0,6) and (8,0) is 62 + 82 = 10, and so the circle has radius 5. The area in question �is composed of half a circle and a right triangle.

1 1 25π Its area is × 8 × 6 + π52 = + 24 square units. 2 2 2

� � � �

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

c. What is the area of the portion of the interior of the circle that lies in the second quadrant? (Give an approximate answer correct to one decimal place.)

We seek the area of the region shown. We have a chord of length 6 in a circle of radius 5.

Label the angle x as shown and distance d. By the Pythagorean theorem, 3 d = 4. We also have that sin(x) = , and so x ≈ 36.9°. 5

The shaded area is the difference of the area of a sector and of a triangle. We have

DRAFT2x 2 1 73.8 area = π5 ×6×4 ≈ × 25π 12 ≈ 4.1 square units. 360 2 360 � � − � � � � −

d. What is the length of the arc of the circle that lies in the first quadrant with endpoints on the axes? (Give an exact answer in terms of .)

𝜋𝜋

This arc is a semicircle and so is half the circumference of the circle in length: 1 × 2π × 5 = 5π units. 2

e. What is the length of the arc of the circle that lies in the second quadrant with endpoints on the axes? (Give an approximate answer correct to one decimal place.)

2x 73.8 Using the notation of part (c) this length is 2π 5 ≈ × 10π ≈ 6.4 units. 360 360 ⋅ ⋅

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

f. A line of slope 1 is tangent to the circle with point of contact in the first quadrant. What are the coordinates of that point of contact? − Draw a radius from the center of the circle (4, 3) to the point of contact, which we will denote (x, y).

This radius is perpendicular to the tangent line and so

y 3 has slope 1. Consequently, =1; that is, y 3 = x 4. x − 4 − Also, since (x,y) lies on the circle we have − −

2 (x 4)2 + y 3 = 25.

For both equations to− hold �we− must� have (x 4)2 + DRAFT2 5 5 (x 4) = 25 giving x = 4 + or x = 4 . It is clear from the diagram that the point 2 2 − of −contact we seek has x-coordinate√ to− √the right of the x-coordinate of the center of 5 the circle. So choose x = 4 + . The matching y-coordinate is y = x 4 + 3 = x 1 = 3 + 2 5 5 5 , and the point of contact √thus has coordinates 4 + , 3 + . 2 2 2 − − √ � √ √ � g. Describe a sequence of transformations that show circle is similar to a circle with radius one centered at the origin. 𝐶𝐶 Circle C has center (4,3) and radius 5.

First translate the circle four units to the left and three units downward. This gives a congruent circle with the origin as its center. (Radius is still 5.)

1 Perform a dilation from the origin with scale factor . This will produce a similar circle 5 centered at the origin with radius 1.

Translations and dilations map straight lines to straight lines. Thus the tangent line will still be mapped to a straight line. The mappings will not alter the fact that the circle and the line touch at one point. Thus the image will again be a line tangent to the circle.

5 5 Under the translation described in part (i) the point of contact 4 + , 3 + is 2 2 5 5 1 1 mapped to , . Under the dilation described, this is then mapped√ to √ , . 2 2 � 2� 2 �√ √ � �√ √ �

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

2. In the figure below the circle with center circumscribes .

a. Find two angles in the diagram that are congruent, and explain why they are congruent.

Draw two radii as shown. Let m BAC = x. Then by the inscribed/central angle theorem∠ we have m BOC = 2x.

Since∠ BOC is isosceles, it follows that 1 m OCB = (180° 2x) = 90° x. △ 2

By∠ the radius/tangent− theorem,− m OCP = 90°, and so m BCP = x. ∠

We have∠ BAC BCP because they intercept the same arc ∠and have≅ ∠ the same measure.

b. If is the midpoint of and = 7, what is ? �� By𝐵𝐵 the previous question,𝐴𝐴𝐴𝐴 triangles𝑃𝑃𝑃𝑃 ACP and𝑃𝑃𝑃𝑃 CBP each have an angle of measure x and share the angle at P. Thus, they are similar triangles.

Since triangles ACP and CBP are similar, matching sides come in the same ratio. Thus, PB PC PB 7 7 = . Now, AP = 2 PB and PC = 7, so = . This gives PB = . PC AP 7 2PB 2 ⋅ √

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

c. If = 50°, and the measure of the arc is 130°, what is ?

𝑚𝑚∠𝐵𝐵𝐵𝐵𝐵𝐵 By the inscribed/central𝐴𝐴𝐴𝐴 𝑚𝑚∠𝑃𝑃angle theorem, arc BC has measure 100°. By the secant/tangent angle theorem 130° 100° m P = = 15°. 2 − ∠ (One can also draw in radii and chase angles in triangles to obtain the same result.)

DRAFT 3. The circumscribing circle and the inscribed circle of a triangle have the same center.

a. By drawing three radii of the circumscribing circle, explain why the triangle must be equiangular, and hence equilateral.

Draw the three radii as directed and label six angles a, b, c, d, e, and f as shown.

We have a = f because they are base angles of an isosceles triangle. (We have congruent radii.) In the same way, b = c and d = e.

From the construction of an inscribed circle, we know that each radius drawn is an angle bisector of the triangle. Thus, we have a = b, c = d, and e = f.

It now follows that a = b = c = d = e = f. In particular, a + b = c + d = e + f, and the triangle is equiangular (hence, they are equilateral).

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

b. Prove again that the triangle must be equilateral, but this time by drawing three radii of the inscribed circle.

By the construction of the circumscribing circle of a triangle each radius in this picture is the perpendicular bisector of a side of the triangle. If we label the lengths a, b, c, d, e, and f as shown, it follows that b = c, d = e, and a = f.

By the two-tangents theorem, we also have a = b, c = d, and e = f.

Thus, a = b = c = d = e = f and, in particular, b + c = d + e = a + f, and the triangle is equilateral. DRAFT c. Describe a sequence of straightedge and compass constructions that allow you to draw a circle inscribed in a given equilateral triangle.

The center of an inscribed circle lies at the point of intersection of any two angle bisectors of the equilateral triangle.

To construct an angle bisector:

1. Draw a circle with center at one vertex P of the triangle intersecting two sides of the triangle. Call those two points of intersection A and B.

2. Setting the compass at a fixed width, draw two congruent intersecting circles, one centered at A and one centered at B. Call a point of intersection of these two circles Q. (We can assume Q is different from P.)

3. The line through P and Q is an angle bisector of the triangle.

Indeed, construct two such angle bisectors and call their point of intersection O. This is the center of the inscribed circle. Now draw a line through O perpendicular to one side of the triangle. To do this:

1. Draw a circle centered at O that intersects one side of the triangle at two points. Call those points C and D.

2. Draw two congruent intersecting circles, one with center C and one with center D.

3. Draw the line through the points of intersection of those two congruent circles. This is a line through O perpendicular to the side of the triangle.

Suppose this perpendicular line through O intersects the side of the triangle at the point R. Set the compass to have width equal to OR. This is the radius of the inscribed circle and so drawing a circle of this radius with center O produces the inscribed circle.

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

4. a. Show that ( 2)( 6) + ( 5)( + 11) = 0

is the equation of a circle. What𝑥𝑥 −is the𝑥𝑥 center − of𝑦𝑦 this − circle?𝑦𝑦 What is the radius of this circle?

We have

(x 2)(x 6) + y 5 y + 11 = 0 2 2 x − 8x +−12 + y� +−6y�� 55 = 0� 2 2 x − 8x + 16 + y + 6y +−9 = 4 + 64 2 (x −4)2 + y + 3 = 68 DRAFT− � � This is the equation of a circle with center 4,-3 and radius 68.

� � √ b. A circle has diameter with endpoints ( , ) and ( , ). Show that the equation of this circle can be written as ( )( 𝑎𝑎 𝑏𝑏 ) + ( 𝑐𝑐 𝑑𝑑 )( ) = 0.

a + c b + d The midpoint of the diameter,𝑥𝑥 which −𝑎𝑎 𝑥𝑥 is −𝑏𝑏 ,𝑦𝑦 −𝑐𝑐 is 𝑦𝑦the center −𝑑𝑑 of the circle, and half the 2 2 1 2 distance between the endpoints, which is� (c �a)2 + d b , is the radius of the circle. 2 Thus the equation of the circle is � − � − �

2 2 a + c b + d 1 2 x + y = (c a)2 + d b . 2 2 4 � − � � − � � − � − � � Multiplying through by 4 gives

2 2 (2x a c)2 + 2y b d = (c a)2 + d b .

This becomes − − � − − � − � − �

2 2 2 2 4x2 + a2 + c2 4xa 4xc + 2ac + 4y2 + b + d 4yb 4yd + 2bd = c2 + a2 2ac + d + b 2bd.

That is − − − − − −

4x2 4xa 4xc + 4ac + 4y2 4yb 4yd + 4bd = 0.

Dividing through by 4:− − − −

x2 xa xc + ac + y2 yb yd + bd = 0.

That is, − − − −

(x a)(x c) + y b y d = 0,

as desired. − − � − �� − �

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NYS COMMON CORE MATHEMATICS CURRICULUM End-of-Module Assessment Task M5 GEOMETRY

5. Prove that opposite angles of a cyclic quadrilateral are supplementary.

Consider a cyclic quadrilateral with two interior opposite angles of measures x and y as shown.

The vertices of the quadrilateral divide the circle into four arcs. Suppose these arcs have measures a, b, c, and d as shown.

By the inscribed/central angle theorem we have a + d = 2y and b + c = 2x. So, a + b + c + d = 2(x + y).

But, a + b + c + d = 360°. Thus, it follows that DRAFT360° x + y = = 180°. 2

By analogous reasoning, the angles in the second pair of interior opposite angles are supplementary as well. (This also follows from the fact that the interior angles of a quadrilateral sum to 360°: The second pair of interior angles have measures summing to 360° x y = 360° 180° = 180°.)

− − −

Module 5: Circles With and Without Coordinates Date: 8/29/14 18