Geometry: All-In-One Answers Version A

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Lesson 1-1 Patterns and Inductive Reasoning 3 Finding a Counterexample Find a counterexample for the conjecture. Since 32 42 52, the sum of the squares of two consecutive numbers is Lesson Objective NAEP 2005 Strand: Geometry the square of the next consecutive number. 1 Use inductive reasoning to make Topic: Mathematical Reasoning 5 conjectures Begin with 4. The next consecutive number is 4 1 . Local Standards: ______The sum of the squares of these two consecutive numbers is 42 52 16 25 41 . Vocabulary. The conjecture is that 41 is the square of the next consecutive number. Inductive reasoning is reasoning based on patterns you observe. The square of the next consecutive number is (5 1)2 62 36 . So, since 42 52 and 62 are not the same, this conjecture is false . A conjecture is a conclusion you reach using inductive reasoning. 4 Applying Conjectures to Business The price of overnight shipping was $8.00 in A counterexample is an example for which the conjecture is incorrect. 2000, $9.50 in 2001, and $11.00 in 2002. Make a conjecture about the price in 2003. Write the data in a table. Find a pattern.

2000 2001 2002 Examples. $8.00 $9.50 $11.00

1 Finding and Using a Pattern Find a pattern for the sequence. 384 192 96 48 Use the pattern to find the next two terms in the sequence. Each year the price increased by $1.50 . A possible conjecture is that 384, 192, 96, 48,... the price in 2003 will increase by $1.50 . If so, the price in 2003 would 2 2 2 Each term is half the preceding term. The next two be $11.00 $ 1.50 $.12.50 terms are 48 224and 24 2 12 . Quick Check. 2 Using Inductive Reasoning Make a conjecture about the sum of the cubes of the ﬁrst 25 counting numbers. 1. Find the next two terms in each sequence. Find the ﬁrst few sums. Notice that each sum is a perfect square and that a. 1,2,4,7,11,16,22,29 ,37 ,... the perfect squares form a pattern.

13 1 12 12 b. Monday,Tuesday,Wednesday,Thursday , Friday ,... 13 23 9 32 (1 2)2 3 3 3 2 2 1 2 3 36 6 (1 2 3) c. 3 3 3 3 2 2 1 2 3 4 100 10 (1 2 3 4) , ,... 13 23 33 43 53 225 152 (1 2 3 4 5)2 Answers may vary. Sample: The sum of the first two cubes equals the square of the sum of the first two counting numbers. The sum of the first three cubes equals the © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. square of the sum of the first three counting numbers. This pattern continues for the fourth and fifth rows. So a conjecture might be that the sum of the cubes of the first 25 counting numbers equals the square of the sum of the first 25 counting numbers, or (1 2 3 . . . 25)2.

2 3 Geometry Lesson 1-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-1

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2. Make a conjecture about the sum of the ﬁrst 35 odd numbers. Use your Lesson 1-2 Drawings, Nets, and Other Models calculator to verify your conjecture. Lesson Objectives 1 1 12 NAEP 2005 Strand: Geometry 1 Make isometric and orthographic Topic: Dimension and Shape 1 3 4 22 drawings 2 Draw nets for three-dimensional Local Standards: ______13 5 9 32 figures

1 3 5 7 16 42 Vocabulary. 1 3 5 7 9 25 52 An isometric drawing of a three-dimensional object shows a corner view of the figure drawn on isometric dot paper. The sum of the first 35 odd numbers is 352, or 1225. An orthographic drawing is the top view, front view, and right-side view of a three-dimensional figure. A foundation drawing shows the base of a structure and the height of each part. 3. Finding a Counterexample Find a counterexample for the conjecture. Some products of 5 and other numbers are shown in the table. A net is a two-dimensional pattern you can fold to form a three-dimensional figure. 5 7 35 5 13 65 5 3 15 5 9 45 5 11 55 5 25 125

Therefore, the product of 5 and any positive integer ends in 5. Examples. Isometric Drawing Since 5 2 10, not all products of 5 end in 5. 1 Make an isometric drawing of (But they all either end in 5 or 0.) the cube structure at right. An isometric drawing shows three sides of a figure from a corner view. Hidden edges are not shown, so all edges are solid lines.

4. Suppose the price of two-day shipping was $6.00 in 2000, $7.00 in 2001, and $8.00 in 2002. Make a conjecture about the price in 2003.

Each year the price increased by $1.00. A possible conjecture is that the price in 2003 will increase by $1.00. If so, the price in 2003 would be $8.00 $1.00 $9.00. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

4 5 Geometry Lesson 1-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-2

1 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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2 Orthographic Drawing Make an orthographic drawing of the 4 Identifying Solids from Nets Is the pattern a net for a cube? isometric drawing at right. If so, name two letters that will be opposite faces. A

Orthographic drawings ﬂatten the depth of a ﬁgure. An orthographic The pattern is a net because you can Front BDEF drawing shows three views. Because Right fold it to form a cube. Fold squares A and C up to form the back no edge of the isometric drawing is hidden in the top, front, and front of the cube. Fold D up to form a side. Fold E over to C and right views, all lines are solid. form the top. Fold F down to form another side. After the net is folded, the following pairs of letters are on opposite faces: A and C are the back and front faces.

B and E are the bottomand top faces.

DFand are the right and left side faces. Front Top Right

5 Drawing a Net Draw a net for the ﬁgure with a square base 3 Foundation Drawing Make a foundation drawing for the and four isosceles triangle faces. Label the net with its dimensions. 10 cm isometric drawing. Think of the sides of the square base as hinges, and “unfold” the To make a foundation drawing, use the top view of the orthographic drawing. Front ﬁgure at these edges to form a net. The base of each of the four Right isosceles triangle faces is a side of the square . Write in the 8 cm known dimensions.

Top

0 cm Because the top view is formed from 3 squares, show 3 squares in the 8 cm 1 foundation drawing. Identify the square that represents the tallest part. Write the number 2 in the back square to indicate that the back section is 2 cubes high. Write the number 1 in each of the two front squares to indicate Quick Check. that each front section is 1 cube high. 1. Make an isometric drawing of the cube structure below.

6 7 Geometry Lesson 1-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-2

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2. Make an orthographic drawing from this isometric drawing. 5. The drawing shows one possible net for the Graham Crackers box. 14 cm 7 cm

Front Right 20 cm GRAHAM GRAHAM 20 cm Front Top Right CRACKERS CRACKERS 7 cm 14 cm

Draw a different net for this box. Show the dimensions in your diagram. Answers may vary. Example: 14 cm 3. a. How many cubes would you use to make the structure in Example 3? 4

20 cm

b. Critical Thinking Which drawing did you use to answer part (a), the foundation drawing or the isometric drawing? Explain. 7 cm

Answers may vary. Sample: the foundation drawing; you can just add the three numbers.

4. Sketch the three-dimensional ﬁgure that corresponds to the net. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

8 9 Geometry Lesson 1-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-2

2 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 1-3 Points, Lines, and Planes A plane is a ﬂat surface that has no thickness. B A C Lesson Objectives NAEP 2005 Strand: Geometry Two points or lines are coplanar if they lie on the same plane. Plane ABC 1 Understand basic terms of geometry Topic: Dimension and Shape 2 Understand basic postulates of Local Standards: ______geometry A postulate or axiom is an accepted statement of fact.

Vocabulary and Key Concepts. Examples. 1 Identifying Collinear Points In the ﬁgure at right, X Postulate 1-1 name three points that are collinear and three points Through any two points there is exactly one line. that are not collinear. t Points Y ,Z , and W lie on a line, so they are B Line t is the only line that passes through points A and B . A collinear. mYWZ Any other set of three points in the figure do not lie on a line, Postulate 1-2 so no other set of three points is collinear. For example, X, Y, If two lines intersect, then they intersect in exactly one point. and Z form a triangle and are not collinear. A B C * ) * ) R U AE and BD intersect at C 2 Naming a Plane Name the plane shown in two different ways. D E . You can name a plane using any three or more points on that plane that are not collinear. Postulate 1-3 If two planes intersect, then they intersect in exactly one line. Some possible names for the plane shown are the following: S T plane RST , plane RSU , plane RTU , R * ) ST plane STU , and plane RSTU . T W Plane RST and plane STW intersect in ST . S 3 Finding the Intersection of Two Planes Use the diagram at right. H G What is the intersection of plane HGC and plane AED? E F Postulate 1-4 As you look at the cube, the front face is on plane AEFB, the back face C Through any three noncollinear points there is exactly one plane. is on plane HGC, and the left face is on plane AED. The back and left D HD A B faces of the cube* intersect) at . Planes HGC and AED intersect A point is a location. vertically at HD . Space is the set of all points. A line is a series of points that extends in two opposite directions t Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. without end. B A Collinear points are points that lie on the same line.

10 11 Geometry Lesson 1-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-3

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4 Using Postulate 1-4 Shade the plane that contains X, Y, and Z. Z Lesson 1-4 Segments, Rays, Parallel Lines and Planes

Points X, Y, and Z are the vertices of one of the four triangular Lesson Objectives NAEP 2005 Strand: Geometry 1 Identify segments and rays faces of the pyramid. To shade the plane, shade the interior of Topic: Relationships Among Geometric Figures Y 2 Recognize parallel lines the triangle formed by X ,Y , and Z . X Local Standards: ______

V W Vocabulary. Quick Check. 1. Use the ﬁgure in Example 1. A segment is the part of a line consisting of two Segment AB AB a. Are points W, Y, and X collinear? endpoints and all points between them. A B no Endpoint Endpoint

b. Name line m in three different ways. * ) * ) * ) A ray is the part of a line consisting of one endpoint ) Ray YX Answers may vary. Sample: ZW , WY , YZ . YX and all the points of the line on one side of the endpoint. XY Endpoint c. Critical Thinking Why do you think* ) arrowheads are used when drawing a line or naming a line such as ZW ? Opposite rays are two collinear rays with the same Arrowheads are used to show that the line extends in QR S opposite directions without end. endpoint. ) ) RQ and RS are opposite rays.

Parallel lines are coplanar lines that do not intersect. 2. List three different names for plane Z. H G Z Answers may vary. Sample: HEF, HEFG, FGH. Skew lines are noncoplanar; therefore, they are not parallel and do not intersect. E F * ) 3. Name two planes that intersect in BF . H D C ←→ ←→ G AB is parallel to EF. ABF and CBF ←→ ←→ E F A B AB and CG are skew lines. H G

D C E F A B Parallel planes are planes that do not intersect. 4. a. Shade plane VWX. b. Name a point that is coplanar Z with points V, W, and X. G

V W

12 13 Geometry Lesson 1-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-4

3 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Examples. 2. Use the diagram in Example 2. a. Name all labeled segments that are parallel to GF . 1 Naming Segments and Rays Name the segments and rays in the figure. A HE, CB, DA The labeled points in the figure are A, B, and C. A segment is a part of a line consisting of two endpoints and all points GF between them. A segment is named by its two endpoints. So the b. Name all labeled segments that are skew to . segments are BA (or AB) and BC (or CB) . AB, DC, AE, DH B A ray is a part of a line consisting of one endpoint and all the points of C the line on one side of that endpoint. A ray is named by its endpoint first, followed ) ) c. Name another pair of parallel segments and another pair of skew segments. by any other point on the ray. So the rays are BA andBC . Answers may vary. Sample: CG, BF; EF, DH

2 Identifying Parallel and Skew Segments Use the ﬁgure at right. D C Name all segments that are parallel to AE . Name all segments that are skew to AE . 3. Use the diagram to the right. A B a. Name three pairs of parallel planes. S Q Parallel segments lie in the same plane, and the lines that contain them H G do not intersect. The three segments in the ﬁgure that are parallel to PSWT RQVU, PRUT SQVW, PSQR TWVU P R AE are BF ,CG and DH . E F W V

Skew segments are segments that do not lie in the same plane. The four T U segments in the figure that do not lie in the same plane as AE are BC , CD ,FG and GH .

3 Identifying Parallel Planes Identify a pair of parallel planes in your classroom. * ) b. Name a line that is parallel to PQ . Planes are parallel if theydo not intersect . If the walls of your classroom * ) are vertical,opposite walls are parts of parallel planes. If the ceiling and TV floor of the classroom are level, they are parts of parallel planes.

Quick Check. ) ) 1. Critical Thinking Use the ﬁgure in Example 1.CB and BC form a line. Are they opposite rays? Explain. c. Name a line that is parallel to plane QRUV. No; they do not have the same endpoint. * ) Answers may vary. Sample: PS © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

14 15 Geometry Lesson 1-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-4

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Lesson 1-5 Measuring Segments Examples. Lesson Objectives NAEP 2005 Strand: Measurement 1 Comparing Segment Lengths Find AB and BC. 1 Find the lengths of segments Topic: Measuring Physical Attributes AB C Are AB and AC congruent? Local Standards: ______43 2 1 01 2345 AB = ∆-5 -(–1) «=∆-4«= 4 BC = ∆–1 - 4«=»–5» =5 Vocabulary and Key Concepts. AB BC so AB and AC are not congruent. . Postulate 1-5: Ruler Postulate The points of a line can be put into one-to-one 2 Using the Segment Addition Postulate If AB 25, 2x 6 x 7 ﬁnd the value of x. Then ﬁnd AN and NB. correspondence with the real numbers so A NB that the distance between any two points is the absolute value of the difference of the Use the Segment Addition Postulate (Postulate 1-6) to write an equation. corresponding numbers. AN NB AB Segment Addition Postulate ()2x 6 ()x 7 25 Substitute. Postulate 1-6: Segment Addition Postulate 3x 1 25 Simplify the left side. If three points A, B, and C are collinear and B A 3x 24 Subtract 1 from each side. B is between A and C, then AB BC AC. C x 8 Divide each side by 3 . AN 2x 6 2()8 6 10 Substitute 8 for x. NB x 7 ()8 7 15 A coordinate is a point’s distance and direction from zero on a number line. AN 10 and NB 15 , which checks because the sum of the segment lengths equals 25. the length of AB A B R a b AB 5 u a b u Qa coordinate of A coordinate of B

Congruent () segments are segments with the same length.

2 cm A B A B AB CD AB CD 2 cm C D C D © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

16 17 Geometry Lesson 1-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-5

4 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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3 Using the Midpoint M is the midpoint of RT . 5x 9 8x 36 2. EG 100. Find the value of x. Then ﬁnd EF and FG. Find RM, MT, and RT. RMT 4x – 20 2x + 30 Use the deﬁnition of midpoint to write an equation. E FG

RM MT Definition of midpoint 5x 98 x 36 Substitute. x 15, EF 40; FG 60

5x 45 8x Add 36 to each side. 45 3 x Subtract 5x from each side. 15 x Divide each side by 3 . RM 5x 9 5()15 9 84 Substitute 15 for x. XY MT 8x 36 8()15 36 84 3. Z is the midpoint of , and XY 27. Find XZ. RT RM MT 168 Segment Addition Postulate 13.5 RM and MT are each 84 , which is half of 168 , the length of RT .

Quick Check. 1. Find AB. Find C, different from A, such that AB and AC are congruent. AB 7 6 543 2 1 01 2345

AB |7 (2)| |5| 5 We also want BC 5. That means that C must lie 5 units on the other side of B from A. So C must lie at –2 5 3. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

18 19 Geometry Lesson 1-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-5

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Lesson 1-6 Measuring Angles An angle () is formed by two rays with the same endpoint. The rays B are the sides of the angle and the endpoint is the vertex of the angle. Lesson Objectives NAEP 2005 Strand: Measurement T 1 Q 1 Find the measures of angles Topic: Measuring Physical Attributes TBQ 2 Identify special angle pairs Local Standards: ______

x° Vocabulary and Key Concepts. x° x° x°

Postulate) 1-7: Protractor) Postulate ) ) acute angle right angle obtuse angle straight angle OA OB OA OB 0 , x , Let and be opposite rays in a plane.* ,) , and all the rays with 90 x 5 90 90 , x , 180 x 5 180 endpoint O that can be drawn on one side ofAB can be paired with the real numbers) from 0 to 180 so that) OA OB a. is) paired with 0 and ) is paired with 180 . An acute angle has a measurement between 0 and 90. b. If OC is paired with x and OD is paired with y, then mlCOD 5 u x 2 y u . A right angle has a measurement of exactly 90. An obtuse angle has a measurement between 90 and 180. 80 100 1 70 90 10 C 100 80 12 60 10 70 0 D 1 6 1 A straight angle has a measurement of exactly 180 . 20 0 3 50 1 0 0 50 1 13 4 0 0 4 0 4 Congruent angles are 0 two angles with the same measure. 4 1 1 5 0 3 0 3 0 x y 0 5 1 1 6 0 2 0 0 2 0 6 1 1

7 1

0 0 0 0

1 7 A 1 B Examples. 0 180 O 1 Naming Angles Name the angle at right in four ways. C l3 Postulate 1-8: Angle Addition Postulate The name can be the number between the sides of the angle: . lG If point B is in the interior of AOC, then The name can be the vertex of the angle: . 3 Finally, the name can be a point on one side, the vertex, and a point m AOB m BOC m AOC. GA on the other side of the angle:lAGC or lCGA . A B / O C 2 Measuring and Classifying Angles Find the measure of each AOC. Classify each as acute, right, obtuse, or straight. If AOC is a straight angle, then a. b. mAOB mBOC 180 . C 80 100 0 90 110 80 100 7 00 80 1 0 90 110 0 0 1 70 20 B 7 00 80 1 6 11 0 0 1 70 20 0 60 1 6 11 0 12 30 0 60 1 5 5 0 12 3 0 0 1 5 0 13 4 0 5 0 0 C 3 0 1 4 4 1 4 0 0 0 0 4 1 4 4 1 5 A OC 0 0 0 4 1 0 1 3 5 3 0 x y 0 0 5 3 0 1 3 0 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson x All rights reserved. y 0 6 5 0 2 1 1 0 0 0 6 2 6 0 2 0 0 1 2 0 1 6 7 1 1 1 0 0 0 0

1 0 0 0 0

1 7

1 A B A B 0 180 0 180 O O 60, acute 150, obtuse

20 21 Geometry Lesson 1-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-6

5 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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3 Using the Angle Addition Postulate Suppose that m1 42 and m ABC 88. Find m 2. A Quick Check. Use the Angle Addition Postulate (Postulate 1-8) to solve. 1. a. Name CED two other ways. m 1 m 2 mABC Angle Addition Postulate l2, lDEC 42 m2 88 Substitute 42 for m1 and 88 1 for mABC. 2 b. Critical Thinking Would it be correct to name any of the m2 46 Subtract 42 from each side. B C angles E? Explain. No, 3 angles have E for a vertex, so you 4 Identifying Angle Pairs In the diagram identify pairs of need more information in the name to numbered angles that are related as follows: distinguish them from one another. a. complementary Є3 and Є4 1 2 3 b. supplementary 5 2. Find the measure of the angle. Classify it as 4 Є1 and Є2; Є2 and Є3 acute, right, obtuse, or straight. 60; acute c. vertical angles Є1 and Є3 3. If mDEG 145, ﬁnd mGEF. G 35 5 Making Conclusions From a Diagram Can you make each A DE F conclusion from the diagram? a. A C 4. Name an angle or angles in the diagram supplementary to A Yes; the markings show they are congruent. each of the following: E a. DOA B C D b. EOB

b. B and ACD are supplementary a. ЄAOB DO B b. ЄBOC or ЄDOE No; there are no markings. C

5. Can you make each conclusion from the information in the diagram? Explain. c. m( BCA) m( DCA) 180 a. 1 3 Yes; you can conclude that the angles are b. 4 and 5 are supplementary 5 1 supplementary from the diagram. c. m(1) m(5) 180 4 2 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. a. Yes; the markings show they are congruent. d. AB BC b. Yes; you can conclude the angles are adjacent and supplementary from the diagram. Yes; the markings show they are congruent. c. Yes; you can conclude that the angles are supplementary from the diagram and their sum is 180 by the Angle Addition Postulate.

22 23 Geometry Lesson 1-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-6

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Lesson 1-7 Basic Constructions 2 Constructing Congruent Angles Construct Y so that Y G. Step 1 Draw a ray with endpoint Y. Lesson Objectives NAEP 2005 Strand: Geometry 1 Use a compass and a straightedge to Topic: Relationships Among Geometric Figures construct congruent segments and congruent angles Y 75 2 Use a compass and a straightedge to G bisect segments and angles Local Standards: ______

Step 2 With the compass point on point G, draw an arc that intersects E Vocabulary. both sides of G. Label the points of intersection E and F. Construction is using a straightedge and a compass to draw a geometric ﬁgure.

A straightedge is a ruler with no markings on it. 75 G A compass is a geometric tool used to draw circles and parts of circles called arcs. F

Step 3 With the same compass setting, put the compass point on Perpendicular lines are two lines that intersect to form right angles. A D point Y. Draw an arc that intersects the ray. Label the point of intersection Z. C B A perpendicular bisector of a segment is a line, segment, or ray that is perpendicular to the segment at its midpoint, thereby bisecting the segment into two congruent segments. Y Z An angle bisector is a ray that divides an angle into two congruent J K coplanar angles. N L Step 4 Open the compass to the length EF. Keeping the same X Examples. compass setting, put the compass point on Z. Draw an arc Constructing Congruent Segments TW that intersects with the arc you drew in Step 3. Label the 1 Construct congruent KM to KM. point of intersection X.

Step 1 Draw a ray with endpoint T. T

Step 2 KM Open the compass the length of . Y Z

Step 3 With the same compass setting, put the compass point ) T Step 5 Draw YX to complete Y. X on point T. Draw an arc that intersects the ray. Label the W

point of intersection W. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. KM TW

Y G Y Z

24 25 Geometry Lesson 1-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-7

6 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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3 Constructing the Perpendicular Bisector Quick Check. Given: AB. A B * ) * ) XY RS Construct: XY so that XY AB at the midpoint M of AB. 1. Use a straightedge to draw . Then construct so that RS 2XY.

Step 1 Put the compass point on point A and draw a long arc. greater 1 Be sure that the opening is than 2AB. XYR S A B

2. a. Construct F with mF 2mB. Step 2 With the same compass setting, put the compass X point on point B and draw another long arc. Label the points where the two arcs intersect as X and Y. A B Y B BF * ) * ) Step 3 Draw XY . The point of intersection of AB and XY is M, ) X the midpoint of AB. b. Explain how you can use your protractor to check that YP is the angle bisector of XYZ. * ) * ) A B XY AB at the midpoint of AB , so XY is the M Measure ЄXYP and ЄPYZ to see that they Y are congruent. perpendicular bisector of AB. XP

* ) 4 Finding Angle Measures WR bisects AWB so that mAWR x A Y Z and m BWR 4x 48. Find m AWB. R x ST 4x 48 3. Draw . Construct its perpendicular bisector. W B

m AWR m BWR Definition of angle bisector

Substitutex for mЄAWR and4x 48 for x 4x 48 ST mЄBWR.

3x 48 Subtract 4x from each side. x 16 Divide each side by 3 . ) mAWR 16 4. If mJKN 50°, then find mNKL and mJKL.KN bisects JKL. Substitute 16 for x. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. mBWR 4()16 48 16 J K mNKL 50° , mJKL 100° mAWB m AWR m BWR Angle Addition Postulate L mAWB 16 16 32 Substitute 16 for mЄAWR and for mЄBWR. N

26 27 Geometry Lesson 1-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-7

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2 Finding the Midpoint AB has endpoints (8, 9) and (6, 3). Find the Lesson 1-8 The Coordinate Plane coordinates of its midpoint M.

Lesson Objectives NAEP 2005 Strand: Measurement Use the Midpoint Formula. Let (x1, y1) be (8, 9) and (x2, y2) be 1 Find the distance between two points Topic: Measuring Physical Attributes (6, 3) . in the coordinate plane 2 Find the coordinates of the midpoint Local Standards: ______The midpoint has coordinates of a segment in the coordinate plane x1 x2 y1 y2 , . Midpoint Formula 22 Key Concepts. ( ) 8 62 () Formula: The Distance Formula The x-coordinate is 1 Substitute 8 for x1 and 6 for x2. Simplify. 22 The distance d between two points A(x1, y1) and B(x2, y2) is 9 ()36 d (x x )2 (y y )2 The y-coordinate is 3 Substitute 9 for y and 3 for y . Simplify. 2 1 2 1 . 22 1 2

Formula: The Midpoint Formula The coordinates of the midpoint M are (1, 3) . AB The coordinates of the midpoint M of with endpoints A(x1, y1) and B(x , y ) are the following: 2 2 3 Finding an Endpoint The midpoint of DG is M(1, 5). One endpoint is D(1, 4). Find the coordinates of the other endpoint G. x1 x2 y1 y2 M(), . 2 2 Use the Midpoint Formula. Let (x1, y1) be (1, 4) and the midpoint

x1 1 x2 y1 1 y2 a 2 , 2 b be (1, 5) . Solve for x2 and y2 ,the coordinates of G. Examples. Find the x-coordinate of G. 1 Applying the Distance Formula How far is the subway y 6 ride from Oak to Symphony? Round to the nearest tenth. Each unit North represents 1 mile. 4 Jackson 1 x2 4 y2 1 d Use the Midpoint Formula. S 5 Central (2, 4) 2 2 Oak has coordinates (1, 2) . Let (x1, y1) represent Oak. 2 Cedar Symphony d S 2 1 x2 Multiply each side by2 . 10 4 y2 Symphony has coordinates (1, 2) . Let (x2, y2) represent 4 2 City Plaza x 3 x d Simplify. S 6 x Symphony. Oak (0, 0) 2 2 (1, 2) 4 South (3, 6) "(x 2 x )2 1 (y 2 y )2 Elm The coordinates of G are . d 2 1 2 1 Use the Distance Formula. 6 d %( 1 ())1 2 ( 2 ())2 2 Substitute. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. d % 242 2 Simplify within parentheses.

20 4.472135955 Use a calculator.

To the nearest tenth, the subway ride from Oak to Symphony is 4.5 miles.

28 29 Geometry Lesson 1-8 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-8

7 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

Name______Class______Date ______Name______Class______Date______

Quick Check. Lesson 1-9 Perimeter, Circumference, and Area 1. a. Using the map in Example 1, find the distance between Elm and Symphony. Lesson Objectives about 8.9 miles NAEP 2005 Strand: Measurement 1 Find perimeters of rectangles and Topic: Measuring Physical Attributes squares, and circumferences of circles 2 Find areas of rectangles, squares, and Local Standards: ______circles

Key Concepts.

Perimeter and Area b. Maple is located 6 miles west and 2 miles north of City Plaza. Find the b distance between Cedar and Maple. s r hh d O about 3.2 miles s b C Square with side length s. Rectangle with base b and Circle with radius r and height h. diameter d. Perimeter P 4s Perimeter P 2b 2h Circumference C pd 2 Area A s p Area A bh or C 2 r Area pr2 2. Find the coordinates of the midpoint of XY with endpoints X(2, 5) and Y(6, 13).

(4, 4) Postulate 1-9 If two ﬁgures are congruent, then their areas are equal .

Postulate 1-10 The area of a region is the sum of the areas of its non-overlapping parts.

3. The midpoint of XY has coordinates (4, 6). X has coordinates (2, 3). Find the coordinates of Y. Examples.

(6, 9) 1 Finding Circumference G has a radius of 6.5 cm. Find the circumference of G in terms of π. Then ﬁnd the circumference to the nearest tenth. 6.5 cm G C 2 p r Formula for circumference of a circle © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. C 2π()6.5Substitute 6.5 for r. C 13 π Exact answer

C 13 40.840704 Use a calculator.

The circumference of G is 13p , or about 40.8 cm.

30 31 Geometry Lesson 1-8 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-9

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Finding Area of an Irregular Shape Find the area of the figure 2 Finding Perimeter in the Coordinate Plane y B C 4 15 ft Quadrilateral ABCD has vertices A(0, 0), B(9, 12), 12 to the right.

C(11, 12), and D(2, 0). Find the perimeter. 10 ft 10 Draw a horizontal line to separate the figure into three 5 ft 10 ft Draw and label ABCD on a coordinate plane. non-overlapping figures: one rectangle and two squares . 5 ft 8 5 ft Find each area. Then add the areas. 5 ft 5 ft 6 d Use the formulas. S 2 AR bh AS s 4 2 ()()15 5 d Substitute. S ()5 2 75 d Simplify. S 25 A D x O 2 4 6810 12 A 75 25 25 Add the areas. Find the length of each side. Add the lengths to find the perimeter. A 125 Simplify. 2 2 www2wwww w2 2 AB %( 9 0 ) ( 12 0 ) % 9 12 Use the Distance Formula. The area of the figure is125 ft . wwwww AB % 81 144% 225 15

BC u 11 9u uu 2 2 Ruler Postulate Quick Check. π 2 2 1. a. Find the circumference of a circle with a radius of 18 m in terms of . CD %( 211) ( 0 12 ) Use the Distance Formula. 36p m %( 9 12 )2 wwwww % 81 144% 225 15 b. Find the circumference of a circle with a diameter of 18 m to the nearest tenth.

DA u 20 u uu 2 2 Ruler Postulate 56.5m

Perimeter AB BC CD DA 2. Graph quadrilateral KLMN with vertices K(3, 3), L(1, 3), y M(1, 4) 4 Perimeter 152152 M(1, 4), and N(3, 1). Find the perimeter of KLMN. 2 Perimeter 34 20 units N(3, 1) O x The perimeter of quadrilateral ABCD is 34 units. 4 2 2 4 3. You are designing a rectangular banner for the front of 2 a museum. The banner will be 4 ft wide and 7 yd high. 3 Finding Area of a Circle Find the area of B in terms of π. How much material do you need in square yards? K( 3, 3)4 L(1, 3) In B, r 1.5 yd. 1 2 B Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 9yd3 All rights reserved. A π r2 Formula for the area of a circle © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. π 2 1.5 yd A ()1.5 Substitute 1.5 for r. 4. Copy the ﬁgure in Example 4. Separate it in a different way. Find the area. A 2.25 π 125 ft2 The area of B is 2.25p yd2.

32 33 Geometry Lesson 1-9 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-9

8 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

Name______Class______Date______Name______Class______Date ______

Lesson 2-1 Conditional Statements 2 Finding a Counterexample Find a counterexample to show that this conditional is false: If x2 0, then x 0. Lesson Objectives NAEP 2005 Strand: Geometry true 1 Recognize conditional statements Topic: Mathematical Reasoning A counterexample is a case in which the hypothesis is and 2 Write converses of conditional the conclusion is false . This counterexample must be an example statements Local Standards: ______in which x2 0 (hypothesis true) and x 0 or x 0 (conclusion false). Because any negative number has a positive square, one possible Vocabulary and Key Concepts. counterexample is 1. 2 Conditional Statements and Converses Because ( 1) 1, which is greater than 0, the hypothesis is true . Statement Example Symbolic Form You read it Because 1 0, the conclusion is false . Conditional If an angle is a straight angle, p S q If p , The counterexample shows that the conditional is false . then its measure is 180. then q . Converse If the measure of an angle is q S p If q , 3 Writing the Converse of a Conditional Write the converse of the following conditional. 180, then it is a straight angle. then p . If x 9, then x 3 12.

A conditional is an if-then statement. The converse of a conditional exchanges the hypothesis and the conclusion. Conditional Converse The hypothesis is the part that follows if in an if-then statement. Hypothesis Conclusion Hypothesis Conclusion x 9 x 3 12 x 3 12 x 9 The conclusion is the part of an if-then statement (conditional) that follows then. So the converse is: If x 3 12, then x 9. The truth value of a statement is “true” or “false” according to whether the statement is true or false, respectively. The converse of the conditional “if p, then q” is the conditional “if q, then p.” Quick Check. 1. Identify the hypothesis and the conclusion of this conditional statement: If y 3 5, then y 8.

Examples. Hypothesis: y 3 5

1 Identifying the Hypothesis and the Conclusion Identify the hypothesis Conclusion: y 8 and conclusion: If two lines are parallel, then the lines are coplanar. In a conditional statement, the clause after if is the hypothesis and the 2. Show that this conditional is false by ﬁnding a counterexample: clause after then is the conclusion. If the name of a state includes the word “New,” then the state borders an ocean. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Counterexample: Hypothesis: Two lines are parallel. New Mexico does not border an ocean. It is a counterexample, so the conditional is false. Conclusion: The lines are coplanar.

34 35 Geometry Lesson 2-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-1

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Example. Lesson 2-2 Biconditionals and Deﬁnitions

4 Finding the Truth Value of a Converse Write the converse of the Lesson Objectives NAEP 2005 Strand: Geometry conditional. Then determine the truth value of each. 1 Write biconditionals Topics: Dimension and Shape; Mathematical Reasoning 2 2 Recognize good definitions If a 25, then a 5. Local Standards: ______

Conditional: If a2 25, then a 5. Vocabulary and Key Concepts. The converse exchanges the hypothesis and the conclusion . Converse: If a 5, then a2 25. Biconditional Statements A biconditional combines p S q and q S p as p 4 q. The conditional is false . A counterexample is a 5: Statement Example Symbolic Form You read it (5)2 25, and 5 5. Biconditional An angle is a straight angle if p 4 q p if and Because 52 25, the converse is true . and only if its measure is 180°. only if q .

A biconditional statement is the combination of a conditional statement and its converse. Quick Check. A biconditional contains the words “if and only if.” 3. Write the converse of the following conditional: If two lines are not parallel and do not intersect, then they are skew. Examples. If two lines are skew, then they are not parallel and do not intersect. 1 Writing a Biconditional Consider the true conditional statement. Write its converse. If the converse is also true, combine the statements as a biconditional. Conditional: If x 5, then x 15 20. To write the converse, exchange the hypothesis and conclusion. Converse: If x 15 20, then x 5. 4. Write the converse of each conditional. Determine the truth value of the When you subtract 15 from each side to solve the equation, you get x 5. Because conditional and its converse. (Hint: One of these conditionals is not true.) both the conditional and its converse are true , you can combine them in a. If two lines do not intersect, then they are parallel. Converse: a biconditional using the phrase if and only if . If two lines are parallel, then they do not intersect. Biconditional: x 5 if and only if x 15 20.

2 Separating a Biconditional into Parts Write the two statements that form this biconditional. The conditional is false and the converse is true . Biconditional: Lines are skew if and only if they are noncoplanar. b. If x 2, then u x u 2.

Converse: Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson Write a biconditional All rights reserved. as two conditionals that are converses of each other.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. If lines are skew, then they are noncoplanar. If u x u 2, then x 2. Conditional:

Converse: If lines are noncoplanar, then they are skew. The conditional is true and the converse is false .

36 37 Geometry Lesson 2-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-2

9 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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3 Writing a Definition as a Biconditional Show that this definition of 2. Write two statements that form this biconditional about integers greater than 1: triangle is reversible. Then write it as a true biconditional. A number is prime if and only if it has only two distinct factors, 1 and itself. Definition: A triangle is a polygon with exactly three sides. Statement:

The original conditional is true . If a number is prime, then it has only two distinct factors, 1 and itself. Conditional: If a polygon is a triangle, then it has exactly three sides. The converse is also true . Converse: If a polygon has exactly three sides, then it is a triangle. Statement: Because both statements are true , they can be combined to form a biconditional. If a number has only two distinct factors, 1 and itself, then it is prime. Biconditional: A polygon is a triangle if and only if it has exactly three sides.

4 Identifying a Good Definition Is the following statement a good 3. Show that this definition of right angle is reversible. Then write it as a true definition? Explain. biconditional. An apple is a fruit that contains seeds. Definition: A right angle is an angle whose measure is 90 . The statement is true as a description of an apple. Conditional: Exchange “An apple” and “a fruit that contains seeds.” The converse reads: If an angle is a right angle, then its measure is 90°. A fruit that contains seeds is an apple.

There are many fruits that contain seeds but are not apples, such as lemons and peaches. These are counterexamples , so the converse of the Converse:

statement is false . If an angle has measure 90°, then it is a right angle. The original statement is not a good definition because the statement is not reversible. The two statements are true .

Quick Check. Biconditional: 1. Consider the true conditional statement. Write its converse. If the converse is also true, combine the statements as a biconditional. An angle is a right angle if and only if its measure is 90°. Conditional: If three points are collinear, then they lie on the same line. Converse: If three points lie on the same line, then they are collinear.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 4. Is the following statement All rights reserved. a good definition? Explain. A square is a figure with four right angles. The converse is true . It is not a good deﬁnition because a rectangle has four right angles Biconditional: and is not necessarily a square. Three points are collinear if and only if they lie on the same line.

38 39 Geometry Lesson 2-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-2

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Lesson 2-3 Deductive Reasoning 3 Using the Law of Syllogism Use the Law of Syllogism to draw a conclusion from the following true statements: Lesson Objectives NAEP 2005 Strand: Geometry If a quadrilateral is a square, then it contains four right angles. 1 Use the Law of Detachment Topic: Mathematical Reasoning If a quadrilateral contains four right angles, then it is a rectangle. 2 Use the Law of Syllogism Local Standards: ______The conclusion of the first conditional is the hypothesis of the second conditional. This means that you can apply the Law of Syllogism . Vocabulary and Key Concepts. The Law of Syllogism: If p S q and q S r are true statements, S Law of Detachment then p r is a true statement. If a conditional is true and its hypothesis is true, then its conclusion is true. So you can conclude: In symbolic form: If a quadrilateral is a square, then it is a rectangle. If p S q is a true statement and p is true, then q is true. 4 Drawing Conclusions Use the Laws of Detachment and Syllogism to Law of Syllogism draw a possible conclusion. If the circus is in town, then there are tents at the fairground. If there are If p S q and q S r are true statements, then p S r is a true statement. tents at the fairground, then Paul is working as a night watchman. The circus is in town. a process of reasoning logically from given facts to a conclusion. Deductive reasoning is Because the conclusion of the first statement is the hypothesis of the second statement, you can apply the Law of Syllogism to write a new conditional: Examples. If the circus is in town, then Paul______is working as a night watchman. 1 Using the Law of Detachment A gardener knows that if it rains, the garden will be watered. It is raining. What conclusion can he make? The third statement means that the hypothesis of the new conditional is true. The first sentence contains a conditional statement. The hypothesis is You can use the Law of Detachment to form the conclusion: it rains. Paul is working as a night watchman.

Because the hypothesis is true, the gardener can conclude that the garden will be watered. Quick Check. 1. Suppose that a mechanic begins work on a car and finds that the car will not 2 Using the Law of Detachment For the given statements, what can you start. Can the mechanic conclude that the car has a dead battery? Explain. conclude? No, there could be other things wrong with the car, such as a faulty starter. Given: If A is acute, then mA 90°. A is acute.

A conditional and its hypothesis are both given as true. 2. If a baseball player is a pitcher, then that player should not pitch a complete © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. By the Law of Detachment , you can conclude that the game two days in a row.Vladimir Nuñez is a pitcher. On Monday, he pitches a complete game. What can you conclude? Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. conclusion of the conditional, mA 90°, is true . Answers may vary. Sample: Vladimir Nuñez should not pitch a complete game on Tuesday.

40 41 Geometry Lesson 2-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-3

10 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

Name______Class______Date ______Name______Class______Date ______

3. If possible, state a conclusion using the Law of Syllogism. If it is not possible Lesson 2-4 Reasoning in Algebra to use this law, explain why. a. If a number ends in 0, then it is divisible by 10. Lesson Objective NAEP 2005 Strand: Algebra and Geometry If a number is divisible by 10, then it is divisible by 5. 1 Connect reasoning in algebra and Topics: Algebraic Representations; Mathematical geometry Reasoning If a number ends in 0, then it is divisible by 5. Local Standards: ______

Key Concepts.

b. If a number ends in 6, then it is divisible by 2. Properties of Equality If a number ends in 4, then it is divisible by 2. Addition Property If a b, then a c b c . Not possible; the conclusion of one statement is not the hypothesis Subtraction Property If a b, then a c b c . of the other statement. Multiplication Property If a b, then a ccb . a b Division Property If a b and c 0, then . cc

a 4. Use the Law of Detachment and the Law of Syllogism to draw conclusions. Reﬂexive Property a . The Volga River is in Europe. Symmetric Property If a b, then b a . If a river is less than 2300 miles long, it is not one of the world’s ten longest Transitive Property If a b and b c, then a c . rivers. If a river is in Europe, then it is less than 2300 miles long. Substitution Property If a b, then b can replace a in any expression. Conclusion: Distributive Property a(b c) ab ac . The Volga River is less than 2300 miles long.

Properties of Congruence Reﬂexive Property AB AB A ЄA

Conclusion: Symmetric Property If AB CD, then CD AB . Є The Volga River is not one of the world’s ten longest rivers. If A B, then B A .

42 43 Geometry Lesson 2-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-4

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Examples. Example. 1 Justifying Steps in Solving an Equation Justify each step used to solve 3 Using Properties of Equality and Congruence Name the property that 5x 12 32 x for x. justifies each statement. Given: 5x 12 32 x a. If x y and y 4 3x, then x 4 3x. The conclusion of the conditional statement is the same as the equation 5x 44 x Addition Property of Equality y 4 3x (given) after x has been substituted for y (given). The 4x 44 Subtraction Property of Equality property used is the Substitution Property of Equality. x 11 Division Property of Equality b. If x 4 3x, then 4 2x.

2 Justifying Steps in Solving an Equation Suppose that 15 x 4 2x The conclusion of the conditional statement shows the result after x is points A, B, and C are collinear with point B between A BC subtracted from each side of the equation in the hypothesis. points A and C. Solve for x if AC 21, AB 15 x, and The property used is the Subtraction Property of Equality. BC 4 2x. Justify each step. c. If P Q, Q R, and R S, then P S. AB BC AC Segment Addition Postulate Use the Transitive Property of Congruence for (15 x) (4 2x) 21 Substitution Property of Equality the first two parts of the hypothesis: 19 x 21 Simplify. Є ഡ Є x 2 Subtraction Property of Equality If P Q and Q R, then P R .

Use the Transitive Property of Congruence for P R and the third part of the hypothesis: Quick Check. If P R and R S, then ЄP ഡ ЄS . 1. Fill in each missing reason. ) M The property used is the Transitive Property of Congruence. Given: LM bisects KLN. (2x40) 4x ) K Quick Check. LM bisects KLN Given L N 3. Name the property of equality or congruence illustrated. mMLN mKLM Deﬁnition of Angle Bisector a. XY XY 4x 2x 40 Substitution Property of Equality Reﬂexive Property of Congruence 2x 40 Subtraction Property of Equality x 20 Division Property of Equality

2. Recall that the length of line AC is 21. 15 x 4 2x b. If mA 45 and 45 mB, then mA mB © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A BC Transitive or Substitution Property of Equality © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Find the lengths of AB and BC by substituting x 2 into the expressions in the diagram. Check that AB BC 21.

AB 13 BC 8 AB BC 13 8 21

44 45 Geometry Lesson 2-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-4

11 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Lesson 2-5 Proving Angles Congruent complementary angles supplementary angles

Lesson Objectives NAEP 2005 Strand: Geometry 1 Prove and apply theorems about Topic: Relationships Among Geometric Figures angles 2 Local Standards: ______1 3 4

Є1 andЄ2 are complementary angles. Є3 and Є4 are supplementary angles. Vocabulary and Key Concepts. Two angles are complementary angles if Two angles are supplementary angles if Theorem 2-1: Vertical Angles Theorem the sum of their measures is 90°. the sum of their measures is 180°. 1 Vertical angles are congruent . 3 4 2 1 2 and 3 4

Theorem 2-2: Congruent Supplements Theorem If two angles are supplements of the same angle (or of congruent angles), A theorem is a conjecture that is proven. then the two angles are congruent .

Theorem 2-3: Congruent Complements Theorem A paragraph proof is a convincing argument that uses deductive reasoning in which If two angles are complements of the same angle (or of congruent angles), statements and reasons are connected in sentences. then the two angles are congruent .

Theorem 2-4 Examples. All right angles are congruent. 1 Using the Vertical Angles Theorem Find the value of x. Theorem 2-5 The angles with labeled measures are vertical angles. Apply the If two angles are congruent and supplementary, then each is a right angle. Vertical Angles Theorem to find x. (2x 3) 4x 101 2x 3 Vertical Angles Theorem (4x 101) 4x 2x 104 Addition Property of Equality vertical angles adjacent angles 2x 104 Subtraction Property of Equality 1 x 52 Division Property of Equality 3 4 3 2 1 2 4 2 Proving Theorem 2-2 Write a paragraph proof of Theorem 2-2 using the diagram at the right. 1 Є1 andЄ2 are vertical angles, Є1 and Є2 are adjacent angles, 2 Start with the given: 1 and 2 are supplementary, 3 and 2 are as are Є3 and Є4 . as are Є3 and Є4 . 3 supplementary. By the definition of supplementary angles , © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Vertical angles are two angles whose Adjacent angles are two coplanar angles m1 m2 180 and m3 m2 180. By substitution, sides form two pairs of opposite rays. that have a common side and a common m1 m2 mЄ2 mЄ3 . Using the vertex but no common interior points. Subtraction Property of Equality , subtract mЄ2 from each side. You get m1 mЄ3 , or Є1 Є3 .

46 47 Geometry Lesson 2-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-5

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Lesson 3-1 Properties of Parallel Lines Quick Check. 1. Refer to the diagram for Example 1. Lesson Objectives NAEP 2005 Strand: Measurement 1 a. Find the measures of the labeled pair of vertical angles. Identify angles formed by two lines Topic: Measuring Physical Attributes and a transversal 107° 2 Prove and use properties of parallel Local Standards: ______lines

b. Find the measures of the other pair of vertical angles. Vocabulary and Key Concepts. 73°

Postulate 3-1: Corresponding Angles Postulate t / If a transversal intersects two parallel lines, then corresponding 1 c. Check to see that adjacent angles are supplementary. congruent 2 angles are . m 107° 73° 180° Є1 Є2

2. Recall the proof of Theorem 2-2. Does the size of the angles in the diagram Theorem 3-1: Alternate Interior Angles Theorem a t affect the proof? Would the proof change if 1 and 3 were acute rather If a transversal intersects two parallel lines, then alternate interior 4 3 2 than obtuse? Explain. angles are congruent . b 1 5 6 No, the size of the angles does not affect the proof or the truth of Є1 Є3 the theorem. Theorem 3-2: Same-Side Interior Angles Theorem If a transversal intersects two parallel lines, then same-side interior angles are supplementary . m1 m2 180

Theorem 3-3: Alternate Exterior Angles Theorem If a transversal intersects two parallel lines, then alternate exterior angles are congruent . Є4 Є5

Theorem 3-4: Same-Side Exterior Angles Theorem If a transversal intersects two parallel lines, then same-side exterior angles are supplementary .

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. m4 m6 180 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

48 49 Geometry Lesson 2-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-1

12 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

Name______Class______Date ______Name______Class______Date ______

a line that intersects two coplanar lines at two p A transversal is / t 2 Finding Measures of Angles distinct points. 5 6 In the diagram at right, ᐉm and pq. Find m1 and m2. q 1 3 86 42 m 4 1 and the 42 angle are corresponding angles . / 2 7 7 8 ᐉ 42 Because m, m 1 by the 4 2 1 Alternate interior angles are nonadjacent interior angles that lie m Corresponding Angles Postulate . 5 3 on opposite sides of the transversal. Because 1 and 2 are adjacent angles that form a Є1 and Є2 are 180 Angle Addition Postulate Same-side interior angles are interior angles that lie on the same side alternate interior angles. straight angle, m 1 m 2 by the . Є of the transversal. If you substitute 42 for m 1, the equation becomes 42 m 2 180 . Є1 and Є4 are Subtract 42 from each side to find m2 138 . same-side interior angles. Corresponding angles are angles that lie on the same side of the Є1 and Є7 3 Using Algebra to Find Angle Measures / transversal and in corresponding positions relative to the coplanar a c In the diagram, ᐉm. Find the values of a, b, and c. b lines. corresponding angles. A two-column proof is a display that shows the steps to prove a theorem. The first a 65 Alternate Interior Angles Theorem 65 40 m column shows the steps and the second column shows the reason for each step. c 40 Alternate Interior Angles Theorem a b c 180 Angle Addition Postulate Examples. 65 b 40 180 Substitution Property of Equality 1 Applying Properties of Parallel Lines In the diagram of b 75 Subtraction Property of Equality Lafayette Regional Airport, the black segments are runways and the gray areas are taxiways and terminal buildings. Quick Check. Compare 2 and the angle vertical to 1. Classify the angles 2 1. Use the diagram in Example 1. Classify 2 and 3 as alternate interior angles, as alternate interior angles, same-side interior angles, or 3 corresponding angles. 1 same-side interior angles, or corresponding angles. same-side interior angles The angle vertical to 1 is between the runway segments. 2 is between the runway segments and on the opposite side of the 2. Using the diagram in Example 2 find the measure of each angle. Justify each answer. transversal runway. Because alternate interior angles are not a. 3 42; Vertical angles are congruent

adjacent and lie between the lines on opposite sides of the transversal, b. 4 138; Corresponding angles are congruent 2 and the angle vertical to 1 arealternate interior angles . c. 5 138; Same-side interior angles are supplementary d. 6 42; Corresponding angles are congruent

f. 8 138; Vertical angles are congruent

3. Find the values of x and y. Then ﬁnd the measures of the four angles 2x° y° in the trapezoid. x 45, y 115; 90, 90, 115, 65 (y 50)°

50 51 Geometry Lesson 3-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-1

Name______Class______Date______Name______Class______Date ______

Lesson 3-2 Proving Lines Parallel A ﬂow proof uses arrows to show the logical connections between the statements.

Lesson Objectives NAEP 2005 Strand: Measurement 1 Use a transversal in proving lines Topic: Measuring Physical Attributes Reasons are written below the statements. parallel Local Standards: ______

Vocabulary and Key Concepts. Examples.

/ 1 Using Postulate 3-2 Postulate 3-2: Converse of the Corresponding Angles Postulate 1 Use the diagram at the right.Which lines, if any, must be parallel If two lines and a transversal form corresponding angles that are congruent, m 2 if 3 and 2 are supplementary? Justify your answer with 3 then the two lines are parallel. a theorem or postulate. E C ᐉ m . It is given that 3 and 2 are supplementary.The diagram D 1 4 K 2 shows that Є4 and Є2 are supplementary. Because congruent Theorem 3-5: Converse of the Alternate Interior Angles Theorem / 56 supplements of the same angle are 1 4 (Congruent Supplements Theorem), Є3 Є4 . Because Є3 If two lines and a transversal form alternate interior angles that are m 2 ) 3 ) congruent, then the two lines are parallel. and Є4 are congruent corresponding angles, EC DK by the If 1 2, Converse of the Corresponding Angles Postulate. then ᐉ m .

2 Using Algebra Theorem 3-6: Converse of the Same-Side Interior Angles Theorem If 2 and 4 are Find the value of x for which ᐉm. If two lines and a transversal form same-side interior angles that are supplementary, The labeled angles are alternate interior angles . ᐉ / supplementary, then the two lines are parallel. then m . (14 3x) If ᐉm, the alternate interior angles are congruent , (5x 66) m Theorem 3-7: Converse of the Alternate Exterior Angles Theorem If 3 5, and their measures are equal . Write and solve the If two lines and a transversal form alternate exterior angles that are then ᐉ m . equation 5x 66 14 3x. congruent, then the two lines are parallel. 5x 66 14 3x 5x 80 3x Add 66 to each side. Theorem 3-8: Converse of the Same-Side Exterior Angles Theorem If 3 and 6 are 2x 80 Subtract 3x from each side. If two lines and a transversal form same-side exterior angles that are supplementary, 40 2 supplementary, then the two lines are parallel. then ᐉ m . x Divide each side by . © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

52 53 Geometry Lesson 3-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-2

13 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Quick Check. Lesson 3-3 Parallel and Perpendicular Lines 1. Use the diagram from Example 1. Which lines, if any, must be parallel if Lesson Objectives NAEP 2005 Strand: Geometry 3 4? Explain. 1 Relate parallel and perpendicular Topic: Relationships Among Geometric Figures u u ʈ lines EC DK; Converse of Corresponding Angles Postulate Local Standards: ______

Key Concepts. 2. Find the value of x for which ab. Explain how you can check your answer. a x 18; 7(18) 8 118°, and 62° 118° 180° (7x 8) b 62 Theorem 3-9 a If two lines are parallel to the same line , then b they are parallel to each other. c a b .

t Theorem 3-10 m In a plane, if two lines are perpendicular to the same line , n then they are parallel to each other. m n .

Theorem 3-11 In a plane, if a line is perpendicular to one of two n ᐉ parallel lines, then it is also perpendicular to the other. m n ⊥ m

Examples. 1 Real-World Connection A picture frame is assembled as shown. Given the book’s corners cut explanation for why the outer edges on opposite sides of to form 45 the frame are parallel, why must the inner edges on opposite angles sides be parallel, too? © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

54 55 Geometry Lesson 3-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-3

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The sides are constructed so that the inner and outer edges of each Parallel Lines and the Triangle Angle-Sum Theorem side are parallel. And if an inner edge on one side is parallel to the Lesson 3-4 outer edge on the same side, and at the same time the outer Lesson Objectives NAEP 2005 Strand: Geometry edge is parallel to the outer edge on the opposite side, then by 1 Classify triangles and ﬁnd the Topic: Relationships Among Geometric Figures measures of their angles Theorem 3-9 outer each inner edge is parallel to the edge on the 2 Use exterior angles of triangles Local Standards: ______opposite side. But that outer edge is parallel to the inner edge on the same side, so again by Theorem 3-9 , inner edges on opposite sides are Vocabulary and Key Concepts. parallel.

Theorem 3-12: Triangle Angle-Sum Theorem 2 Using Theorem 3-11 C 1 The sum of the measures of the angles of a triangle is 180 . Write a paragraph proof. k ᐉ mA mB mC 180 A B Given: In a plane, k l and k m. Also, m1 90°. m Prove: The transversal is perpendicular to line m. Theorem 3-13: Triangle Exterior Angle Theorem The measure of each exterior angle of a triangle equals the sum of the 2 Since mЄ1 90° , the transversal is perpendicular to line k . 1 3 measures of its two remote interior angles. Since k ʈ m , by Theorem 3-11 the transversal is also perpendicular to line m.

Quick Check.

1. Can you assemble the framing at the right into a frame 60 60 with opposite sides parallel? Explain. 30 An acute A right An obtuse An equiangular 30 triangle has triangle has triangle has triangle has 60 60 Yes; 30° 60° 90° ______three acute angles. ______one right angle. ______one obtuse angle. ______three congruent angles. ______

2. From what is given in Example 2, can you also conclude that the transversal is perpendicular to line l? An equilateral An isosceles A scalene Yes, by Theorem 3-11 and the fact that k ʈ l triangle has triangle has triangle has ______three congruent sides. ______at least two congruent sides. ______no congruent sides. ______

An exterior angle of a polygon is an angle formed by a side © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. and an extension of an adjacent side. 1 Exterior

angle Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 3 2 Remote interior angles are the two nonadjacent interior angles corresponding to each exterior angle of a triangle. Remoteinterior angles

56 57 Geometry Lesson 3-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-4

14 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Examples. Quick Check. 1 Applying the Triangle Angle-Sum Theorem C 1. Critical Thinking Describe how you could use either CAB or CDB to In triangle ABC, ACB is a right find the value of b in Example 1. 70 c angle, and CD AB. Find the values of a and c. For ᭝CAB, (70 20) 70 b 180. Then 160 b 180 and b 20. For ᭝CDB, 70 90 b 180. Then 160 b 180 and b 20. Find c first, using the fact that ACB is a right angle. a b A B mACB 90 Definition of a right angle D c 70 90 Angle Addition Postulate c 20Subtract 70 from each side. 2. a. Find m3. To find a, use ADC. 90 45 a mADC c 180 Triangle Angle-Sum Theorem 45 3 mADC 90 Definition of perpendicular lines b. Critical Thinking Is it true that if two acute angles of a triangle are complementary, then the triangle must be a right triangle? Explain. a 90 20 180 Substitute 90 for mЄADC and 20 for c. a 110 180 Simplify. True, because the measures of the two complementary angles add to 90, leaving 90 for the third angle. a 70 Subtract 110 from each side.

2 Applying the Triangle Exterior Angle Theorem Explain what happens to the angle formed by the back of the chair and the armrest as you make a lounge chair recline more.

Back x x Arm (180180 x) (180180 x)

The exterior angle and the angle formed by the back of the chair and the armrest are adjacent angles , which together form a straight angle. As one measure increases , the other measure decreases . The angle formed by the back of the chair and the armrest increases as you make a lounge chair recline more. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

58 59 Geometry Lesson 3-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-4

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Lesson 3-5 The Polygon Angle-Sum Theorems An equilateral polygon has all sides congruent. An equiangular polygon has all angles congruent. Lesson Objectives NAEP 2005 Strand: Geometry 1 Classify polygons Topic: Relationships Among Geometric Figures A regular polygon is both equilateral and equiangular. 2 Find the sums of the measures of the interior and exterior angles of Local Standards: ______polygons Examples. 1 Classifying Polygons Classify the polygon at the right by its sides. Identify it Vocabulary and Key Concepts. as convex or concave. Starting with any side, count the number of sides clockwise around the ﬁgure. Theorem 3-14: Polygon Angle-Sum Theorem Because the polygon has 12 sides, it is a dodecagon. Think of the The sum of the measures of the angles of an n-gon is (n 2)180 . polygon as a star. Draw a diagonal connecting two adjacent points of the star.

Theorem 3-15: Polygon Exterior Angle-Sum Theorem That diagonal lies outside the polygon, so the dodecagon is concave . The sum of the measures of the exterior angles of a polygon, 3 one at each vertex, is 360 . 2 4 1 2 Finding a Polygon Angle Sum Find the sum of the measures of the angles For the pentagon, m 1 m 2 m 3 m 4 m 5 360 . 5 of a decagon. A decagon has 10 sides, so n 10 . A polygon is a closed plane ﬁgure with at least three sides that are segments. The sides Sum (n 2)(180) Polygon Angle-Sum Theorem intersect only at their endpoints, and no two adjacent sides are collinear. ( 10 2)(180) Substitute 10 for n. B B B 8 ? 180 Subtract. A A A C C C 1440 Simplify.

E D E D DE 3 Using the Polygon Angle-Sum Theorem Find mX in quadrilateral Y A polygon Not a polygon ; Not a polygon ; Z XYZW. not a closed ﬁgure two sides intersect The ﬁgure has 4 sides, so n 4 . between endpoints. 100° X W A convex polygon does not have diagonal points outside A D of the polygon. R Y A convex m X m Y m Z m W (4 2)(180) Polygon Angle-Sum Theorem T polygon m X m Y 90 100 360 Substitute.

Diagonal Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A concave polygon has at least one diagonal m X m Y 190 360 Simplify. M P

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 170 190 with points outside of the polygon. K m X m Y Subtract from each side. W G Q A concave mX mЄX 170 Substitute mЄX for mЄY. polygon Є S 2m X 170 Simplify. mX 85 Divide each side by 2 .

60 61 Geometry Lesson 3-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-5

15 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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4 Applying Theorem 3-15 A regular hexagon is inscribed in a rectangle. 2 2 Lesson 3-6 Lines in the Coordinate Plane Explain how you know that all the angles labeled 1 have equal measures. 11 Lesson Objectives NAEP 2005 Strand: Algebra congruent The hexagon is regular, so all its angles are . An 1 Graph lines given their equations Topics: Patterns, Relations, and Functions; exterior angle is the supplement of a polygon’s angle because they 2 Write equations of lines Algebraic Representations are adjacent angles that form a straight angle. Because supplements Local Standards: ______1 1 of congruent angles are congruent , all the angles marked 1 2 2 have equal measures. Vocabulary. Slope-intercept form The slope-intercept form of a linear equation is y Quick Check. y mx b. y 2x 3 1. Classify each polygon by its sides. Identify each as convex or concave. The standard form of a linear equation is a. b. Ax By C. 1 Standard 4 O 4 x The point-slope form for a nonvertical line is form 2x y 1 y y m(x x ). 1 1 3 y 2 2(x 1)

hexagon; convex octagon; concave Point-slope form

2. a. Find the sum of the measures of the angles of a 13-gon. Examples. 1980 1 Graphing Lines Using Intercepts Use the x-intercept and y-intercept to graph 5x 6y 30. To find the x-intercept, substitute To find the y-intercept, substitute b. Critical Thinking The sum of the measures of the angles of a given 0 for y and solve for x. 0 for x and solve for y. polygon is 720. How can you use Sum (n 2)180 to find the number 5x 6y 30 5x 6y 30 of sides in the polygon? 5x 6()0 30 5()0 6y 30 You can solve the equation (n 2)180 720. 5x 0 30 0 6y 30 5x 30 6y 30 x 6 y 5 3. Pentagon ABCDE has 5 congruent angles. Find the measure of each angle. The x-intercept is 6 . The y-intercept is 5 . 108 A point on the line is ( 6 ,0). A point on the line is (0, 5 ). Plot (6, 0) and (0, 5). Draw the line containing the two points. y 4. a. In the figure from Example 4, find m1 by using the Polygon Exterior 2 Angle-Sum Theorem. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. (6, 0) x 60 2 O 2468 2 b. Find m2. Is 2 an exterior angle? Explain. 4 30; no, it is not formed by extending one side of the polygon. (0, 5) 6

62 63 Geometry Lesson 3-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-6

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2 Transforming to Slope-Intercept Form Transform the equation Example. 6x 3y 12 to slope-intercept form. Then graph the resulting equation. 4 Writing an Equation of a Line Given Two Points Write an equation in Step 1 Transform the equation to slope-intercept form. point-slope form of the line that contains the points G(4, 9) and H(1, 1). 6x 3y 12 Step 1 Find the slope. 3y 6x 12 Add 6x to each side. y2 y1 3y 6x 12 m Use the formula for slope. Divide each side by 3 . x2 x1 333

y 2x 4 Simplify. 1 9 () m Substitute (4, 9) for (x1, y1) and ( 1, 1) for (x2, y2). The y-intercept is 4 and the slope is 2 . 14

Step 2 Use the y-intercept and the slope to plot two points and draw y 10 6 the line containing them. (1, 6) Simplify. 5 4 (0, 4) 2 2 Step 2 Select one of the points. Write the equation in point-slope form. 2 O 2 4 x y y1 m (x x1 ) Point-slope form. 2 y ()9 24(x ) Substitute 2for m and (4, 9) for (x1, y1). y 9 2 x 4 Simplify. 3 Using Point-Slope Form Write an equation in point-slope form ( ) of the line with slope 8 that contains P(3, 6).

y y1 m (x x1 ) Use point-slope form. Quick Check. 6 83 8 (3, 6) y () (x ) Substitute for m and for (x1, y1). 2. Write an equation of the line with slope 1 that contains the point P(2, 4). y 6 8 (x 3 ) Simplify. y 4 1(x 2)

Quick Check. 1. Graph each equation. 1 a. y 2x 2 b. 2x 4y 8 c. 5x y 3 y O y y O x x 2 1 1 2 4 3. Write an equation of the line that contains the points P(5, 0) and Q(7, 3). O 2 x 3 3 3 y 0 2(x 5) or y 3 2(x 7)

2 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

64 65 Geometry Lesson 3-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-6

16 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 3-7 Slopes of Parallel and Perpendicular Lines 2 Writing Equations of Parallel Lines Write an equation in point-slope form for the line parallel to 6x 3y 9 that contains (5, 8). Lesson Objectives NAEP 2005 Strand: Measurement Step 1 To ﬁnd the slope of the line, rewrite the equation in slope-intercept form. 1 Relate slope and parallel lines Topic: Measuring Physical Attributes 2 Relate slope and perpendicular lines Local Standards: ______6x 3y 9 3 y 6x 9 Subtract 6x from each side. Key Concepts. y 2x 3 Divide each side by 3 . The line 6x 3y 9 has slope 2 . Slopes of Parallel Lines Step 2 Use point-slope form to write an equation for the new line. If two nonvertical lines are parallel, their slopes are equal. y x If the slopes of two distinct nonvertical lines are equal, the lines are parallel. y 1 m (x 1 ) Any two vertical lines are parallel. y ()8 2 (x ())5 Substitute 2 for m and ( 5, 8) for (x1, y1). y 8 2 x 5 Simplify. Slopes of Perpendicular Lines ( ) slopes 1. If two nonvertical lines are perpendicular, the product of their is Quick Check. If the slopes of two lines have a product of 1, the lines are perpendicular. 1. Are the lines parallel? Explain. Any horizontal line and vertical line are perpendicular. 1 1 a. y 2x 5 and 2x 4y 9 b. y 2x 5 and 2x 4y 20

Yes; Each line has slope 1 and the No; the lines have the same slope and Examples. 2 y-intercept, so they are the same line. y-intercepts are different. 1 Determining Whether Lines are Parallel Are the lines y 5x 4 and x 5y 4 parallel? Explain.

The equation y 5x 4 is in slope-intercept form. Write the equation x 5y 4 in slope-intercept form.

x 5y 4 x 4 5y Subtract 4 from each side.

1 4 x y Divide each side by 5 . 5 5

1 4 2. Write an equation for the line parallel to y x 4 that contains (2, 5). y x 5 5 y 5 (x 2) © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 1 The line x 5y 4 has slope 5 .

The line y 5x 4 has slope 5 .

The lines are not parallel because their slopes are not equal .

66 67 Geometry Lesson 3-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-7

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Lesson 3-8 Constructing Parallel and Perpendicular Lines Example. Lesson Objectives NAEP 2005 Strand: Geometry 3 Writing Equations for Perpendicular Lines Write an equation for a line 1 perpendicular to 5x 2y 1 that contains (10, 0). Construct parallel lines Topic: Relationships Among Geometric Figures 2 Construct perpendicular lines Step 1 To ﬁnd the slope of the given line, rewrite the equation in Local Standards: ______slope-intercept form. 5x 2y 1 2y 55x 1 Subtract x from each side. Example. Constructing ᐉm 5 1 1 Examine the diagram at right. Explain y x Divide each side by 2 . how to construct 1 congruent to H. Construct the angle. 2 2 N 1 Use the method learned for constructing congruent angles. m 5 / The line 5x 2y 1 has slope 2 . Step 1 With the compass point on point H, draw an arc that intersects the sides of ЄH . H J Step 2 Find the slope of a line perpendicular to 5x 2y 1. Let m be the slope Step 2 With the same compass setting, put the compass point of the perpendicular line. on point N . Draw an arc. 5 m 1 The product of the slopes of perpendicular lines is 1 . Step 3 Put the compass point below point N where the arc 2 * ) intersects HN . Open the compass to the length where 2 2 the arc intersects lineᐉ . Keeping the same compass m 1 ? Multiply each side by . ()5 5 setting, put the *compass) point above point N where the arc intersects HN . Draw an arc to locate a point. 2 m Simplify. Step 4 Use a straightedge to draw line m through the point 5 you located and point N.

Step 3 Use point-slope form, y y1 m(x x1 ), to write an equation for the new line. Quick Check. 2 2 1. Use Example 1. Explain why lines ᐉ and m must be parallel. y 0 (x 10 ) Substitute for m and (10, 0) for (x1, y1). 5 5 If corresponding angles are congruent, the lines are parallel by the Converse of Corresponding Angles Postulate. 2 y (x 10 ) Simplify. 5

Quick Check.

3. Write an equation for the line perpendicular to 5y x 10 that contains (15, 4). Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. y 4 5(x 15)

68 69 Geometry Lesson 3-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-8

17 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Example. Example. 2 Constructing a Special Quadrilateral Construct a quadrilateral with both 3 Perpendicular From a Point to a Line *Examine) the R pairs of sides parallel. construction. At what special point does RG meet line ᐉ? Step 1 Draw point A and two rays with endpoints at A. Label point B on / Point R is the same distance from pointE as it is from point F one ray and point C on the other ray. E F ) because the arc was made with one compass opening. Step 2 Construct a ray parallel to AC through point B. Point G is the same distance from pointEF as it is from point G because both arcs were made with the same compass opening. B * ) This *means) that RG intersects line ᐉ at the midpoint of EF , and that RG is the perpendicular bisector of EF . A C ) Step 3 Construct a ray parallel to AB through point C. Quick Check. * ) * ) * ) * ) ) 3. a. Use a straightedge to draw EF . Construct FG so that FG EF at Step 4 Label point D where the ray parallel to AC intersects the ray ) point F. parallel to AB . Quadrilateral ABDC has both pairs of opposite sides parallel.

D B G

F A C E

Quick Check. 2. Draw two segments. Label their lengths c and d. Construct a quadrilateral with one pair of parallel sides of lengths c and 2d. * ) * ) * ) * ) * ) b. Draw a line CX and a point Z not on CX . Construct ZB so that ZB CX .

c d d

70 71 Geometry Lesson 3-8 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-8

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Lesson 4-1 Congruent Figures 3 Finding Congruent Triangles Can you conclude that ABC CDE? E List corresponding vertices in the same order. 3 Lesson Objective NAEP 2005 Strand: Geometry If ABC CDE, then BAC DCE . B 4 1 Recognize congruent ﬁgures and their Topic: Transformation of Shapes and Preservation of C 4 D corresponding parts Properties The diagram above shows BAC DEC , not DCE. 3 The statement ABC CDE is not true. Local Standards: ______A Notice that BC DC , BA DE , and AC EC . Vocabulary and Key Concepts. Also, CBA CDE and BAC DEC . Using Theorem 4-1, you can conclude that ECD ACB . Theorem 4–1 D Since all of the corresponding sides and angles are congruent, the triangles If two angles of one triangle are congruent to A are congruent. The correct way to state this is ABC EDC . two angles of another triangle, then the third angles are congruent. B C EF Quick Check. C F 1. WYS MKV. List the congruent corresponding parts. Use three letters for each angle. Congruent polygons are polygons that have corresponding H D Sides: WY O MK WS O MV YS O KV sides congruent and corresponding angles congruent. I Angles: lWSY OlMVK lSWY OlVMK lWYS OlMKV G E F DEF GHI 2. It is given that WYS MKV. If m Y 35, what is m K? Explain. Examples. mK 35 1 Naming Congruent Parts ABC QTJ. List the Q T Corresponding angles of congruent triangles are congruent and congruent corresponding parts. B have the same measure.

List the corresponding sides and angles in the same order. Angles: A Q B TC J A C Sides: AB > QT BC TJ AC QJ J 3. Can you conclude that JKL MNL? Justify your answer. J N 2 Using Congruency XYZ KLM, mY 67, and mM 48. Find mX. Use the Triangle Angle-Sum Theorem and the deﬁnition of congruent L polygons to ﬁnd mX. K mX mY mZ 180 Triangle Angle-Sum Theorem M

Є Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. m Z m M Corresponding angles of congruent triangles are congruent. No. The corresponding sides are not necessarily equal.

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18 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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G Triangle Congruence Example. Lesson 4-2 by SSS and SAS 4 Proving Triangles Congruent Lesson Objective NAEP 2005 Strand: Geometry Given: CG DG, CN DN, C D, GN # CD 1 Prove two triangles congruent using Topic: Transformation of Shapes and Preservation of the SSS and SAS Postulates Prove: CNG DNG Properties Local Standards: ______C N D Key Concepts. Statements Reasons > 1.CG DG 1. Given Postulate 4-1: Side-Side-Side (SSS) Postulate 2.CN > DN 2. Given If the three sides of one triangle are H Q 3.GN > GN 3. Reﬂexive Property of Congruence congruent to the three sides of another the two triangles are 4. C D 4. triangle, then G Given P congruent. 5. CNG DNG 5. Right angles are congruent F 6.ЄCGN ഡ ЄDGN 6. If two angles of one triangle are congruent to two GHF PQR R angles of another triangle, then the third angles are congruent(Theorem 4-1 ). Postulate 4-2: Side-Angle-Side (SAS) Postulate F If two sides and the included angle of one triangle 7. CNG DNG 7. Deﬁnition of congruent triangles are congruent to two sides and the included angle D of another triangle, then the two triangles B E are congruent. Quick Check. C BCA FDE A 4. Given: A D, E C, AE DC, EB CB, BA BD Prove: AEB DCB C Examples. A B 1 Proving Triangles Congruent A D Given: M is the midpoint of XY , AX > AY Prove: AMX AMY E Write a paragraph proof. Statements Reasons You are given that M is the midpoint of XY , and AX AY . 1. ЄA ഡ ЄD; ЄE ഡ ЄC 1. Given > Midpoint M implies that MX MY .AM AM by the X Y 2. ЄABE ഡ ЄDBC 2. M Vertical angles are congruent. Reflexive Property of Congruence , so © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. AMX AMY by the SSS Postulate . 3. AE O CD; AB O BD; EB O BC 3. Given

4. ᭝AEB ഡ ᭝DCB 4. Deﬁnition of Congruent Triangles

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Triangle Congruence 2 Using SAS AD > BC. What other information do you A B Lesson 4-3 by ASA and AAS need to prove ADC BCD by SAS? Lesson Objective NAEP 2005 Strand: Geometry It is given that AD > BC . Also,DC > CD by the 1 Prove two triangles congruent using Topic: Transformation of Shapes and Preservation Reflexive Property of Congruence . the ASA Postulate and the AAS of Properties D C Theorem You now have two pairs of corresponding congruent sides. Therefore, if Local Standards: ______you know ЄADC Х ЄBCD , you can prove ADC BCD by SAS . Key Concepts. R 3 Are the Triangles Congruent? From the information given, Postulate 4-3: Angle-Side-Angle (ASA) Postulate can you prove RSG RSH? Explain. If two angles and the included side of one triangle are congruent P Given: RSG RSH, SG > SH to two angles and the included side of another triangle, then > B It is given that RSG RSH and SG SH. K N G H the two triangles are congruent . RS RS by the Reflexive Property of Congruence . HGB NKP G H Two pairs of corresponding sides and their included angles are S congruent, so RSG RSH by the SAS Postulate.

FJ Theorem 4-2: Angle-Angle-Side (AAS) Theorem Quick Check. If two angles and a nonincluded side of one triangle are congruent C 1. Given: HF HJ, FG JK, to two angles and the corresponding nonincluded side of another M T H is the midpoint of GK. GHK triangle, then the triangles are congruent . D Prove: FGH JKH CDM XGT G Statements Reasons X

1. HF > HJ, FG > JK 1. Given 2. H is the midpoint of GK. 2. Given Example. > D 3.GH HK 3. Deﬁnition of Midpoint 1 Using ASA Suppose that F is congruent to C O C 4.᭝FGH Х ᭝JKH 4. SSS Postulate and I is not congruent to C. Name the triangles that are congruent by the ASA Postulate. F 2. What other information do you need to prove ABC CDA by SAS? A The diagram shows N A D and G N 7 T 8 FN > CA > GD. ЄDCA ഡ ЄBAC B A 6 I D If F C, then F C G . 7 C Therefore, FNI CAT GDO by ASA . 3. From the information given, can you prove AEB DBC? Explain. A > > B Quick Check. Given: EB CB, AE DB E C © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 1. Using only the information in the diagram, can you conclude that INF is No; you don’t know that ЄAEB Х ЄDBC or that AB O DC . D © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. congruent to either of the other two triangles? Explain. No; Only one angle and one side are shown to be congruent. At least one more congruent side or angle is necessary to prove congruence with SAS, ASA, or AAS.

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Examples. Quick Check. 2 Writing a Proof Using ASA Write a paragraph proof. A 2. Write a proof. LP Given: /A > /B, AP > BP Y Given: NM > NP, M P Prove: APX BPY N P Prove: NML NPO MO It is given that A B and AP > BP. APX BPY by It is given that NM O NP and ЄM ഡ ЄP. the Vertical Angles Theorem. Because X ЄMNL ഡ ЄPNO because vertical angles are two pairs of corresponding angles and their included sides are B congruent. ᭝NML ഡ ᭝NPO by ASA. congruent, APX BPY by ASA .

3 Planning a Proof Write a plan for a proof that uses AAS. 3. Write a ﬂow proof. A B P Given: B D, AB6CD RP bisects SRQ Prove: ABC CDA Given Because AB6CD, BAC DCA by the Alternate Interior S Q CDA C Angles Theorem. Then ABC if a pair of D R SRP ЄQRP corresponding sides are congruent. By the Reﬂexive Property, Given: S Q,RP bisects SRQ. Definition of angle bisector AC AC , so ABC CDA by AAS . Prove: SRP QRP

4 Writing a Proof Write a two-column proof that uses AAS. A B S Q RP RP Given: B D, AB6CD Given Reflexive Property of Congruence Prove: ABC CDA

C k Ok D SRP QRP AAS Statements Reasons 1. B D, AB6CD 1. Given 4. Recall Example 4. Explain how you could prove ABC CDA using ASA. 2.lBAC OlDCA 2. If lines are parallel, then alternate interior l Ol l Ol 2. angles are congruent. It is given that B D . You know that BAC DCA by the Alternate Interior Angles Theorem. By Theorem 4-1, lBCA OlDAC . 3.AC > AC 3. Reﬂexive Property of Congruence By the Reﬂexive Property of Congruence, AC O AC . Therefore, kABC OkCDA by ASA. 4. ABC CDA 4. AAS Theorem © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Lesson 4-4 Using Congruent Triangles: CPCTC 2 Using Right Triangles According to legend, one of Napoleon’s followers used congruent triangles to estimate the width of a river. On the riverbank, the ofﬁcer Lesson Objective NAEP 2005 Strand: Geometry stood up straight and lowered the visor of his cap until the farthest thing he could 1 Use triangle congruence and CPCTC Topic: Transformation of Shapes and Preservation of see was the edge of the opposite bank. He then turned and noted the spot on his to prove that parts of two triangles are Properties side of the river that was in line with his eye and the tip of his visor. congruent Local Standards: ______

G Vocabulary. F D CPCTC stands for: CPoforresponding arts Congruent E Triangles are Congruent .

Examples. Given: DEG and DEF are right angles; EDG EDF. 1 Congruence Statements The diagram shows the frame of an umbrella. The officer then paced off the distance to this spot and declared that distance to be the width of the river! What congruence statements besides 3 4 can you C prove from the diagram, in which SL > SR and The given states that DEG and DEF are right angles . 1 2 are given? 34 Rib What conditions must hold for that to be true? SC SC by the Reflexive Property of Congruence, L R DEG and DEF are the angles that the officer makes with the ground. and LSC RSC by SAS . 3 4 because 5 6 So the officer must stand perpendicular to the ground, and the corresponding parts of congruent triangles are congruent. 1 2 ground must be level or flat . Stretcher When two triangles are congruent, you can form congruence S statements about three pairs of corresponding angles and Quick Check. three pairs of corresponding sides. List the congruence Shaft statements. 1. Given: Q R, QPS RSP Prove: SQ > PR Sides: PQ Є ഡ Є Є ഡ Є O SL > SR Given It is given that Q R, QPS RSP. PS SP by the Reflexive Property of ഡ. kQPS OkRSP by ASA. SQ O PR by CPCTC. SC SC Reflexive Property of Congruence RS

CL CR Other congruence statement

Angles: 1 2 Given

3 4 Corresponding Parts of Congruent Triangles Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. lCLS OlCRS Other congruence statement 2. Recall Example 2. About how wide was the river if the ofﬁcer paced off 1 20 paces and each pace was about 22 feet long? The congruence statements that remain to be proved are 50 feet lCLS OlCRS and CL O CR.

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Lesson 4-5 Isosceles and Equilateral Triangles The legs of an isosceles triangle The vertex angle of an isosceles are the congruent sides. triangle is formed by the two Lesson Objective NAEP 2005 Strand: Geometry Vertex Angle congruent sides (legs). 1 Use and apply properties of isosceles Topic: Relationships Among Geometric Figures triangles Local Standards: ______The base angles of an isosceles The base of an isosceles triangle Legs triangle are formed by the base is the third, or non-congruent, Base Vocabulary and Key Concepts. and the legs. side. Base Angles Theorem 4-3: Isosceles Triangle Theorem C If two sides of a triangle are congruent, then Example. the angles opposite those sides are congruent. 1 Using the Isosceles Triangle Theorems Explain why ABC is A B A B isosceles. X A Theorem 4-4: Converse of Isosceles Triangle Theorem ABC and XAB are alternate interior angles formed C ←→ ←→ * ) ←→ ←→ 6 If two angles of a triangle are congruent, then by XA, BC, and the transversal AB . Because XA BC, the sides opposite those sides are congruent. ABC XAB . BC The diagram shows that XAB ACB.By AC BC A B the Transitive Property of Congruence , ABC ACB . Theorem 4-5 C You can use the Converse of the Isosceles Triangle Theorem to The bisector of the vertex angle of an isosceles triangle is the conclude that AB > AC . perpendicular bisector of the base. By the deﬁnition of an isosceles triangle, ABC is isosceles. CD ' AB and CD bisects AB. AD B T Quick Check. U Corollary to Theorem 4-3 1. Can you deduce that RUV is isosceles? Explain. Y W lRUV lRVU R If a triangle is equilateral, then No; neither nor can be shown to be congruent to lR . the triangle is equiangular. V S XY Z X Z

Corollary to Theorem 4-4 Y Examples. M If a triangle is equiangular, then 2 Using Algebra Suppose that mL y. Find the values of x and y. the triangle is equilateral. MO ' LN The bisector of the vertex y

XY YZ ZX X Z Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. angle of an isosceles triangle is the perpendicular bisector of the base. x A corollary is a statement that follows immediately from a theorem. x 90 Deﬁnition of perpendicular LO N mN mL Isosceles Triangle Theorem

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mL y Given Lesson 4-6 Congruence in Right Triangles mN y Transitive Property of Equality Lesson Objective NAEP 2005 Strand: Geometry mN mNMO mMON 180 Triangle-Sum Theorem 1 Prove triangles congruent using the Topic: Transformation of Shapes and Preservation of y y 90 180 Substitute. HL Theorem Properties 2y 90 180 Simplify. Local Standards: ______2y 90 Subtract 90 from each side. y 45 Divide each side by 2 . Vocabulary and Key Concepts. Therefore, x 90 and y 45 . Theorem 4-6: Hypotenuse-Leg (HL) Theorem If the hypotenuse and a leg of one right triangle are congruent to the 3 Using Isosceles Triangles The garden shown at the right is in the shape of hypotenuse and leg of another right triangle, then a regular hexagon. Suppose that a segment is drawn between the endpoints of the angle marked x. Find the angle measures of the triangle that is formed. the triangles are congruent. The measure of x is 120.

Because the garden is a regular hexagon, the sides have The hypotenuse of a right triangle is its longest side, or the side opposite the right angle. raised equal length, so the triangle is isosceles . x bed By the Isosceles TriangleTheorem, the unknown angles garden The legs of a right triangle are the two shortest sides, or the sides that are not opposite the arecongruent . y right angle. The measure of the angle marked x is 120, and the sum railroad z Hypotenuse of the angle measures of a triangle is 180 . tie steps Label each unknown angle y: x y y 180 Legs 120 2y 180 flagstone walk 2y 60 Examples. y 30 1 Proving Triangles Congruent One student wrote “CPA MPA A So the angle measures of the triangle are 120 , 30 , and 30 . by the HL Theorem” for the diagram. Is the student correct? Explain. The diagram shows the following congruent parts. Quick Check. CA > MA M 2. Suppose mL 43. Find the values of x and y. CPA lMPA x 90, y 47 PA y PA CPM

Since AC is the hypotenuse and PA is a leg 3. Use the diagram from Example 3. The path around the N Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. garden is made up of rectangles and equilateral triangles. x of right triangle CPA, and MA is the hypotenuse and PA is a © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. O What is the measure of z, the angle at the outside corner 63 leg of right triangle MPA, the triangles are congruent by L of the path? the HL Theorem . 150 The student is correct.

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2 Flow Proof—Using the HL Theorem XYZ is isosceles. From vertex X, Quick Check. a perpendicular is drawn to YZ , intersecting YZ at point M. Explain why 1. R 3 S XMY XMZ. L O X 3 3 5 5 M 5 Q

P N T YMZ Which two triangles are congruent by the HL Theorem? Write a correct congruence statement. XMY and XMY and ᭝ ഡ ᭝ XM ⊥ YZ XMZ are XMZ are LMN OQP right angles right triangles Given Definition of Definition of right triangle perpendicular lines 2. Write a paragraph proof of Example 2. XY XZ XMY XMZ It is given that XM is perpendicular to YZ . This means that lYMX and ЄZMX are right angles, since that is the definition of perpendicular lines. Definition of HL Theorem It follows, then, that ᭝YMX and ᭝ZMX are right triangles, since they isosceles XM XM each contain a right angle. It is given that ᭝XYZ is isosceles, so XY O XZ by the definition of isosceles. XM O XM by the Reflexive Property of Reflexive Property of Congruence Congruence. Therefore, since the hypotenuses are congruent and one leg is congruent, ᭝XMY ഡ ᭝XMZ by the HL Theorem.

3 Two-Column Proof—Using the HL Theorem C D Given: ABC and DCB are right angles, AC > DB E Prove: ABC DCB

Statements Reasons B A 1. ABC and DCB are 1. Given right angles 2. ABC and DCB 2. Deﬁnition of Right Triangle are right triangles 3. You know that two legs of one right triangle are congruent to two legs of AC > DB 3. 3. Given another right triangle. Explain how to prove the triangles are congruent.

4.BC O CB 4. Reﬂexive Property of Congruence Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Since all right angles are congruent, the triangles are congruent by SAS. DCB 5. ABC 5. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the triangles are congruent (HL Theorem).

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Using Corresponding Parts Lesson 4-7 of Congruent Triangles Examples. 3 Using Two Pairs of Triangles Write a paragraph proof. Lesson Objectives NAEP 2005 Strand: Geometry X Y 1 Identify congruent overlapping Topic: Relationships Among Geometric Figures Given: XW > YZ, XWZ and YZW are right angles. triangles Prove: XPW YPZ 2 Local Standards: ______Prove two triangles congruent by ﬁrst lZYW proving two other triangles congruent Plan: XPW YPZ by AAS if WXZ . These angles P are congruent by CPCTC if XWZ YZW. These triangles are Examples. congruent by SAS . Proof: You are given XW > YZ . Because XWZ and YZW are W Z 1 Identifying Common Parts Name the parts of the sides that G DFG and EHG share. right angles , XWZ YZW.WZ > ZW by the Identify the overlapping triangles. H F Reﬂexive Property of Congruence . Therefore, XWZ YZW by Parts of sides DG and EG are shared by DFG and EHG. SAS. WXZ ZYW by CPCTC, and XPW YPZ D E vertical angles These parts are HG and FG , respectively. because are congruent.Therefore, XPW YPZ by AAS .

2 Using Common Parts Write a plan for a proof that Z Y does not use overlapping triangles. M 4 Separating Overlapping Triangles C Given: ZXW YWX, ZWX YXW Given: CA > CE, BA > DE > Prove: ZW YX W X Write a two-column proof to show that CBE CDA. B D X Label point M where ZX intersects WY. Plan: CBE CDA by CPCTC if CBE CDA. This congruence holds by SAS if CB > CD E ZW > YX by CPCTC if ᭝ZWM ഡ ᭝YXM . . A You can prove these triangles congruent using ASA as follows: Proof: Look at MWX.MW > MX by the Statement Reason Converse of the Isosceles Triangle Theorem . 1. BCE DCA 1. Reﬂexive Property of Congruence ZMW YMX because vertical angles are congruent.By the 2. CA CE , BA DE 2. Given Angle Addition Postulate, ZWM YXM. So ZWM YXM by ASA . 3. CA CE , BA DE 3. Congruent sides have equal measure. So ZW > YX by CPCTC . 4. CA BA CE DE 4. Subtraction Property of Equality 5. CA BA CB 5. Segment Addition Postulate Quick Check. CE DE CD 6. CB CD 6. Substitution 1. The diagram shows triangles from the scaffolding that workers used A C F CB > CD when they repaired and cleaned the Statue of Liberty. 7. 7. Definition of Congruence a. Name the common side in ACD and BCD. B D 8. CBE CDA 8. SAS CD Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 9. CBE CDA All rights reserved. 9. CPCTC E © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. b. Name another pair of triangles that share a common side. Name the common side.

Answers may vary. Sample: ᭝ABD and ᭝CBD; BD

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Quick Check. AB Lesson 5-1 Midsegments of Triangles 2. Write a paragraph proof. Lesson Objective NAEP 2005 Strand: Geometry Given: ACD BDC E 1 Use properties of midsegments to Topics: Relationships Among Geometric Figures Prove: CE > DE CD solve problems Local Standards: ______It is given that ᭝ACD ഡ ᭝BDC, so lADC OlBCD by CPCTC. This means that ᭝CED is isosceles by the Converse of Isosceles Triangle Vocabulary and Key Concepts. Theorem. Then CE O DE by the deﬁnition of isosceles triangle.

Theorem 5-1: Triangle Midsegment Theorem If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half its length. Q 3. Write a two-column proof. P R Given: PS > RS, PSQ RSQ A midsegment of a triangle is a segment connecting the midpoints of two sides. Prove: QPT QRT T

S A coordinate proof is a form of proof in which coordinate geometry and algebra are used Statement Reason to prove a theorem. 1.PS O RS, lPSQ OlRSQ 1. Given 2.QS O QS 2. Reflexive Property of Congruence Examples. 3.᭝PSQ ഡ ᭝RSQ 3. SAS 1 Finding Lengths In XYZ, M, N, and P are midpoints. 4.PQ O RQ 4. CPCTC The perimeter of MNP is 60. Find NP and YZ. 5.lPQT OlRQT 5. CPCTC 6.QT O QT 6. Reflexive Property of Congruence Because the perimeter of MNP is 60, you can find NP. 22 P M ᭝ ഡ ᭝ 7.QPT QRT 7. SAS NP MN MP 60 Definition of perimeter 24 C Substitute 24 for MN and Z 4. Write a two-column proof. B NP 24 22 60 Y N 22 for MP. Given: CAD EAD, C E BD > FD A Prove: D NP 46 60 Simplify. F E NP 14 Subtract 46 from each side. Statement Reason Use the Triangle Midsegment Theorem to find YZ. 1.lCAD OlEAD, lC OlE 1. Given 1 AD O AD 2. 2. Reflexive Property of Congruence MP YZ Triangle Midsegment Theorem

3.᭝ACD ഡ ᭝AED 3. AAS Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 2 All rights reserved. 4.CD O ED 4. CPCTC 22 1 YZ Substitute 22 for MP. 5.lBDC OlFDE 5. Vertical Angles are Congruent 2 6.᭝BDC ഡ ᭝FDE 6. ASA 44 YZ Multiply each side by 2 . 7.BD O FD 7. CPCTC

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2 Identifying Parallel Segments Find mAMN and mANM. MN and BC A Quick Check. are cut by transversal AB , so AMN and B are corresponding 1. AB 10 and CD 18. Find EB, BC, and AC. angles.MN6BC by the Triangle Midsegment Theorem, so A AMN B by the Corresponding Angles Postulate. E N M B mAMN 75 because congruent angles have the same measure. D C In AMN, AM AN , so mANM mЄAMN by the EB 9; BC 10; AC 20 Isosceles TriangleTheorem. mANM 75 75 CB by substitution.

3 Applying the Triangle Midsegment Theorem Dean plans to swim the length of the lake, as shown in the photo. Explain how Dean could use the Triangle Midsegment Theorem to measure the length of the lake. 2. Critical Thinking Find mVUZ. Justify your answer. To ﬁnd the length of the lake, Dean starts at the middle X of the left edge of the lake and paces straight along that end of the 65 U lake. He counts the number of strides (35). Where the lake ends, he sets a stake. He paces the same number of strides (35) in the YZ V same direction and sets another stake. The ﬁrst stake marks the midpoint of one side of a triangle. Dean paces from the 65; UV6XY so ЄVUZ and ЄYXZ are corresponding and congruent. second stake straight to the middle of the other end of the lake. He counts the number of his strides (236).

Dean ﬁnds the midsegment of the second side by pac- ing exactly half the number of strides back 35 236 toward the second stake. He paces 118 strides. 35 From this midpoint of the second side of the triangle, ? 3. CD is a new bridge being built over a lake as shown. Find the length of he returns to the midpoint of the ﬁrst side, counting the the bridge. number of strides (128). Dean has paced a triangle. He 963 ft has also formed a midsegment of a triangle C 2640 ft Bridge whose third side is the length of the lake. 963 ft

By the Triangle Midsegment D 118 Theorem, the segment connecting the two midpoints 128 1320 ft © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. is half the distance across the lake. So, Dean

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. ? multiplies the length of the midsegment by 2 to ﬁnd the length of the lake.

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Lesson 5-2 Bisectors in Triangles Examples. 1 Applying the Perpendicular Bisector Theorem Use the map of Lesson Objective NAEP 2005 Strand: Geometry Washington, D.C. Describe the set of points that are equidistant from the 1 Use properties of perpendicular Topics: Relationships Among Geometric Figures bisectors and angle bisectors Lincoln Memorial and the Capitol. Local Standards: ______

Vocabulary and Key Concepts.

Theorem 5-2: Perpendicular Bisector Theorem If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

Theorem 5-3: Converse of the Perpendicular Bisector Theorem If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. The Converse of the Perpendicular Bisector Theorem states: If a point is equidistant from the endpoints of a segment, then it is on Theorem 5-4: Angle Bisector Theorem the perpendicular bisector of the segment. If a point is on the bisector of an angle, then it is A point that is equidistant from the Lincoln Memorial and equidistant from the sides of the angle. the Capitol must be on the perpendicular bisector Theorem 5-5: Converse of the Angle Bisector Theorem of the segment whose endpoints are the Lincoln Memorial and the Capitol. If a point in the interior of an angle is equidistant from the sides of the 2 Using the Angle Bisector Theorem C angle, then it is on the angle bisector. Find x, FB, and FD in the diagram at right.

Angle Bisector B D FD FB The distance from a point to a line is the length of the perpendicular Theorem 2x 5 7x 37 segment from the point to the line. 7x 37 2x 5 Substitute. A F E

* ) * ) 7x 2x 42 Add 37 to each side.

6 D is 3 in. from AB and AC .

5 5x 2x

4 C 42 Subtract from each side. 3

2 1

0 x 8.4 Divide each side by 5 . D B Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson FB 2(8.4 All rights reserved. ) 5 21.8 Substitute. A FD 7(8.4 ) 37 21.8 Substitute.

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Concurrent Lines, Quick Check. Lesson 5-3 Medians, and Altitudes * ) 1. Use the information given in the diagram.CD is the perpendicular C Lesson Objective NAEP 2005 Strand: Geometry 5 bisector of AB . Find CA and DB. Explain your reasoning. 1 Identify properties of Topics: Relationships Among Geometric Figures * ) A B perpendicular bisectors and CA 5; DB 6; CD is the perpendicular bisector of AB ; therefore angle bisectors Local Standards: ______CA CB and DA DB. 6 2 Identify properties of medians D and altitudes of a triangle

) ) Vocabulary and Key Concepts. 2. a. According to the diagram, how far is K from EH ? From ED ? D 10; 10 2x Theorem 5-6 K C E The perpendicular bisectors of the sides of a triangle are concurrent at a point equidistant from the vertices. 10 (x 20) H Theorem 5-7 ) The bisectors of the angles of a triangle are b. What can you conclude about EK ? concurrent at a point equidistant from the sides. ) EK is the angle bisector of lDEH. Theorem 5-8 The medians of a triangle are concurrent at a point that is two-thirds the distance from each vertex to the midpoint of the opposite side.

2 2 2 D DC DJ EC EG FC FH H 3 3 3 C G E c. Find the value of x. Theorem 5-9 J 20 The lines that contain the altitudes of a triangle are concurrent. F

Concurrent lines are three or more lines that meet in one point. The point of concurrency is the point at which concurrent lines intersect. A circle is circumscribed about a polygon when the vertices Circumscribed circle of the polygon are on the circle. d. Find mDEH. S The circumcenter of a triangle is the point of concurrency of QC SC RC 80 the perpendicular bisectors of a triangle. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. C

A median of a triangle is a segment that Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Median has as its endpoints a vertex of the Q R triangle and the midpoint of the opposite circumcenter side.

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if the sides of the polygon are A circle is inscribed in a polygon Inscribed circle 2 Applying Theorem 5-7 City planners want to locate a tangent to the circle. fountain equidistant from three straight roads that enclose T XI YI ZI Mariposa Boulevard a park. Explain how they can ﬁnd the location. The incenter of a triangle is the point of currency of the angle Y X bisectors of the triangle. I The roads form a triangle around the park. the perpendicular segment from a An altitude of a triangle is Theorem 5-7 states that the Highway 101 Park U Z V vertex to the line containing the opposite side. bisectors of the angles of a triangle are incenter Andover Road concurrent at a point equidistant from the sides.

The city planners should ﬁnd the point of concurrency of the angle bisectors of the triangle formed by the three roads and locate the fountain there. Acute Triangle Right Triangle Obtuse Triangle Altitude is Altitude is Altitude is inside . a side . outside . 3 Finding Lengths of Medians M is the centroid of WOR, and WM 16. W Find WX. The centroid of a triangle is the point of intersection of the medians of the triangle. The centroid is the point of concurrency of the medians Z Y The orthocenter of a triangle is the point of intersection of the lines containing the altitudes of a triangle. M of the triangle. The medians of a triangle are concurrent at a point that is O R X two-thirds the distance from each vertex to the midpoint Examples. of the opposite side. (Theorem 5-8) 1 Finding the Circumcenter Find the center of the circle that y Because M is the centroid of WOR, WM 2WX . circumscribes XYZ. 3 8 Y (1, 1) Because X has coordinates and Y has 2 6 coordinates (1, 7) ,XY lies on the vertical line x 1 . WM WX Theorem 5-8 (3, 4) 3 The perpendicular bisector of XY is the horizontal line that 4 1 7 passes through (1,2 ) or (1, 4) , so the 2 2 16 16 equation of the perpendicular bisector of XY is y 4 . WX Substitute for WM. X Z 3 O 2 46x Because X has coordinates (1, 1) and Z has 24 3 WX Multiply each side by 2 . coordinates (5, 1) ,XZ lies on the horizontal line

y 1 . The perpendicular bisector of XZ is the vertical line that passes through1 5, 1 or (3, 1) , so the equation

( 2 ) Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. of the perpendicular bisector of XZ is x 3 .

Draw the lines y 4 and x 3 . They intersect at the point (3, 4) . This point is the center of the circle that circumscribes XYZ.

98 99 Geometry Lesson 5-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 5-3

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Inverses, Contrapositives, Quick Check. Lesson 5-4 and Indirect Reasoning 1. a. Find the center of the circle that you can y Lesson Objectives NAEP 2005 Strand: Geometry circumscribe about the triangle with 1 Write the negation of a statement Topics: Mathematical Reasoning vertices (0, 0), (8, 0), and (0, 6). and the inverse and contrapositive Local Standards: ______of a conditional statement 5 2 Use indirect reasoning (4, 3) Vocabulary and Key Concepts.

55x Negation, Inverse, and Contrapositive Statements Statement Example Symbolic Form You Read It S 5 Conditional If an angle is a straight angle, p qp,q.If then then its measure is 180.

b. Critical Thinking In Example 1, explain why it is not necessary to ﬁnd Negation (of p) An angle is not a straight angle. pp.Not the third perpendicular bisector. Theorem 5-6: All the perpendicular bisectors of the sides of a Inverse If an angle is not a straight angle, p S qp,q.If not then not triangle are concurrent. then its measure is not 180. Contrapositive If an angle’s measure is not 180, q S pq,p.If not then not then it is not a straight angle.

2. a. The towns of Adamsville, Brooksville, and Cartersville want to Writing an Indirect Proof Adamsville build a library that is equidistant from the three towns. Show on Step 1 State as an assumption the the diagram where they should build the library. Explain. Brooksville opposite (negation) of what you want to prove. Draw segments connecting the towns. Build the library at the intersection point of the perpendicular bisectors of the segments. Step 2 Show that this assumption leads to a contradiction. Step 3 Conclude that the assumption must be Cartersville false and that what you want to prove must be true. b. What theorem did you use to ﬁnd the location? The perpendicular bisectors of the sides of a triangle are concurrent at a point equidistant from the vertices (Theorem 5-6). The negation of a statement has the opposite truth value from that of the original statement.

The inverse of a conditional statement negates both the hypothesis and the conclusion. 3. Using the diagram in Example 3, find MX. Check that WM MX WX. K 8 The contrapositive of a conditional switches the hypothesis and the conclusion and negates both. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A conditional and its contrapositive always have the same truth value. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson 4. Is MY a median, an altitude, or neither? Explain. All rights reserved. Y Equivalent statements have the same truth value. MY median; is a segment drawn from vertex M to the midpoint In indirect reasoning all possibilities are considered and then all but one are proved false. of the opposite side. The remaining possibility must be true. L M X An indirect proof is a proof involving indirect reasoning.

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Examples. 4 Identifying Contradictions Identify the two statements that contradict each other. 1 Writing the Negation of a Statement Write the negation of “ABCD is I. P, Q, and R are coplanar. not a convex polygon.” II. P, Q, and R are collinear. The negation of a statement has the opposite truth value. The III. m PQR 60 negation of is not in the original statement removes the word not . Two statements contradict each other when they cannot both be true at the same time. Examine each pair of statements to see whether they The negation of “ABCD is not a convex polygon” is contradict each other. “ABCD is a convex polygon.” I and II I and III II and III 2 Writing the Inverse and Contrapositive Write the inverse and contrapositive P, Q, and R are coplanar and P, Q, and R are coplanar, and P, Q, and R are collinear, and of the conditional statement “If ABC is equilateral, then it is isosceles.” collinear. mPQR 60. mPQR 60. To write the inverse of a conditional, negate both the hypothesis and the Three points that lie on the Three points that lie on an If three distinct points are conclusion. same line are both coplanar angle are coplanar, so these collinear, they form a straight Hypothesis Conclusion and collinear, so these two two statements do not angle, so mPQR cannot TT statements do not contradict each other. equal 60. Statements II and Conditional: If ABC is equilateral, then it is isosceles. contradict each other. III do contradict T Negate both. T each other. Inverse: If ABC is not equilateral , then it is not isosceles . 5 Indirect Proof Write an indirect proof. To write the contrapositive of a conditional, switch the hypothesis and Prove: ABC cannot contain two obtuse angles. conclusion, then negate both. Hypothesis Conclusion Step 1 Assume as true the opposite of what you want to prove. T T That is, assume that ABC contains two obtuse angles. Conditional: If ABC is equilateral, then it is isosceles. Let lA and lB be obtuse . Switch and negate both. Step 2 If A and B are obtuse , mA mB mC 180 .

Contrapositive: If ABC is not isosceles , then it is not equilateral . This contradicts the Triangle Angle-Sum Theorem, which states that mA mB mC 180 .

3 The First Step of an Indirect Proof Write the first step of an indirect proof. Step 3 The assumption in Step 1 must be false . ABC cannot Prove: A triangle cannot contain two right angles. contain two obtuse angles. In the first step of an indirect proof, you assume as true the negation of what you want to prove. Quick Check. Because you want to prove that a triangle cannot contain two right angles, 1. Write the negation of each statement. can you assume that a triangle contain two right angles. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson a. mXYZ 70. All rights reserved. The first step is “Assume that a triangle contains two right angles.” The measure of ЄXYZ is not more than 70.

b. Today is not Tuesday. Today is Tuesday.

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2. Write (a) the inverse and (b) the contrapositive of Maya Angelou’s Lesson 5-5 Inequalities in Triangles statement “If you don’t stand for something, you’ll fall for anything.” a. Lesson Objectives NAEP 2005 Strand: Geometry If you stand for something, you won’t fall for anything. 1 Use inequalities involving angles of Topics: Relationships Among Geometric Figures triangles 2 Use inequalities involving sides of Local Standards: ______triangles b. If you won’t fall for anything, then you stand for something. Key Concepts.

Comparison Property of Inequality 3. You want to prove each statement true. Write the ﬁrst step of an indirect proof. If a b c and c 0, then a b . a. The shoes cost no more than $20. The shoes cost more than $20.

Corollary to the Triangle Exterior Angle Theorem The measure of an exterior angle of a triangle is greater 3 b. mA mB than the measure of each of its remote interior angles. mlA # mlB 2 1 mЄ1 m2 and mЄ1 m3

4. Identify the two statements that contradict each other. Explain. Theorem 5-10 I.FG6KL II.FG ' KL III. FG > KL If two sides of a triangle are not congruent, then Y I and II; the larger angle lies opposite the longer side. If XZ XY, then mY mZ. Two segments cannot be parallel as well as perpendicular. I and II XZ contradict each other. Theorem 5-11 Two segments can be parallel and congruent. I and III do not contradict B each other. If two angles of a triangle are not congruent, then C the longer side lies opposite the larger angle. Two segments can be perpendicular and congruent. II and III do not contradict each other. If mA mB, then BC AC. A Theorem 5-12: Triangle Inequality Theorem L The sum of the lengths of any two sides of a triangle is greater than the length of the third side. MN LN 5. Critical Thinking You plan to write an indirect proof showing that X is an LM MN © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. obtuse angle. In the ﬁrst step you assume that X is an acute angle. What MN LN LM have you overlooked? LN LM MN Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. ЄX could be a right angle.

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26 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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4 Finding Possible Side Lengths In FGH, FG 9 m and GH 17 m. Examples. Describe the possible lengths of FH . B 1 Applying the Corollary Explain why m 4 m 5. The Triangle Inequality Theorem: The sum of the lengths of any two sides of 5 greater 4 is an exterior angle of XYC. The Corollary Y a triangle is than the length of the third side. 2 to the Exterior Angle Theorem states that the measure Solve three inequalities. of an exterior angle of a triangle is greater FH 9 17 FH 17 9 9 17 FH 4 1 3 A C than the measure of each of its remote interior angles. X FH 8 FH 8 26 FH The remote interior angles are Є2 and Є3 , so FH must be longer than 8 m and shorter than 26 m. m4 m2. 2 and 5 are corresponding angles formed by transversal BC . Because AB6XY , you can conclude that 2 Є5 by the Corresponding Angles Postulate. Thus, Quick Check. m2 mЄ5 . Substituting mЄ5 for m2 in the 1. Use the diagram in Example 1. If CY XY, explain why m4 m1. Є Є inequality m4 m2 produces the inequality m 4 m 5 . mЄ1 mЄ3 by the Isosceles Triangle Theorem mЄ4 mЄ3 by the Corollary to the Triangle Exterior Angle Theorem mЄ4 mЄ1 by Substitution 2 Applying Theorem 5-10 In RGY, RG 14, GY 12, and RY 20. List the angles from largest to smallest. Theorem 5-10: If two sides of a triangle are not congruent, then G 2. List the angles of ABC in order from smallest to largest. the larger angle lies opposite the longer side. A 12 14 ЄA, ЄC, ЄB 27 ft No two sides of RGY are congruent, so the largest angle lies opposite the longest side. C Y R 21 ft Find the angle opposite each side. 20 18 ft The longest side is 20 . The opposite angle, ЄG , Є is the largest. The shortest side is 12 . The opposite angle, R , is the B smallest. From largest to smallest, the angles are ЄG ,,.ЄY ЄR 3. Can a triangle have sides with the given lengths? Explain. a. 2 m, 7 m, and 9 m 3 Using the Triangle Inequality Theorem Can a triangle have sides with the given lengths? Explain. no; 2 7 is not greater than 9 a. 2 cm, 2 cm, 4 cm

According to the Triangle Inequality Theorem, the b. 4 yd, 6 yd, and 9 yd sum of the lengths of any two sides of a triangle is greater yes; 4 6 9, 6 9 4, and 4 9 6 than the length of the third side.

2 2 Ѐ 4 The sum of 2 and 2 is not greater than 4. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 4. A triangle has sides of lengths 3 in. and 12 in. Describe the possible lengths No cannot , a triangle have sides with these lengths. of the third side. b. 8 in., 15 in., 12 in. 9 x 15 8 15 12 8 12 15 15 12 8 Yes , a triangle can have sides with these lengths.

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Lesson 6-1 Classifying Quadrilaterals Examples. Classifying by Coordinate Methods Determine the most precise name Lesson Objective NAEP 2005 Strand: Geometry 1 for the quadrilateral with vertices Q(4, 4), B(2, 9), H(8, 9), and A(10, 4). 1 Deﬁne and classify special types of Topic: Relationships Among Geometric Figures quadrilaterals Graph quadrilateral QBHA. Local Standards: ______y Find the slope of each side. BH 9 4 5 Slope of QB 2 Key Concepts. 2 ( 4) 8 9 9 Slope of BH 0 Special Quadrilaterals 6 8 ( 2 ) QuadrilateralQuadrilateral Parallelogram QA4 9 Slope of HA 25 1 pair of 2 pairs of 2 o pairs of 10 8 N parallel sides parallel sides Rectangle Rhombus 2 parallel sides 4 4 Kite Trapezoid Slope of QA 0 4 22468O x 4 10 Isosceles Square Trapezoid BH is parallel to QA because their slopes are equal . HA is not parallel to QB because their slopes are not equal . One pair of opposite sides is parallel, so QBHA is a trapezoid . Next, use the distance formula to see whether any pairs of sides are congruent. A parallelogram Akite is a 2 2 QB (2 (4)) 9 4 4 25 "29 is a quadrilateral with quadrilateral with two pairs ( ) both pairs of opposite of ______adjacent sides 2 2 HA (10 8) 4 9 425 "29 sides ______parallel congruent and no ______opposite ( ) ______. ______sides __ 2 2 BH (8 (2)) 9 9 100 0 10 congruent. ( ) 22 QA (4 10 ))( 44 196 0 14

Because QB HA , QBHA is an isosceles trapezoid . A rhombus A rectangle A square U is a parallelogram with is a parallelogram with is a parallelogram with 2 Using the Properties of Special Quadrilaterals In parallelogram RSTU, four______congruent______sides four_____right______angles four______congruent sides mR 2x 10 and mS 3x 50. Find x. T ______. ______. ______and four RSTU is a parallelogram. Given ___right______angles . (2x 10) (3x 50) R ST6 RU Deﬁnition of parallelogram S If lines are parallel, then interior m R m S 180 angles on the same side of a transversal are supplementary . © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A trapezoid An ______isosceles trapezoid (2x 10) (3x 50) 180 Substitute 2x 10 for mЄR and 3x 50 for mЄS.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson is a quadrilateral with All rights reserved. ______ ______exactly one pair of is a trapezoid whose 5x 40 180 Simplify. parallel sides ______nonparallel sides are 5x 140 Subtract 40 from each side. ______. congruent. x 28 Divide each side by 5.

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Quick Check. Lesson 6-2 Properties of Parallelograms 1. a. Graph quadrilateral ABCD with vertices A(3, 3), B(2, 4), y B Lesson Objectives NAEP 2005 Strand: Geometry C(3, 1), and D(2, 2). A 1 Use relationships among sides and Topic: Relationships Among Geometric Figures b. Classify ABCD in as many ways as possible. among angles of parallelograms 2 Local Standards: ______ABCD is a quadrilateral because it has four sides. It is a x Use relationships involving diagonals parallelogram because both pairs of opposite sides are of parallelograms and transversals parallel. It is a rhombus because all four sides are congruent. C It is a rectangle because it has four right angles and its D opposite sides are congruent. It is a square because it has four right angles and its sides are all congruent. Vocabulary and Key Concepts.

Theorem 6-1 B Opposite sides of a parallelogram are congruent. C

c. Which name gives the most information about ABCD? Explain. Theorem 6-2 A D Square; the name square means a figure is both a rectangle Opposite angles of a parallelogram are congruent . and a rhombus, since its sides are all congruent and it has R S four right angles. Theorem 6-3 U The diagonals of a parallelogram bisect each other. T

Theorem 6-4 If three (or more) parallel lines cut off congruent segments on one A B 2. Find the values of the variables in the rhombus. Then ﬁnd the L 3b 2 N transversal, then they cut off congruent segments on every lengths of the sides. C D transversal. a 2, b 4, LN ST NT SL 14 5a 4 E F 3a 8 BD DF

S 4b 2 T

Consecutive angles of a polygon share a common side.

J K

M L In JKLM, J and M are consecutive angles, as are J and KL . J and are not consecutive. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

Examples

110 111 Geometry Lesson 6-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 6-2

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1 Using Algebra Find the value of x in ABCD. Then find mA. A B off congruent segments on one transversal, then they cut off congruent Opposite angles of a (x 15) segments on every transversal. x 15 135 x parallelogram are Explain how to divide a blank card into five equal rows using Theorem 6-4 congruent . and a sheet of lined paper. 2x 15 135 Add x to each side. Place a corner of the top of the card on the first line of the lined paper. 2x 120 Subtract 15 from each side. (135 x) Place a corner that forms a consecutive angle with the first corner on the D C sixth line. x 60 Divide each side by 2 . Mark the points where the lines intersect one side of the card. mB 60 15 75 Substitute 60 for x. Mark the points where the lines intersect the opposite side of the card. Connect the marks on opposite sides using a straightedge. mA mB 180 Consecutive angles of a parallelogram are supplementary . mA 75 180 Substitute 75 for mB. Quick Check. mA 105 Subtract 75 from each side. 1. Find the value of y in EFGH. Then find m E, m F, m G, and mH. F G (3y 37) 2 Using Algebra Find the values of x and y in KLMN. y 5 11; mlE 5 70, mlF 5 110, mlG 5 70, and mlH 5 110 (6y 4) 2x 5 L E H K x A 7y 16 5y M N

The diagonals of a parallelogram x 7y 16 bisect each other. 2. Find the values of a and b. X Y 2x 5 5y a 16, b 14 a 8 a Substitute for x in 2 7y 16 7y 16 b 2()5 5y 10 the second equation to solve for y. b 2

14y 32 5 5y Distribute. W Z

14y 27 5y Simplify.

27 9y Subtract 14y from each side. ←→ ←→ ←→ ←→ 3. In the ﬁgure, DH 6 CG 6 BF 6 AE, AB BC CD 2, and 3 y Divide each side by 9 . EF 2.5. Find EH. DH Substitute 3 for y in the ﬁrst equation 2 x 7()3 16 7.5 to solve for x. CG 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. x 5 Simplify. BF

3 Using Theorem 6-4 Theorem 6-4 states If three (or more) parallel lines cut

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Proving That a Quadrilateral Lesson 6-3 Is a Parallelogram 2 Is the Quadrilateral a Parallelogram? Can you prove the quadrilateral is a parallelogram from what is given? Explain. Lesson Objective NAEP 2005 Strand: Geometry > 1 Determine whether a quadrilateral is Topic: Geometry a. Given: AB CD a parallelogram Prove: ABCD is a parallelogram. Local Standards: ______All you know about the quadrilateral is that only BC one pair of opposite sides is congruent . Key Concepts. Therefore, you cannot conclude that the quadrilateral Theorem 6-5 is a parallelogram. If both pairs of opposite sides of a quadrilateral are congruent, then A D the quadrilateral is a parallelogram.

Theorem 6-6 b. Given: mE mG 75°, mF 105° If both pairs of opposite angles of a quadrilateral are congruent, Prove: EFGH is a parallelogram. then the quadrilateral is a parallelogram. F G The sum of the measures of the angles of a polygon is 105 75 (n 2)180 where n represents the number of sides, Theorem 6-7 so the sum of the measures of the angles of a If the diagonals of a quadrilateral bisect each other, quadrilateral is 4 2 180 360 . then the quadrilateral is a parallelogram. ( ) If x represents the measure of the unmarked angle, Theorem 6-8 75 x 75 105 75 360 , so x = 105 . EH If one pair of opposite sides of a quadrilateral is both congruent and parallel , then the quadrilateral is a parallelogram. Theorem 6-6 states If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Examples. A B 1 Finding Values for Parallelograms Find values for x and y Because both pairs of opposite angles are congruent, the quadrilateral for which ABCD must be a parallelogram. 8x 12 is a parallelogram by Theorem 6-6 . y 9 If the diagonals of quadrilateral ABCD bisect each other, then ABCD is a parallelogram by Theorem 6-7 . 2y 80 G Write and solve two equations to ﬁnd values of x and y for 10x 24 which the diagonals bisect each other. D C Diagonals of parallelograms 10x 24 8x 12 2y 80 y 9 bisect each other. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 2x 24 12 Collect the variables on one side. y 80 9

2x 36 Solve. y 89 x 18

If x 18 and y 89 , then ABCD is a parallelogram.

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Quick Check. Lesson 6-4 Special Parallelograms 1. Find the values of a and c for which PQRS must be a Q R Lesson Objectives parallelogram. (a 40) a NAEP 2005 Strand: Geometry 1 Use properties of diagonals of Topic: Geometry rhombuses and rectangles a 70, c 2 3c 3 c 1 2 Determine whether a parallelogram is Local Standards: ______a rhombus or a rectangle P S Key Concepts.

Rhombuses Theorem 6-9 2. Can you prove the quadrilateral is a parallelogram? Explain. B 3 C Each diagonal of a rhombus bisects two angles of the rhombus. 4 a. Given: PQ > SR, PQ 6SR AC bisects BAD, so 1 2 1 Prove: PQRS is a parallelogram. 2 AC bisects BCD , so 3 4 A D Yes; a pair of opposite sides are parallel QR and congruent (Theorem 6-8). Theorem 6-10 B C x x The diagonals of a rhombus are perpendicular . AC ' BD PS A D

Rectangles > , > b. Given: DH GH EH FH Theorem 6-11 A D Prove: DEFG is a parallelogram. The diagonals of a rectangle are congruent . DE B C No; the diagonals do not necessarily bisect AC BD each other—the ﬁgure could be a trapezoid, with DG and EF being parallel but of unequal length. H Parallelograms GF Theorem 6-12 If one diagonal of a parallelogram bisects two angles of the parallelogram, then the parallelogram is a rhombus.

Theorem 6-13 If the diagonals of a parallelogram are perpendicular, then © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson the parallelogram All rights reserved. is a rhombus. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Theorem 6-14 If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.

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Examples. Example. 1 Finding Angle Measures Find the measures of the 3 Identifying Special Parallelograms The diagonals of ABCD are such that numbered angles in the rhombus. AC 16 cm and BD 8 cm. Can you conclude that ABCD is a rhombus or B C 1 a rectangle? Explain. Theorem 6-9 states that each diagonal of a rhombus 78 bisects two angles of the rhombus, so m1 78. AC BD 3 2 ABCD cannot be a rectangle, because and the diagonals of a Theorem 6-10 states that 4 A rectangle are congruent (Theorem 6-11). ABCD may be a rhombus, the diagonals of the rhombus are perpendicular , D but it may also not be one, depending on whether AC ' BD (Theorem 6-10). so m2 90 . Because the four angles formed by the diagonals all must have measure 90, 3 and ABD must be complementary . Because mABD 78, Quick Check. m3 90 78 12 . 2. Find the length of the diagonals of rectangle GFED if FD 5y 9 F E and GE = y 5. Finally, because BC DC, the Isosceles Triangle Theorem 81 allows you to conclude that 1 4. So m4 78 . 2 G D 2 Finding Diagonal Length One diagonal of a rectangle has length 8x 2. The other diagonal has length 5x 11. Find the length of each diagonal. By Theorem 6-11, the diagonals of a rectangle are congruent .

5x 11 8x 2 Diagonals of a rectangle are congruent.

11 3x 2 Subtract 5x from each side. 9 3x Subtract 2 from each side. 3 x Divide each side by 3 . 8x 2 8()3 2 26 Substitute. 5x 11 5()3 11 26 Substitute. The length of each diagonal is 26 . 3. A parallelogram has angles of 30°, 150°, 30°, and 150°. Can you conclude that it is a rhombus or a rectangle? Explain.

Quick Check. No; there is not enough information to conclude that the parallelogram is a rhombus. It cannot be a rectangle because it has 1. Find the measures of the numbered angles in the rhombus. 50 no right angles. mЄ1 90, mЄ2 50, mЄ3 50, mЄ4 40 1

3 4 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Lesson 6-5 Trapezoids and Kites Examples. 1 Finding Angle Measures in Trapezoids WXYZ is XY Lesson Objective NAEP 2005 Strand: Geometry an isosceles trapezoid, and mX 156. Find mY, 156 1 Verify and use properties of Topic: Relationships Among Geometric Figures mZ, and mW. trapezoids and kites W Z Local Standards: ______Two angles that share mX mW 180 a leg of a trapezoid Vocabulary and Key Concepts. are supplementary .

156 mW 180 Substitute. Trapezoids Theorem 6-15 mW 24 Subtract 156 from each side.

The base angles of an isosceles trapezoid are congruent . Because the base angles of an isosceles trapezoid are congruent , mY mX 156 and mZ mЄW 24 . Theorem 6-16 The diagonals of an isosceles trapezoid are congruent. A 2 Using Isosceles Trapezoids Half of a spider’s web is shown at the right, B formed by layers of congruent isosceles trapezoids. Find the measures C D of the angles in ABDC. The base angles of a trapezoid are two angles that share a base of the trapezoid. Trapezoid ABDC is part of an isosceles triangle whose vertex is at the center of the web. Because there are 6 adjacent congruent vertex angles at the center of the web, together forming a 180 straight angle, each vertex angle measures 6 , or 30 . Base angles Leg Leg By the Triangle Angle-Sum Theorem, mA mB 30 180 , Base angles so mA mB 150 . Because ABDC is part of an isosceles triangle, mA mB, so 2(mA) 150 and mA mB 75 . Another way to ﬁnd the measure of each acute angle is to divide the Kites difference between 180 and the measure of the vertex angle by 2: Theorem 6-17 180 2 30 2 75 . The diagonals of a kite are perpendicular . Because the bases of a trapezoid are parallel, the two angles that share a leg are supplementary, so mC mD 180 75 = 105 .

Quick Check. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 1. In the isosceles trapezoid, mS 70. Find mP, mQ, and mR. P Q 110, 110, 70 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

70 S R

120 121 Geometry Lesson 6-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 6-5

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Placing Figures in Example. Lesson 6-6 the Coordinate Plane 3 Finding Angle Measures in Kites Find m1, m2, and m3 in the kite. Lesson Objective NAEP 2005 Strand: Geometry S 1 Name coordinates of special figures Topic: Position and Direction 1 by using their properties Local Standards: ______D R T 3 2 72 Example. U 1 Proving Congruency Show that TWVU is a parallelogram by y A W(a c, b d) proving pairs of opposite sides congruent. m2 90 Diagonals of a kite are perpendicular . C If both pairs of opposite sides of a quadrilateral are RU RS Deﬁnition of a kite T(a, b) congruent , then the quadrilateral is a V(c e, d) m1 72 Isosceles Triangle Theorem parallelogram by Theorem 6-5 . x Triangle Angle-Sum Theorem m 3 m RDU 72 180 O U(e, 0) E You can prove that TWVU is a parallelogram by showing that mRDU 90 Diagonals of a kite are perpendicular. TW VU and WV TU . Use the distance formula. m3 90 72 180 Substitute. m3 162 180 Simplify. Use the coordinates T( a , bde), W( a c , b dc), V( e , ), and U( , 0 ). m3 18 Subtract 162 from each side. 2 2 TW ( a c a ) ()b d b c2 d2

Quick Check. 2 2 2 2 VU ( c ee) ()d 0 c d 2. The middle ring of the piece of ceiling shown is made from congruent isosceles 2 2 trapezoids. Imagine a circular glass ceiling made from the ceiling pieces with WV ( a c c e ) ( b d d ) (a e)2 b2 18 angles meeting at the center. What are the measures of the two sets of base 2 2 2 2 angles of the trapezoids in the middle ring? TU ()a e ()b 0 (a e) b

The measure of both inner angles is 85. Because TW VU and WV TU, TWVU is a parallelogram . The measure of both outer angles is 95.

Quick Check. 1. Use the diagram above. Use a different method: Show that TWVU is a parallelogram by finding the midpoints of the diagonals.

TV 5 aa 1 c 1 e , b 1 db UW Midpoint of 2 2 midpoint of . Thus, the 3. Find m1, m2, and m3 in the kite. diagonals bisect each other, and TWVU is a parallelogram.

mЄ1 90, mЄ2 46, and mЄ3 44 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 46 1 2 3

122 123 Geometry Lesson 6-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 6-6

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Example. Lesson 6-7 Proofs Using Coordinate Geometry 2 Naming Coordinates Use the properties of parallelogram y Lesson Objective NAEP 2005 Strand: Geometry OCBA to ﬁnd the missing coordinates. Do not use any new (p, q) A 1 Prove theorems using figures in the Topics: Position and Direction; Mathematical Reasoning variables. B coordinate plane The vertex O is the origin with coordinates O (0, 0) . Local Standards: ______Because point A is p units to the left of point O, point B is also p units to the left of point C , because OCBA is a Vocabulary and Key Concepts. parallelogram. So the ﬁrst coordinate of point B is p x . x Theorem 6-18: Trapezoid Midsegment Theorem Because AB 6 CO and CO is horizontal,isAB C (x, 0) O (1) The midsegment of a trapezoid is parallel to the bases. also horizontal . So point B has the same second coordinate,q , as point A. (2) The length of the midsegment of a trapezoid is half the sum of the The missing coordinates are O (0, 0)and B (p x, q) . lengths of the bases .

The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel Quick Check. opposite sides of the trapezoid. 2. Use the properties of parallelogram OPQR to ﬁnd the missing y R(b, c) Q(?, ?) coordinates. Do not use any new variables. RA Q(s b, c) MN

TP O P(s, 0) x 1 MN 6 TP ,MN 6 RA , and MN 2( TP RA )

Examples. 1 Planning a Coordinate Geometry Proof Examine trapezoid y TRAP. Explain why you can assign the same y-coordinate to points R(2b, 2c) A(2d, 2c) R and A. MN In a trapezoid, only one pair of sides is parallel. In TRAP, x TP 6 RA . Because TP lies on the horizontal x-axis,RA also T P(2a, 0) must be horizontal . The y-coordinates of all points on a horizontal line are the same, so

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson points R and A All rights reserved. have the same y-coordinates. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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2 Using Coordinate Geometry Use coordinate geometry to prove that the Quick Check. quadrilateral formed by connecting the midpoints of rhombus ABCD is 1. Complete the proof of Theorem 6-18. y a rectangle. MN R(2b, 2c) A(2d, 2c) Draw quadrilateral XYZW by connecting the midpoints of ABCD. Given: is the midsegment of trapezoid TRAP. 1 Prove: MN 6 TP,MN 6 RA , and MN = 2(TP + RA) MN y x a. Find the coordinates of midpoints M and N. How do the multiples T P(2a, 0) D(0, 2b) of 2 help? Z(a, b) W(a, b) M(b, c) and N(a d, c); starting with multiples of 2 eliminates C( 2a, 0) A(2a, 0) fractions when using the midpoint formula. x Y(a, b) X(a, b) B(0, 2b)

b. Find and compare the slopes of MN, TP, and RA . From Lesson 6-6, you know that XYZW is a parallelogram. 0, 0, 0; they are all parallel.

If the diagonals of a parallelogram are congruent , then the parallelogram is a rectangle from Theorem 6–14.

To show that XYZW is a rectangle, find the lengths of its diagonals, and then compare them to show that they are equal . c. Find and compare the lengths MN, TP, and RA. MN d a b, TP 2a, RA 2d 2b. So RA TP 2d 2a 2b XZ (a a)2 (b (b))2 which is twice MN. So the midsegment is half the sum of the bases. ( 2a )2 ( 2b )2 4a2 4b2 d. In parts (b) and (c), how does placing a base along the x-axis help? YW (a a)2 (b b)2 The base along the x-axis allows calculating length by subtracting ( 2a )2 ( 2b )2 x-values. 4a2 4b2

XZ YW

Because the diagonals are congruent, 2. Use the diagram from Example 2. Explain why the proof using A(2a, 0), B(0, 2b), C(2a, 0), and D(0, 2b) is easier than the proof using A(a, 0), parallelogram XYZW is a rectangle . B(0, b), C(a, 0), and D(0, b).

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson Using multiples All rights reserved. of 2 in the coordinates for A, B, C, and D eliminates the use of fractions when finding midpoints, since finding midpoints requires division by 2.

126 127 Geometry Lesson 6-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 6-7

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Lesson 7-1 Ratios and Proportions 2 Properties of Proportions a 12 b Lesson Objective NAEP 2005 Strand: Geometry Complete: If 4 5 b , then 12 5 . 1 Write ratios and solve proportions Topic: Position and Direction ab 5 48 Cross-Product Property Local Standards: ______ab 48 12a 12a Divide each side by 12a . Vocabulary and Key Concepts. b 4 12 5 Simplify. Properties of Proportions a a c b d is equivalent to 3 Solving for a Variable Solve each proportion. d b 1 c 1 d 2 n bc b a a b 5 (1) ad (2) a c (3) c d (4) b d a. 5 35 5n 5 2 ()35 Cross-Product Property A proportion is a statement that two ratios are equal. 5n 5 70 Simplify. n 5 a c 14 Divide each side by 5 . b d and a : b c : d are examples of proportions. x 1 1 x An extended proportion is a statement that three or more ratios are equal. b. 3 5 2 6 4 1 3x 5 2 x 1 1 Cross-Product Property 24 16 4 is an example of an extended proportion. () The Cross-Product Property states that the product of the extremes of a proportion is equal 3x 5 2x 1 2 Distributive Property to the product of the means. x 5 2 Subtract 2x from each side.

means a c b d Quick Check. a : b c : d 1 abd c 1. A photo that is 8 in. wide and 53 in. high is enlarged to a poster that is 2 ft extremes 1 wide and 13 ft high. What is the ratio of the height of the photo to the height A scale drawing is a drawing in which all lengths are proportional to corresponding actual of the poster? lengths. 1 : 3 A scale is the ratio of any length in a scale drawing to the corresponding actual length. The lengths may be in different units.

Examples. m n 1 Finding Ratios A scale model of a car is 4 in. long. The actual car is 15 ft 2. Write two expressions that are equivalent to 4 5 11 . long. What is the ratio of the length of the model to the length of the car? Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 4 11 m 1 4 n 1 11 Answers may vary. Sample: m 5 n , 5 Write both measurements in the same units. 4 11 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 15 ft 15 12 in. 180 in. 1 length of model 5 4 in. 5 4 in. 5 4 5 length of car 15 ft 180 in. 180 45 The ratio of the length of the scale model to the length of the car is 1 :45 .

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Example. Lesson 7-2 Similar Polygons 1 4 Using Proportions Two cities are 3 in. apart on a map with the scale 2 Lesson Objectives NAEP 2005 Strand: Geometry and Measurement 1 in. 50 mi. Find the actual distance. 1 Identify similar polygons Topics: Transformation of Shapes and Preservation of Let d represent the actual distance. 2 Apply similar polygons Properties; Measuring Physical Attributes . map distance (in ) 5 1 in. actual distance (mi.) 50 mi Local Standards: ______

31 2 1 Substitute. Vocabulary. d 50 Similar ﬁgures have the same shape but not necessarily the same size. Two polygons are d 5 1 50 ()32 Cross-Product Property similar if corresponding angles are congruent and corresponding sides are proportional.

d 5 175 Simplify. The mathematical symbol for similarity is ϳ . The cities are actually 175 miles apart. The similarity ratio is the ratio of the lengths of corresponding sides of similar ﬁgures.

Quick Check. A golden rectangle is a rectangle that can be divided into a square and a rectangle that is 3. Solve each proportion. similar to the original rectangle. 5 20 18 6 a. z 5 3 b. n 1 6 5 n The golden ratio is the ratio of the length to the width of any golden rectangle, about 1.618 : 1. z 0.75 n 3

Example.

1 Understanding Similarity ᭝ABC ᭝XYZ. B Complete each statement. Y a. mЄB A C 78 ЄB Є Y and mЄY 78, so mЄB 78 because congruent angles have the same measure. 42 X Z BC 5 b. YZ XZ 4. Recall Example 4. You want to make a new map with a scale of 1 in. 35 mi. BC AC AC corresponds to XZ , so YZ . Two cities that are actually 175 miles apart are to be represented on your map. XZ What would be the distance in inches between the cities on the new map?

5 inches Quick Check.

AB Є BC m A 42 and YZ XY

130 131 Geometry Lesson 7-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-2

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Examples. Quick Check.

᭝ ᭝ Є Є Є Є Є Є 2 Determining Similarity Determine whether the KL2 2. Sketch XYZ and MNP with X M, Y N, and Z P. parallelograms are similar. Explain. 100 Also, XY 12, YZ 14, ZX 16, MN 18, NP 21, and PM 24. Can Check that corresponding sides are proportional. you conclude that the two triangles are similar? BC1 AB 2 BC 1 JK 5 4 KL 2 120 N 224 4 Y CD 2 DA 1 LM 4 MJ 2 21 18 AD 1 14 12 Corresponding sides of the two parallelograms are proportional. JM 2 Z 16 X PM24 Check that corresponding angles are congruent. ЄB corresponds to Є K , but mЄBm ЄK, so corresponding Yes; corresponding angles are congruent and corresponding sides are angles are not congruent. proportional.

Although corresponding sides are proportional, the 3. Refer to the diagram for Example 3. Find AB. parallelograms are not similar because 9 the corresponding angles are not congruent .

3 Using Similar Figures If ᭝ABC ᭝YXZ, find the value of x. Y Because ᭝ABC ᭝YXZ, you can write and solve A 50 a proportion. x 30 BC AC Corresponding sides are B C XZ YZ 5 12 40 XZ proportional .

12 x 40 Substitute. 50 4. A golden rectangle has shorter sides of length 20 cm. Find the length of the 12 longer sides. x 50 40 Solve for x. about 32.4 cm

x 5 15 Simplify.

4 Using the Golden Ratio The dimensions of a rectangular tabletop are in

the Golden Ratio. The shorter side is 40 in. Find the longer side. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Let ᐍ represent the longer side of the tabletop. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. ᐉ 1.618 40 5 1 Write a proportion using the Golden Ratio . ᐉ 5 64.72 Cross-Product Property

The table is about 65 in. long.

132 133 Geometry Lesson 7-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-2

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Lesson 7-3 Proving Triangles Similar 2 Using Similarity Theorems Explain why the triangles W must be similar. Write a similarity statement. Lesson Objectives NAEP 2005 Strand: Geometry Є > Є 24 YVZ WVX because they are vertical angles. Z 1 Use AA, SAS, and SSS similarity Topic: Transformation of Shapes and Preservation of 18 statements V Properties VY 12 1 VZ 18 1 2 Apply AA, SAS, and SSS similarity VW 5 24 and VX 5 , Local Standards: ______statements 2 36 2 12 36 so corresponding sides are proportional. Y Vocabulary and Key Concepts. X Therefore, ᭝YVZ ᭝WVX Postulate 7-1: Angle-Angle Similarity (AAϳ) Postulate by the Side-Angle-Side Similarity (SAS) Theorem. If two angles of one triangle are congruent to P T two angles of another triangle, then the 3 Using Similar Triangles Joan places a mirror 24 ft from the base of a T triangles are similar. tree. When she stands 3 ft from the mirror, she can see the top of the tree R S L M reﬂected in it. If her eyes are 5 ft above the ground, how tall is the tree? ᭝TRS ᭝ PLM TR represents the height of the tree, Point M represents the mirror, and J point J represents Joan’s eyes. Theorem 7-1: Side-Angle-Side Similarity (SASϳ) Theorem Both Joan and the tree are perpendicular to the ground, 5 If an angle of one triangle is congruent to an angle of a second triangle, so mЄJOM mЄTRM, and therefore lJOM OlTRM . and the sides including the two angles are proportional, then The light reﬂects off the mirror at the same angle at which it hits the O 324M R the triangles are similar. mirror, so ЄJMO Є TMR .

Use similar triangles to ﬁnd the height of the tree. Theorem 7-2: Side-Side-Side Similarity (SSSϳ) Theorem If the corresponding sides of two triangles are proportional, then nJOM , nTRM AA Postulate the triangles are similar. TR RM Corresponding sides of similar triangles are proportional. JO OM Indirect measurement is a way of measuring things that are difﬁcult to measure directly. TR 24 Substitute. 5 3

Examples. 24 TR 5 Solve for TR. ϳ MX ' AB 1 Using the AA Postulate . Explain why the triangles M 3 are similar. Write a similarity statement. Because MX ' AB, ЄAXM and ЄBXK are TR 5 40 right angles , so ЄAXM > Є BXK . K Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

ЄA > Є B because their measures are equal. 58 The tree is 40 feet tall. ᭝AMX ᭝BKX 58 A B by the Angle-Angle Similarity (AA) Postulate. X

134 135 Geometry Lesson 7-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-3

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Quick Check. Lesson 7-4 Similarity in Right Triangles 1. In Example 1, you have enough information to write a similarity statement. Do you have enough information to ﬁnd the similarity ratio? Explain. Lesson Objective NAEP 2005 Strand: Geometry 1 Find and use relationships in similar Topic: Transformation of Shapes and Preservation of No; none of the side lengths are given. right triangles Properties

Local Standards: ______

2. Explain why the triangles at the right must be similar. G Write a similarity statement. Vocabulary and Key Concepts. 8 8 AC 5 CB 5 AB 5 3 EG GF EF 4 , so the triangles are similar by the SSS~ Theorem; Theorem 7-3 nABC , nEFG EF12 The altitude to the hypotenuse of a right triangle divides the triangle into

AB9 two triangles that are similar to the original triangle and similar to each other. 6 6 C Corollary 1 to Theorem 7-3 The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse. 3. In sunlight, a cactus casts a 9-ft shadow.At the same time, Corollary 2 to Theorem 7-3 a person 6 ft tall casts a 4-ft shadow. Use similar triangles to ﬁnd the height of the cactus. The altitude to the hypotenuse of a right triangle separates the hypotenuse so that the length of each leg of the triangle is the geometric mean of the length of the adjacent hypotenuse segment x ft and the length of the hypotenuse .

The geometric mean of two positive numbers a and b is the positive number x such a x that x 5 b . 6 ft Examples. 4 ft 9 ft 1 Finding the Geometric Mean Find the geometric mean of 3 and 12.

3 x 13.5 ft x 5 Write a proportion.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 12 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 2

x 5 !36 Find the positive square root.

x 5 6

The geometric mean of 3 and 12 is 6 .

136 Geometry Lesson 7-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-4 137

34 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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2 Finding Distance At a golf course, Maria drove her ball 192 yd straight Cup Lesson 7-5 Proportions in Triangles y toward the cup. Her brother Gabriel drove his ball straight 240 yd, but not x Lesson Objectives toward the cup. The diagram shows the results. Find x and y, their remaining G M NAEP 2005 Strand: Geometry distances from the cup. 1 Use the Side-Splitter Theorem Topic: Transformation of Shapes and Preservation of Use Corollary 2 of Theorem 7-3 to solve for x. 2 Use the Triangle-Angle-Bisector Properties Theorem x 192 240 Write a proportion. 240 192 Local Standards: ______240 192 2 192 (x 1 192) 5 240 Cross-Product Property Key Concepts. ★ 192x 36,864 57,600 Distributive Property Theorem 7-4: Side-Splitter Theorem 192 x 20,736 Solve for x. If a line is parallel to one side of a triangle and intersects the other two sides, x 5 108 then it divides those sides proportionally. Use Corollary 2 of Theorem 7-3 to solve for y.

x 192 y x Write a proportion. Theorem 7-5: Triangle-Angle-Bisector Theorem y If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that 108 192 y are proportional to the other two sides of the triangle. Substitute 108 for x. y 108

300 y Simplify. Corollary to Theorem 7-4 y 108 If three parallel lines intersect two transversals, then the segments intercepted on the 2 y 32,400 Cross-Product Property transversals are proportional. y !32,400 Find the positive square root. c a 180 a 5 c y bd b d

Maria’s ball is 108 yd from the cup, and Gabriel’s ball is 180 yd from the cup. Examples.

Quick Check. 1 Using the Side-Splitter Theorem Find y. y M 12 C B 1. Find the geometric mean of 15 and 20. 2. Recall Example 2. Find the distance CN 10 CM Side-Splitter between Maria’s ball and Gabriel’s ball. MB Theorem NA N 10"3 144 yd 6

10 y 72 Cross-Product Property

y 7.2 Solve for y.

138 139 Geometry Lesson 7-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-5

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2 Using the Corollary to the Side-Splitter Theorem The segments R S 6 5 Example. joining the sides of trapezoid RSTU are parallel to its bases. Find x and y. x 12.5 3 Using the Triangle-Angle-Bisector Theorem Find the value of x. Because RSTU is a trapezoid,RS 6 UT . 9 y G UT 40 6 5 24 x 5 Corollary to the Side-Splitter Theorem H 12.5 I x K 30

5 x 75 Cross-Product Property IK IG Triangle-Angle-Bisector Theorem. 15 Solve for x. GH x HK 12.5 x 24 9 Corollary to the Side-Splitter Theorem x y Substitute. 40 30 15 12.5 24 9 Substitute 15 for x. ()30 x Solve for x. y 40 x 18 15y 112.5 Cross-Products Property

y 5 7.5 Solve for y. Quick Check. x 5 15 and y 5 7.5 3. Find the value of y. 3.6 5.76 y Quick Check. 5 1. Use the Side-Splitter Theorem to ﬁnd the value of x. 5 8 1.5 x 1.5

x 2.5

2. Solve for x and y. y 225 16.5 x ϭ 13 ; y ϭ 28.6 15 26

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Lesson 8-1 The Pythagorean Theorem and Its Converse Example. 1 Pythagorean Triples Find the length of the hypotenuse of #ABC. Lesson Objectives NAEP 2005 Strand: Geometry B Do the lengths of the sides of #ABC form a Pythagorean triple? 1 Use the Pythagorean Theorem Topic: Relationships Among Geometric Figures 2 2 2 2 Use the Converse of the Pythagorean a 1 b 5 c Use the Pythagorean Theorem. Theorem Local Standards: ______52 122 c2 Substitute5 for a, and12 for b. 12 Vocabulary and Key Concepts. 169 c2 Simplify.

Î 169 c Take the square root. AC Theorem 8-1: Pythagorean Theorem 5 13 c Simplify. In a right triangle, the sum of the squares of the lengths of the legs c a 13 13 is equal to the square of the length of the hypotenuse . The length of the hypotenuse is . The lengths of the sides, 5, 12, and , Pythagorean triple whole a2 b2 c 2 b form a because they are numbers that satisfy a2 1 b2 5 c2. Theorem 8-2: Converse of the Pythagorean Theorem If the square of the length of one side of a triangle is equal to the sum of Quick Check. the squares of the lengths of the other two sides, then the triangle is a 1. A right triangle has a hypotenuse of length 25 and a leg of length 10. Find the right triangle. length of the other leg. Do the lengths of the sides form a Pythagorean triple? 5!21; no

Theorem 8-3 If the square of the length of the longest side of a triangle is greater than B c the sum of the squares of the lengths of the other two sides, the triangle a is obtuse . A C b If c2 a2 b2, the triangle is obtuse .

Theorem 8-4 If the square of the length of the longest side of a triangle is less than B the sum of the squares of the lengths of the other two sides, the triangle a c is acute . A C b If c2 a2 b2, the triangle is acute . © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

A Pythagorean triple is a set of nonzero whole numbers a, b, and c that satisfy the equation a2 ϩ b2 ϭ c2.

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Example. Examples. 2 Using Simplest Radical Form Find the value of b. Leave your 3 Is It a Right Triangle? Is this triangle a right triangle? 4 m answer in simplest radical form 12 2 2 2 8 a2 1 b2 0 c2 a 1 b 5 c Use the Pythagorean Theorem. 42 62 0 72 Substitute 46for a, for b, and 7for c. 7 m 82 b2 122 Substitute8 for a,12 for c. 6 m b 16 360 49 Simplify. 64 b2 144 Simplify. 52 2 49 b2 80 Subract64 from both sides. Because a2 b2 c2, the triangle is not a right triangle. b Î 80 Take the square root. b Î 16 () 5 Simplify. 4 Classifying Triangles as Acute, Obtuse, or Right The numbers 10, 15, and 20 represent the lengths of the sides of a triangle. Classify the triangle b 4!5 as acute, obtuse, or right.

2 2 2 c 0 a 1 b Compare c2 with a2 b2. Quick Check. 202 0 102 152 Substitute the greatest length for c .

2. The hypotenuse of a right triangle has length 12. One leg has length 6. Find the 4000 100 225 Simplify. length of the other leg. Leave your answer in simplest radical form. 400. 325 6!3 Because c2 Ͼ a2 b2, the triangle is a(n) obtuse triangle.

Quick Check.

3. A triangle has sides of lengths 16, 48, and 50. Is the triangle a right triangle? no

4. A triangle has sides of lengths 7, 8, and 9. Classify the triangle by its angles. acute © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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36 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 8-2 Special Right Triangles 2 Applying the 45-45-90 Triangle Theorem The distance from one corner to the opposite corner of a square playground is 96 ft. s Lesson Objectives NAEP 2005 Strand: Geometry To the nearest foot, how long is each side of the playground? 1 Use the properties of 45 -45 -90 Topic: Relationships Among Geometric Figures The distance from one corner to the opposite corner, 96 ft, is the 96 ft triangles length of the hypotenuse of a 45-45-90 triangle. 2 Use the properties of 30-60-90 Local Standards: ______triangles s 96 "2 ? leg hypotenuse "2 ? leg

96 Key Concepts. leg Divide each side by "2 . "2 Theorem 8-5: 45 -45 -90 Triangle Theorem leg 67.882251 Use a calculator. In a 45-45-90 triangle, both legs are congruent and the s 2 45 s Each side of the playground is about 68 ft. length of the hypotenuse is "2 times the length of a leg. 45 "2 ? hypotenuse leg s 3 Using the Length of One Side Find the value of each variable. Use the 30-60-90 Triangle Theorem to ﬁnd the lengths of the legs.

4!3 22? x hypotenuse ? shorter leg 30 Theorem 8-6: 30-60-90 Triangle Theorem 4"3 43 twice In a 30 -60 -90 triangle, the length of the hypotenuse is x Divide each side by 2 . y the length of the shorter leg . The length of the longer leg 2 2s 30 s3 is "3 times the length of the shorter leg . x 2 "3 Simplify. 60 2 ? shorter leg hypotenuse 60 y "3 ? shorter leg 30°-60°-90° Triangle Theorem x "3 ? shorter leg s longer leg y "3 ? 2!3 Substitute 2"3 for shorter leg. y 2 ? "3 ? "3 Simplify. y 6 Examples. The length of the shorter leg is 2"3 , and the length of the longer leg is 6 . 1 Finding the Length of the Hypotenuse Find the value of the variable. Use the 45-45-90 Triangle Theorem to ﬁnd the 45 hypotenuse. Quick Check. h 56 1. Find the length of the hypotenuse of a 45-45-90 triangle with legs of length 5!3 . h "2 ? 5!6 hypotenuse "2 ? leg 5!6 h 5 "12 Simplify. 45 h 5 4 ( 3 ) 56 2. A square garden has sides of 100 ft long. You want to build a brick path along h 5()2 3 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson a diagonal of the All rights reserved. square. How long will the path be? Round your answer to the nearest foot. h 10 "3 141 ft The length of the hypotenuse is 10"3 .

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Example. Lesson 8-3 The Tangent Ratio 4 Applying the 30-60-90 Triangle Theorem A rhombus-shaped 25 ft Lesson Objective NAEP 2005 Strand: Measurement garden has a perimeter of 100 ft and a 60 angle. Find the area of the 1 Use tangent ratios to determine side Topic: Measuring Physical Attributes garden to the nearest square foot. lengths in triangles h 25 Local Standards: ______Because a rhombus has four sides of equal length, each side is 25 ft. 60° Because a rhombus is also a parallelogram, you can use the area formula Vocabulary. A bh . Draw the altitude h, and then solve for h. Є the ratio of the length of the leg opposite The height h is the longer leg of the right triangle. To ﬁnd the height h, you The tangent of acute A in a right triangle is can use the properties of 30-60-90 triangles. Then apply the lA to the length of the leg adjacent to lA . area formula. B 25 2 ? shorter leg hypotenuse 2 ? shorter leg Leg opposite ЄA 25 shorter leg 12.5 Divide each side by 2 . 2 A C Leg adjacent to ЄA h 12.5 "3 longer leg "3 ? shorter leg

A bh Use the formula for the area of a parallelogram . l Є length of leg opposite A tangent of A length of leg adjacent to lA A ()25 ()12.5!3 Substitute 25 for b and 12.5"3 for h. A 541.26588 Use a calculator. opposite You can abbreviate the equation as tan A adjacent . To the nearest square foot, the area is 541 ft2.

Examples. Quick Check. 1 Writing Tangent Ratios Write the tangent ratios for ЄA and ЄB. 3. Find the value of each variable. 8 60 x B x 4, y 4!3 30 y

20 29

C 21 A

opposite BC 20 4. A rhombus has 10-in. sides, two of which meet to form the indicated angle. tan A adjacent AC 21 Find the area of each rhombus. (Hint: Use a special right triangle to ﬁnd height.)

a. a 30 angle b. a 60 angle Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson opposite All rights reserved. AC 21 tan B

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2 Using a Tangent Ratio To measure the height of a tree, Alma walked 125 ft from the tree and measured a 32 angle from the ground to the top Quick Check. of the tree. Estimate the height of the tree. 1. a. Write the tangent ratios for ЄK and ЄJ. J 3 7 7, 3 3 LK7 32 125 ft

The tree forms a right angle with the ground, so you can use the tangent ratio to estimate the height of the tree.

height tan 32 Use the tangent ratio. 125 b. How is tan K related to tan J? They are reciprocals. height 125() tan 32° Solve for height.

125 32 78.108669 Use a calculator. 2. Find the value of w to the nearest tenth. a. b. c. The tree is about 78 feet tall. 10 w w 54 1.0 2.5 28 57 33 3 Using the Inverse of Tangent Find mЄR to the nearest degree. w R

13.8 1.9 3.8 41

S 47 T 47 tan R 41

47 mЄR tan 1 ()41 3. Find mЄY to the nearest degree. P 100 T 47 41 1.14634146 68 Ϫ1 1.14634146 48.900494 41

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2 Using the Cosine Ratio A 20-ft wire supporting a flagpole forms a Sine and Cosine Ratios Lesson 8-4 35 angle with the flagpole. To the nearest foot, how high is the flagpole? 35 Lesson Objective NAEP 2005 Strand: Measurement The flagpole, wire, and ground form a right triangle with the 20 ft 1 Use sine and cosine to determine side Topic: Measuring Physical Attributes lengths in triangles wire as the hypotenuse . Because you know an angle and the Local Standards: ______measure of its hypotenuse, you can use the cosine ratio to find the height of the flagpole. Vocabulary. height The sine of ЄA is the ratio of the length of the leg opposite ЄA to the length of the cos 35 Use the cosine ratio. 20 hypotenuse. height 20 ? cos 35° Solve for height. The cosine of ЄA is the ratio of the length of the leg adjacent to ЄA to the hypotenuse. 20 35 16.383041 Use a calculator.

16 leg opposite lA B The ﬂagpole is about feet tall. sine of ЄA hypotenuse hypotenuse Leg opposite opposite 3 Using the Inverse of Cosine and Sine A right triangle has a leg 1.5 units This can be abbreviated sin A hypotenuse . ЄA long and hypotenuse 4.0 units long. Find the measures of its acute angles to AC leg adjacent to lA the nearest degree. Є Leg cosine of A hypotenuse C adjacent to adjacent Є This can be abbreviated cos A hypotenuse . A 1.5

An identity is an equation that is true for all allowed values of the variable. AB 4.0

Use the inverse of the cosine function to find mЄA. An example of a trigonometric identity is sin xcos(90 x)Њ . 1.5 cos A 0.375 Use the cosine ratio. Examples. 4.0 1 Writing Sine and Cosine Ratios Use the triangle to find sin T, cos T, T mЄA cos 1()0.375 Use the inverse of the cosine. sin G, and cos G. Write your answers in simplest terms. Ϫ1 0.375 67.975687 Use a calculator. opposite 12 3 sin T 16 20 mЄA 68 Round to the nearest degree. hypotenuse 20 5 To find mЄB, use the fact that the acute angles of a right triangle are adjacent 16 4 cos T R G complementary . hypotenuse 20 5 12 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson mЄA mЄB All rights reserved. 90 Deﬁnition of complementary angles opposite 16 4 sin G Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. hypotenuse 68 mЄB 90 Substitute. 20 5 mЄB 22 Simplify. adjacent 12 3 G cos hypotenuse The acute angles, rounded to the nearest degree, measure 68 and 22 . 20 5

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Angles of Elevation Quick Check. Lesson 8-5 and Depression Є Є 1. a. Write the sine and cosine ratios for X and Y. X Lesson Objective NAEP 2005 Strand: Measurement 64 48 48 64 1 Topic: sin X ϭ 80, cos X ϭ 80, sin Y ϭ 80, cos Y ϭ 80 Use angles of elevation and Measuring Physical Attributes 48 80 depression to solve problems Local Standards: ______Z Y 64 Vocabulary.

An angle of elevation is the angle formed by a horizontal line and the line of sight to an object above the horizontal line. An angle of depression is the angle formed by a horizontal line and the line of sight to an b. In general, how are sin X and cos Y related? Explain. object below the horizontal line. sin X ϭ cos Y when lX and lY are complementary. Horizontal line Angle of B Angle of depression elevation A Horizontal line 2. In Example 2, suppose that the angle the wire makes with the ground is 50°. What is the height of the ﬂagpole to the nearest foot?

15 feet 20 ft Examples. 1 Identifying Angles of Elevation and Depression Describe Є1 and Є2 50 as they relate to the situation shown.

3. Find the value of x. Round your answer to the nearest degree. a. b. 1 10 10 6.5 2 x x 3 27

41 68 4

One side of the angle of depression is a horizontal line. Є1 is the angle of

One side of the angle of elevation is a horizontal line. Є2 is the angle of elevation from the building to the airplane .

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2 Surveying A surveyor stands 200 ft from a building to measure its height Top of with a 5-ft tall theodolite. The angle of elevation to the top of the building is building Quick Check. 35 . How tall is the building? Use a diagram to represent the situation. 1. Describe each angle as it relates to the situation in Example 1. x a. Є3 x tan 35 Use thetangent ratio. 35 The angle of depression from the building to the person on the ground 200 Theodolite 5 ft 200 ft x 200 ? tan 35Њ Solve for x.

200 35 140 .041508 Use a calculator. b. Є4 x 140 The angle of elevation from the person on the ground to the building

To ﬁnd the height of the building, add the height of the theodolite, which is 5 feet tall.

The building is about 140 ft 5 ft, or 145 ft tall. 2. You sight a rock climber on a cliff at a 32 angle of elevation. The horizontal ground distance to the cliff is 1000 ft. Find the line-of-sight distance to the 3 Aviation An airplane ﬂying 3500 ft above the ground begins a 2 descent rock climber. to land at the airport. How many miles from the airport is the airplane when Rock it starts its descent? (Note: The angle is not drawn to scale.) climber

2 x 3,500 ft You 32 Ground 1000 ft about 1179 ft Њ 3,500 sin 2 x Use thesine ratio. 3500 x Solve for x. sin 2Њ 3500 2 100287 .9792 Use a calculator. Divide by5280 to convert 100287.9792 5280 18.993935 feet to miles. 3. An airplane pilot sees a life raft at a 26 angle of depression. The airplane’s altitude is 3 km. What is the airplane’s surface distance d from the raft? The airplane is about 19 miles from the airport when it starts its descent. about 6.8 km © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Lesson 8-6 Vectors Examples. y 1 Describing a Vector Describe OM as an ordered pair. Lesson Objectives NAEP 2005 Strand: Geometry 1 Describe vectors Topic: Position and Direction Give coordinates to the nearest tenth. 2 Solve problems that involve vector Use the sine and cosine ratios to ﬁnd the values of x and y. x Local Standards: ______addition 40 O x x y y cos 40 Use sine and cosine. sin 40 80 Vocabulary and Key Concepts. 80 80 M x 80 cos 40 Solve for the variable. y 80 sin 40 Adding Vectors () () x 61.28355545 Use a calculator. y 51.42300878 kx 1 x , y 1 y l For a x1, y1 and c x2, y2 , a c 1 2 1 2 . Because point M is in the third quadrant, both coordinates are negative .To the nearest tenth, OM k261.3, 251.4l . A vector is any quantity with magnitude (size) and direction.

A vector can be represented with an arrow . 2 Describing a Vector Direction A boat sailed 12 mi east and 9 mi south. k12, 29l The magnitude of a vector is its size, or length. The trip can be described by the vector . Use distance and direction to describe this vector a second way. N

Draw a diagram for the situation. The initial point of a vector is the point at which it starts. 12 mi To ﬁnd the distance sailed, use the Distance Formula . W (0, 0) E The terminal point of a vector is the point at which it ends. d ()12 0 2 ()9 0 2 Distance Formula 9 mi d 144 81 Simplify. (12, 9) S A resultant vector is the sum of other vectors. d "225 Simplify. d 15 Take the square root.

The magnitude of vector ON is the distance from its To ﬁnd the direction the boat sails, ﬁnd the angle that the vector forms with the x-axis. initial point O to its terminal point N. The ordered pair 9 notation for the vector is k5, 2l . tan x 0.75 Use the tangent ratio. 12 y 4 x tan 1 (0.75) Find the angle whose tangent is 0.75.

Ϫ1 2 0.75 36.869898 Use a calculator. N © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson The boat sailed All rights reserved. 15 miles at about 37 south of east . 422 O 4 x 2

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y 3 Adding Vectors Vectors v 4, 3 and w 4, 3 are shown at the right. 4 Lesson 9-1 Translations Write the sum of the two vectors as an ordered pair. Then draw s, Lesson Objectives NAEP 2005 Strand: Geometry the sum of v and w. 2 (4, 3) 1 Identify isometries Topic: Transformation of Shapes and Preservation of first To find the first coordinate of s, add the coordinates of 2 Properties; Position and Direction 2468 x Find translation images of figures v and w. 2 (4, 3) Local Standards: ______To find the second coordinate of s, add the second coordinates of v and w. 4 Vocabulary. k l k l s 4, 3 4, 3 A transformation of a geometric figure is a change in its position, shape, or size. y k 4 4 , 3 (3) l Add the coordinates. 4 k l 8 , 0 Simplify. v In a transformation, the preimage is the original image before changes are made. 2 w Draw vector v using the origin as the initial point. Draw vector w 246s8 x In a transformation, the image is the resulting figure after changes are made. using the terminal point of v, (4, 3), as the initial point. Draw the 2 resultant vector s from the initial point of v to An isometry is a transformation in which the preimage and the image are congruent. the terminal point of w. 4

A translation (slide) is a transformation that maps all points the same distance and in the Quick Check. same direction. 1. Describe the vector at the right as an ordered pair. Give the coordinates to y A composition of transformations is a combination of two or more transformations. the nearest tenth. 51 k221.6, 46.2l 65 O x Examples. 1 Identifying Isometries Does the transformation appear to be an isometry?

2. A small airplane lands 246 mi east and 76 mi north of the point from which it Image took off. Describe the magnitude and the direction of its ﬂight vector.

about 257 mi at 17 N of E Preimage

The image appears to be the same as the preimage, but turned . Because the ﬁgures appear to be congruent , the transformation © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 3. Write the sum of the two vectors k2, 3l and k24, 22l as an ordered pair. Hall. Prentice Education, Inc., publishing as Pearson © Pearson appears to be an All rights reserved. isometry. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. k22, 1l

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b. List all pairs of corresponding sides. 2 Naming Images and Corresponding Parts In the diagram, CZ NI and SU; ID and UP; ND and SP ᭝XYZ is an image of ᭝ABC. a. Name the images of ЄB and ЄC. B Y Because corresponding vertices of the preimage and the Є Є image are listed in the same order, the image of B is Y , AX 3. Find the image of ᭝ABC for the translation (x 2, y 1). and the image of ЄC is Є Z . b. List all pairs of corresponding sides. A ( 3, 3), B ( 1, 1), C ( 2, 0) Because corresponding sides of the preimage and the image are listed in the same order, the following pairs are corresponding sides: AB and XY ,AC and XZ ,BC and YZ . Examples. 3 Finding a Translation Image Find the image of ᭝ABC y → under the translation (x, y) (x 2, y 3). A 4 Writing a Rule to Describe a Translation Write a rule to B y ᭝ S ᭝ 4 → 1 2 2 → describe the translation ABC A B C . B A( 1, 2) A (,3) Use the rule (x, y ) (x 2, y 3) B x You can use any point on ᭝ABC and its image on ᭝ABC 2 B(1, 0) → B( 1 2,0 3 ) C A A → to describe the translation. Using A( 4, 1) and its image C C(0, 1) C (,0 2 1 3 ) 4 2 2 x Ϫ ᭝ ᭝ AЈ(2, 0) , the horizontal change is 2 ()4 , or 6 , C The image of ABC is A B C with A (,1 1 ),B (,3 3 ), 2 and the vertical change is 0 1 , or Ϫ1 . and C(,2 4 ). The rule is (x, y) S (x ϩ 6, y Ϫ 1) . 4

Quick Check. 5 Adding Translations Tritt rides his bicycle 3 blocks north and 5 blocks east of a 1. Does the transformation appear to be an isometry? Explain. pharmacy to deliver a prescription. Then he rides 4 blocks south and 8 blocks a. b. west to make a second delivery. How many blocks is he now from the pharmacy? Preimage Preimage Image Image The rule (x, y) → (x ϩ 5, y ϩ 3) represents a ride of 3 blocks north and 5 blocks east.

The rule (x, y) → (x Ϫ 8, y Ϫ 4) represents a ride of 4 blocks south and 8 blocks west. Yes; the figures appear to be Yes; the figures appear to be Tritt’s position after the second delivery is the composition of the two congruent by a flip. congruent by a flip and a slide. translations.

(00 , ) translates to ( 0 5, 0 3 ) or ( 5 ,3 ). Then, ( 5 ,3 ) 2. In the diagram, NID S SUP. translates to ( 5 Ϫ8 , 3 Ϫ4 ) or (Ϫ3 ,Ϫ1 ). a. Name the images of ЄI and point D. U lU, P © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson Tritt is 1 block All rights reserved. south and 3 blocks west of the N P pharmacy. S I

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Quick Check. Lesson 9-2 Reflections 4. Use the rule (x, y) S (x 7, y 1) to ﬁnd the translation image of ᭝LMN. Lesson Objectives NAEP 2005 Strand: Geometry Graph the image ᭝LMN. 1 Find reflection images of figures Topics: Transformation of Shapes and Preservation of y Properties 6 Local Standards: ______4

2 Vocabulary. 6 4 2 246 L N x A reﬂection in line r is a transformation such that if a point A is on line r, then L N 2 the image of A is itself , and if a point B is not on line r, then its image BBr M B is the point such that r is the perpendicular bisector of . 4 M r 6 B A A

Line of 5. Refer to Example 5. Tritt now makes a third delivery. Starting from where reflection B the second delivery was, he goes 5 blocks north and 1 block west. How many blocks is Tritt from the pharmacy? He is 4 blocks north and 4 blocks west of the pharmacy. Example.

1 Finding Reflection Images If point Q(1, 2) is reflected across y x = 1 line x 1, what are the coordinates of its reflection image? Q Q is 2 units to the left of the reflection line, so its image Q is x 2 units to the right of the reflection line. The reflection line is the 22 perpendicular bisector of QQr if Q is at (3, 2) .

Quick Check. 1. What are the coordinates of the image of Q if the reflection line is y 1? (1, 4) © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Examples. Lesson 9-3 Rotations 2 Drawing Reflection Images ᭝XYZ has vertices X(0, 3), Y(2, 0), y Lesson Objective NAEP 2005 Strand: Geometry and Z(4, 2). Draw ᭝XYZ and its reflection image in the line x 4. x 4 4 ‚ 1 Draw and identify rotation images of Topic: Transformation of Shapes and Preservation of First locate vertices X, Y, and Z and draw ᭝XYZ in a coordinate X ‚ X ZZ figures Properties plane. 2 Local Standards: ______Locate points X, Y, and Z such that the line of reflection x 4 O ‚ Y Y x is the perpendicular bisector of XXr, YYr, and ZZr . 268 2 Vocabulary. Draw the reflection image XYZ. A rotation of x about a point R is a transformation for which the following are true: V 3 Congruent Angles D is the reflection of D across ᐉ. D • The image of R is itself (that is, R R ). R Show that PD and PW form congruent angles with line ᐉ. x W R • For any point V, RVRV and mЄVRV x . Because a reflection is an isometry ,PD and V PDr form congruent angles with line ᐉ. The center of a regular polygon is a point that is equidistant P ഞ vertices PDr and PW also form congruent angles with line ᐉ from the polygon’s . because vertical angles are congruent. PD PW ᐉ Therefore, and form congruent angles with line Examples. by the Transitive Property. D 1 Drawing a Rotation Image Draw the image of ᭝LOB under a 60 rotation about C. Step 1 Use a protractor to draw a 60 angle at vertex C with one side CO . Quick Check. Step 2 Use a compass to construct COr CO. 2. ᭝ABC has vertices A(3, 4), B(0, 1), and C(2, 3). Draw ᭝ABC and its Step 3 Locate LЈ and BЈin a similar manner. Then draw ᭝LЈOЈBЈ. reflection image in the line x 3. y 8 O C O C O C A A 60° 4 C C O B B 8 4 4812x L L L 4 B B B 8 O C O C O C 3. In Example 3, what kind of triangle is ᭝DPD? Imagine the image of point W ᐉ ᭝ ᭝ reflected across line . What can you say about WPW and DPD ? B B O © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson O All rights reserved. O isosceles; they are similar by SAS L L L

B B B L L L

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2 Identifying a Rotation Image Regular hexagon ABCDEF is divided A B .Examples. into six equilateral triangles. 3 Finding an Angle of Rotation A regular 12-sided polygon can be formed a. Name the image of B for a 240 rotation about M. F C by stacking congruent square sheets of paper rotated about the same center Because 360 6 60 , each central angle of ABCDEF measures M on top of each other. Find the angle of rotation about M that maps W to B. W B 60° . A 240 counterclockwise rotation about center M moves point B E D Consecutive vertices of the three squares form the outline of a regular M across four triangles. The image of point B is point D . 12-sided polygon. Њ b. Name the image of M for a 60° rotation about F. 360 12 30 , so each vertex of the polygon is a 30 rotation K ᭝AMF is equilateral, so ЄAFM has measure 180 3 60 . A 60 about point M. rotation of ᭝AMF about point F would superimpose FM on FA , so the You must rotate counterclockwise through 7 vertices to map point W to point B, so the angle of rotation is 7 ? 30 210 . image of M under a 60 rotation about point F is point A .

4 Compositions of Rotations Describe the image of quadrilateral XYZW for a composition of a 145 rotation and then a 215 rotation, both about Quick Check. point X. 1. Draw the image of LOB for a 50 rotation about point B. Label the vertices Two rotations of 145 and 215 about the same point make a total rotation of of the image. 145215, or 360 . Because this forms a complete rotation about point X, the image is the preimage XYZW . O Quick Check. L 3. In the ﬁgure from Example 3, ﬁnd the angle of rotation about M that maps O BB B to K. 270° L

4. Draw the image of the kite for a composition rotation of two 90 rotations about point K. 2. Regular pentagon PENTA is divided into 5 congruent triangles. Name the P image of T for a 144 rotation about point X. E A X E © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Examples. Lesson 9-4 Symmetry 1 Identifying Lines of Symmetry Draw all lines of symmetry for the Lesson Objective NAEP 2005 Strand: Geometry isosceles trapezoid. 1 Identify the type of symmetry in Topic: Transformation of Shapes and Preservation of a ﬁgure Properties

Local Standards: ______Draw any lines that divide the isosceles trapezoid so that half of the ﬁgure is a mirror image of the other half.

Vocabulary.

A ﬁgure has symmetry if there is an isometry that maps the ﬁgure onto itself.

There is one line of symmetry. A figure has reflectional symmetry if there is a reflection that maps the figure onto itself. 2 Identifying Rotational Symmetry Judging from appearance, do the Line symmetry is the same as reflectional symmetry. letters V and H have rotational symmetry? If so, give an angle of rotation.

The letter V does not have rotational symmetry because it must be rotated 360 before it is its own image. Line of Symmetry The letter H is its own image after one half-turn, so it has rotational symmetry with a 180 angle of rotation.

3 Finding Symmetry A nut holds a bolt in place. Some nuts have square The heart-shaped figure has reflectional faces, like the top view shown below. Tell whether the nut has rotational ( line ) symmetry. symmetry and/or reﬂectional symmetry. Draw all lines of symmetry.

A ﬁgure has rotational symmetry if it is its own image for some rotation of 180Њ or less.

A ﬁgure has point symmetry if it has 180Њ rotational symmetry.

The ﬁgure is its own image after one half-turn, so it has rotational symmetry with a 180 angle of rotation. The ﬁgure also has point symmetry. The nut has a square outline with a circular opening. The square and circle are concentric.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson The nut is its own All rights reserved. image after one quarter-turn, so it has 90 rotational symmetry. The nut has 4 lines of symmetry.

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Quick Check. Lesson 9-5 Dilations 1. Draw a rectangle and all of its lines of symmetry. Lesson Objective NAEP 2005 Strand: Geometry 1 Locate dilation images of figures Topic: Transformation of Shapes and Preservation of Properties

Local Standards: ______

Vocabulary. A dilation is a transformation with center C and scale factor n for which the following are true: • The image of C is itself ) (that is, C C ). • For any point R, R is on CR and CRnCR . R R C 1 3 RrQr is the image of RQ under a dilation Q 2. a. Judging from appearance, tell whether the figure at the right has with center C and scale factor 3 . rotational symmetry. If so, give the angle of rotation. Q yes; 180° An enlargement is a dilation with a scale factor greater than 1. b. Does the figure have point symmetry? yes A reduction is a dilation with a scale factor less than 1. 3. Tell whether the umbrella has rotational symmetry about a line and/or reflectional symmetry.

Examples. 1 Finding a Scale Factor Circle A with 3-cm diameter and center C is a dilation of concentric circle B with 8-cm diameter. Describe the dilation. The circles are concentric, so the dilation has center C . Because the diameter of the dilation image is smaller, the dilation is a reduction . The umbrella has both rotational and reﬂectional symmetry. diameter of dilation image 5 3 Scale factor: diameter of preimage 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 3

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. The dilation is a reduction with center C and scale factor 8 .

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2 Scale Drawings The scale factor on a museum’s floor plan is 1 : 200. The Quick Check. length and width of one wing on the drawing are 8 in. and 6 in. Find the 1. Quadrilateral JKLM is a dilation image of quadrilateral JKLM. actual dimensions of the wing in feet and inches. Describe the dilation. 1 The floor plan is a reduction of the actual dimensions by a scale factor of . 200 y J Multiply each dimension on the drawing by 200 to find the actual J dimensions. Then write the dimensions in feet and inches. K K 224O x 8 in. 200 1600 in. 133 ft,4 in. M M L 3 6 in. 200 1200 in. 100 ft L

The museum wing measures 133 ft, 4 in. by 100 ft . 1 The dilation is a reduction with center (0, 0) and scale factor 2 . 3 Graphing Dilation Images ᭝ABC has vertices A(2, 3), B(0, 4), and C(6, 12). What are the coordinates of the image of ᭝ABC for a dilation with center (0, 0) and scale factor 0.75? 2. The height of a tractor-trailer truck is 4.2 m. The scale factor for a model 1 A(2, 3) → A(0.75 ? (2) , 0.75 (3) ) The scale factor is 0.75 , so use truck is 54 . Find the height of the model to the nearest centimeter.

B(0, 4) → B(0.75 0 , 0.75 4 ) the rule (x, y) → (0.75x, 0.75y) . 8 cm

C(6, 12) → C(,0.75 6 0.75 (12) )

The vertices of the reduction image of ᭝ABC are A (Ϫ1.5, Ϫ2.25) , B (0, 3) , and C (4.5, Ϫ9) .

1 3. Find the image of ᭝PZG for a dilation with center (0, 0) and scale factor 2 . Draw the reduction on the grid and give the coordinates of the image’s vertices. y 2

1 Z Z P 2 1 1P 2 x 1 G © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 2 G

a2 1, 1b a1, 21b Vertices: P (1, 0) , Z 2 4 , and G 2

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Lesson 9-6 Compositions of Reflections Examples. 1 Composition of Reflections in Intersecting Lines The letter D is Lesson Objectives NAEP 2005 Strand: Geometry reflected in line x and then in line y. Describe the resulting rotation. 1 Use a composition of reflections Topic: Transformation of Shapes and Preservation Find the image of D through a reflection across line x. 2 Identify glide reflections of Properties Find the image of the reflection through another reflection across line y. Local Standards: ______D D 43 Vocabulary and Key Concepts. y A D

Theorem 9-1 A translation or rotation is a composition of two reﬂections . x

The composition of two reflections across intersecting lines is a rotation . Theorem 9-2 The center of rotation is the point where the lines intersect , and the angle A composition of reflections across two parallel lines is a translation . is twice the angle formed by the intersecting lines . So, the letter D Theorem 9-3 is rotated 86 clockwise, or 274 counterclockwise, with the center of A composition of reflections across two intersecting lines is a rotation . rotation at point A .

2 Finding a Glide Reflection Image ᭝ABC has vertices A(4, 5), B(6, 2), and C(0, 0). Find the image of ᭝ABC for a glide reflection where the Theorem 9-4: Fundamental Theorem of Isometries translation is (x, y) → (x, y 2) and the reflection line is x 1. In a plane, one of two congruent figures can be mapped onto the other by a A y A composition of at most three reflections . 6 A First, translate ᭝ABC by (x, y) → (x, y 2). x 1 B Theorem 9-5: Isometry Classification Theorem 4 B (4, 5) S (4 0 ,5 2 ), or (4 ,7 ) There are only four isometries. They are the following: B 2 S C C (6,2) (6 0 ,2 2 ), or (64 , ) (0, 0) S (0 0 ,0 2 ), or (02 , ) Reflection Translation Rotation Glide reflection 4 2 C 246x Then, reflect the translated image across the line x 1 . 2 R R R R R

R R R R The glide reﬂection image ᭝ABC has vertices A (6, 7) , B (Ϫ4, 4) , and C (2, 2) .

A glide reﬂection is the composition of a glide (translation) and a reﬂection across a line Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Quick Check. Lesson 9-7 Tessellations 1. a. Reﬂect the letter R across a and then b. Describe the resulting rotation. a Lesson Objectives NAEP 2005 Strand: Geometry Answers may vary. The result is a clockwise rotation about point c 1 Identify transformation in tessellations, Topic: Geometry R

through an angle of 2mlacb . R b and ﬁgures that will tessellate 2 Identify symmetries in tessellations Local Standards: ______R c Vocabulary and Key Concepts.

Theorem 9-6 b. Use parallel lines ᐉ and m. Draw R between ᐉ and m. Find the image of Every triangle tessellates. R for a reﬂection across line ᐉ and then across line M. Describe the resulting translation. R R Theorem 9-7 Answers may vary. The result is a translation twice the distance R Every quadrilateral tessellates. between ᐉ and m.

A tessellation, or tiling, is a repeating pattern of ﬁgures that completely covers a plane, ഞ m without gaps or overlaps. Translational symmetry is the type of symmetry for which there is a translation that maps a

2. a. Find the image of ᭝TEX under a glide reflection y figure onto itself. where the translation is (x, y) → (x 1, y) and the Glide reflectional symmetry is the type of symmetry for which there is a glide reflection that 4 reflection line is y 2. Draw the translation first, E E maps a figure onto itself. T then the reflection. T 2 b. Would the result of part (a) be the same if you reflected X X x ᭝TEX first, and then translated it? Explain. 6 4 2 O 246 Examples. 2 1 Determining Figures That Will Tessellate Determine whether a regular yes; Order of steps does not matter in a glide y 2 reflection. 15-gon tessellates a plane. Explain. 4 X Because the figures in a tessellation do not overlap or leave gaps, the sum of T 6 the measures of the angles around any vertex must be 360 . Check E to see whether the measure of an angle of a regular 15-gon is a factor of 360.

180 ( n 2 ) a Use the formula for the measure of an angle of a regular polygon. n

180 ( 15 2)

a 156 Simplify.

Because 156 is not a factor of 360, a regular 15-gon will not tessellate a plane.

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2 Identifying the Transformation in a Tessellation Identify the 2. Identify a transformation and outline the smallest repeating ﬁgure in each repeating ﬁgures and a transformation in the tessellation. tessellation below. a. A repeated combination of an octagon and one adjoining square will completely cover the plane without gaps or overlaps. Use arrows to show a translation.

3 Identifying Symmetries in Tessellations List the symmetries in the tessellation.

Starting at any vertex, the tessellation can be mapped onto itself using a 180Њ rotation, so the tessellation has point symmetry centered at any vertex.

translation The tessellation also has translational symmetry, as can be seen by sliding any triangle b. onto a copy of itself along any of the lines.

Quick Check. 1. Explain why you can tessellate a plane with an equilateral triangle.

The interior angles of an equilateral triangle measure 60Њ. Because 60 divides evenly into 360, it will tessellate. Answers may vary. Sample: rotation

3. List the symmetries in the tessellation.

line symmetry, rotational symmetry, glide reﬂectional symmetry, translational symmetry © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Lesson 10-1 Areas of Parallelograms and Triangles Examples. 1 Finding a Missing Dimension A parallelogram has 9-in. and 18-in. sides. Lesson Objectives NAEP 2005 Strand: Measurement The height corresponding to the 9-in. base is 15 in. Find the height 1 Find the area of a parallelogram Topic: Measuring Physical Attributes corresponding to the 18-in. base. 2 Find the area of a triangle Local Standards: ______First ﬁnd the area of the parallelogram using the 9-in. base and its corresponding 15-in height.

Vocabulary and Key Concepts. A bh Area of a parallelogram A 915()Substitute 9for b and 15for h. Theorem 10-1: Area of a Rectangle A 135 Simplify. The area of a rectangle is the product of h The area of the parallelogram is 135 in.2 its base and height . b A bh Use the area to ﬁnd the height corresponding to the 18-in. base. A bh Area of a parallelogram Theorem 10-2: Area of a Parallelogram 135 18 h Substitute 135 for A and 18 for b. The area of a parallelogram is the product of a h baseand the corresponding height . 135 h Divide each side by 18 . b 18 A bh 7.5 h Simplify. Theorem 10-3: Area of a Triangle The height corresponding to the 18-in base is 7.5 in. The area of a triangle is half the product of h a base and the corresponding height . 2 Finding the Area of a Triangle Find the area of ᭝XYZ. 1bh b 1 A 2 A 5 bh Area of a triangle 2 X 1 A 5 2 30 13 Substitute 30 for b and 13 for h. ()() 31 cm 13 cm A base of a parallelogram is any of its sides. A 5 195 Simplify. 2 The altitude of a parallelogram corresponding to a given base is Altitude ᭝XYZ has area 195 cm . Base YZ the segment perpendicular to the line containing that base 30 cm drawn from the side opposite the base. Quick Check. 1. A parallelogram has sides 15 cm and 18 cm. The height corresponding to a 15-cm base is 9 cm. Find the height corresponding to an 18-cm base. The height of a parallelogram is the length of its altitude. 7.5 cm © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 9 cm A base of a triangle is any of its sides. 18 cm The height of a triangle is the length of the altitude to the line h containing that base. b 15 cm

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Example. Lesson 10-2 Areas of Trapezoids, Rhombuses, and Kites 3 Structural Design The front of a garage is a square 15 ft on each side with a triangular roof above the square. The height of the triangular roof Lesson Objectives NAEP 2005 Strand: Measurement is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h 1 Find the area of a trapezoid Topic: Measuring Physical Attributes 2 Find the area of a rhombus or a kite wind blowing directly against the front of the garage? Use the formula Local Standards: ______F 0.004Av 2, where F is the force in pounds, A is the area of the surface in square feet, and v is the velocity of the wind in miles per hour. Use the area formulas for rectangles and triangles to ﬁnd the area of the Vocabulary and Key Concepts. front of the garage. 10.6 ft Theorem 10-4: Area of a Trapezoid Area of the square: bh 15 2 225 ft2 15 ft The area of a trapezoid is half the product of the height and the sum of b1 1bh 1 2 Area of the triangle: 2 2 ()()1510.6 79.5 ft the bases. 15 ft h The total area of the front of the garage is 1 A 2h(b1 1 b2) b 225 79.5 304.5 ft2. 2 Theorem 10-5: Area of a Rhombus or a Kite Find the force of the wind against the front of the garage. The area of a rhombus or a kite is half the product of the lengths of its 2 d d F 0.004Av Use the formula for force. diagonals. 1 2 2 F 0.004()()304.580Substitute 304.5 for A and 80 for v. 1 A 2d1d2 F 7795.2 Simplify. F 7800 Round to the nearest hundred. the perpendicular An 80 mi/h wind exerts a force of about 7800 lb against the front The height of a trapezoid is Base of the garage. distance h between the bases. Leg b1 Leg h Quick Check. b2 2. Find the area of the triangle. 13 cm Base 5 cm 30 cm2 12 cm Examples. 1 Applying the Area of a Trapezoid A car window is 20 in. shaped like the trapezoid shown. Find the area of the window. 3. Critical Thinking Suppose the bases of the square and triangle in Example 3 18 in. are doubled to 30 ft, but the height of each ﬁgure stays the same. How is the force of the wind against the building affected? The force is doubled. 36 in. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson 1 All rights reserved. A 5 2h(b1 1 b2) Area of a trapezoid © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A 5 1 2()(18 20 36) Substitute 18 for h,20 for b1, and 36 for b2. A 5 504 Simplify.

The area of the car window is 504 in.2.

184 Geometry Lesson 10-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-2 185

46 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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2 Finding Area Using a Right Triangle Find the area of AB11 ft Examples. trapezoid ABCD. First draw an altitude from vertex B to DC 3 Finding the Area of a Kite Find the area of kite XYZW. that divides trapezoid ABCD into a rectangle and a right 1 cm Y 13 ft 3 cm triangle. The altitude will meet DC at point X. Find the lengths of the diagonals of kite XYZW. XZ Because opposite sides of rectangle ABXD are congruent, XZ 5 d1 5 3 1 3 5 6 and YW 5 d2 5 141 5 5 1 3 cm DX 11 ft and XC 16 ft 11 ft 5 ft. DC16 ft X A 5 2 d1 d2 Use the formula for the area of a kite. By the Pythagorean Theorem, BX2 XC2 BC2, A 5 1 2 65 Substitute 6 for d1 and 5 for d2. ()() 4 cm 2 132 52 144 so BX . A 5 15 Simplify. W Taking the square root, BX 12 ft. You may remember The area of kite XYZW is 15 cm2. that 5, 12, 13 is a Pythagorean triple . R S A 5 1 4 Finding the Area of a Rhombus Find the area of rhombus RSTU. 2 h ( b1 b2 ) Use the trapezoid area formula. To ﬁnd the area, you need to know the lengths of both diagonals. A 5 1 24 ft 2()(12 11 16) Substitute 12for h, 11 for b1, and 16 for b2. Draw diagonal SU , and label the intersection of the diagonals point X. 13 ft A 5 162 Simplify. ᭝SXT is a right triangle because the diagonals of a rhombus X 2 The area of trapezoid ABCD is 162 ft . are perpendicular. UT The diagonals of a rhombus bisect each other, so TX 12 ft. You can use the Pythagorean triple 5, 12, 13 or the Pythagorean Theorem Quick Check. to conclude that SX 5 ft. 1. Find the area of a trapezoid with height 7 cm and bases 12 cm and 15 cm. SU 10 ft because the diagonals of a rhombus bisect each other. 94.5 cm2 1 A 5 2 d1 d2 Area of a rhombus 1 A 5 24 2 ()()24 10 Substitute for d1 and 10 for d2. A 5 120 Simplify. The area of rhombus RSTU is 120 ft2.

Quick Check. 3. Find the area of a kite with diagonals that are 12 in. and 9 in. long. 2. In Example 2, suppose BX and BC change so that mЄC 60 while bases 54 in2 and angles A and D are unchanged. Find the area of trapezoid ABCD.

67.5!3 or 116.9 ft2

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 4. Critical Thinking All rights reserved. In Example 4, explain how you can use a Pythagorean triple to conclude that XU 5 ft.

52 ϩ 122 ϭ 132

186 187 Geometry Lesson 10-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-2

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Lesson 10-3 Areas of Regular Polygons 2 Finding the Area of a Regular Polygon A library is in the shape 18.0 ft of a regular octagon. Each side is 18.0 ft. The radius of the octagon is Lesson Objective NAEP 2005 Strand: Measurement 23.5 ft. Find the area of the library to the nearest 10 ft2. 1 Find the area of a regular polygon Topic: Measuring Physical Attributes Consecutive radii form an isosceles triangle, so an apothem bisects 23.5 ft the side of the octagon. Local Standards: ______1 To apply the area formula A 2ap, you need to ﬁnd a and p. Step 1 Find the apothem a. Vocabulary and Key Concepts. 9.0 ft 9.0 ft a2 ()9.0 2 ()23.5 2 Pythagorean Theorem Theorem 10-6: Area of a Regular Polygon a2 81 552.25 Solve for a. The area of a regular polygon is half the product of the apothem and the a2 471.25 perimeter. a a < 21.7 1ap p Step 2 Find the perimeter p. A 2 p ns Find the perimeter, where n the number of sides of a regular polygon. The center of a regular polygon is the center of the circumscribed circle. p (8)()18.0 144Substitute 8 for n and 18.0 for s, and simplify. The radius of a regular polygon is the distance from the center to a vertex. Step 3 Find the area A. The apothem of a regular polygon is the perpendicular distance from the center to a side. A 1 ap Area of a regular polygon 2 A 1 21.7 144Substitute 21.7 for a and 144 for p. 2 ()() Center A 1562.4 Simplify. Radius To the nearest 10 ft2, the area is 1560 ft2.

Apothem 3 Applying Theorem 10-6 Find the area of an equilateral triangle with apothem 8 cm. Leave your answer in simplest radical form. This equilateral triangle shows two radii forming an angle that measures Examples. 360 s 3 120. Because the radii and a side form an isosceles triangle, 1 Finding Angle Measures This regular hexagon has an apothem and the apothem bisects the 120 angle, forming two 60Њ angles. 60 radii drawn. Find the measure of each numbered angle. You can use a 30-60-90 triangle to ﬁnd half the length of a side. 8 cm 60 360 1 Є 60 s 5 ? ? m 1 Divide 360 by the number of sides. 1 2 "3 a longer leg "3 shorter leg s 6 2 1 s 5 ? 2 "3 8 Substitute 8 for a. 1 The apothem bisects the vertex angle of 3 mЄ2 mЄ1 s 5 16 "3 Multiply each side by 2 . 2 the isosceles triangle formed by p 5 ns Find the perimeter. the radii. p 5 ()(3 16 "3 ) 48 "3 Substitute 3 for n and 16!3 for s, and simplify. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 1 All rights reserved. 1 A 5 ap Area of a regular polygon 2 Є 60 30 Substitute 60 for mЄ1. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson m 2 () All rights reserved. 1 2 A 5 848"3 Substitute 8 for a and 48!3 for p. 2 ()( ) mЄ3 180 ( 90 30 ) 60 The sum of the measures of the angles of a triangle is180 . A 5 192 "3 Simplify. 2 mЄ1 60 , mЄ2 30 , and mЄ3 60 The area of the equilateral triangle is 192"3 cm .

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Quick Check. Lesson 10-4 Perimeters and Areas of Similar Figures 1. At the right, a portion of a regular octagon has radii and an apothem drawn. Lesson Objective NAEP 2005 Strand: Measurement and Number Properties Find the measure of each numbered angle. 1 Find the perimeters and areas of and Operations ml1 5 45; ml2 5 22.5; ml3 5 67.5 1 similar ﬁgures Topics: Systems of Measurement; Ratios and Proportional 2 Reasoning 3 Local Standards: ______

Key Concepts.

Theorem 10-7: Perimeters and Areas of Similar Figures a If the similarity ratio of two similar ﬁgures is b , then a (1) the ratio of their perimeters is b and

a2 2. Find the area of a regular pentagon with 11.6-cm sides and an 8-cm (2) the ratio of their areas is b2 . apothem.

232 cm2 Examples. 1 Finding Ratios in Similar Figures The triangles at the right are 4 similar. Find the ratio (larger to smaller) of their perimeters and of 6 their areas. 5 6.25 The shortest side of the left-hand triangle has length 4 , and 5 the shortest side of the right-hand triangle has length 5 . 7.5 5 From larger to smaller, the similarity ratio is 4 . By the Perimeters and Areas of Similar Figures Theorem, the ratio of 2 the perimeters is 5 , and the ratio of the areas is 5 , or 25 . 4 42 16 3. The side of a regular hexagon is 16 ft. Find the area of the hexagon. 2 Finding Areas Using Similar Figures The ratio of the length of the 384"3 ft2 8 corresponding sides of two regular octagons is 3 . The area of the larger octagon is 320 ft2. Find the area of the smaller octagon. All regular octagons are similar. Because the ratio of the lengths of the corresponding sides of the regular 8 2 octagons is , the ratio of their areas is 8 , or 64 . 3 32 9

64 A 2880 Use the Cross-Product Property. A 45 Divide each side by 64 . The area of the smaller octagon is 45 ft2.

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3 Using Similarity Ratios Benita plants the same crop in two rectangular 1 fields. Each dimension of the larger field is 32 times the dimension of the Example. smaller field. Seeding the smaller field costs $8. How much money does 4 Finding Similarity and Perimeter Ratios The areas of two similar seeding the larger field cost? pentagons are 32 in.2 and 72 in.2. What is their similarity ratio? What is the The similarity ratio of the fields is 3.5 : 1, so the ratio of the areas of the ratio of their perimeters? fields is (3.5)2 : (1)2, or 12.25 to 1 . Find the similarity ratio a : b.

Because seeding the smaller ﬁeld costs $8, seeding 12.25 times as much a2 32 The ratio of the areas is a2 :b2 . 72 land costs 12.25 $8 . b2

Seeding the larger ﬁeld costs $98 . 2 a 16 Simplify. b2 36

Quick Check. a 4 2 Take the square root. 1. Two similar polygons have corresponding sides in the ratio 5 : 7. b 6 3 a. Find the ratio of their perimeters. b. Find the ratio of their areas. 5 : 7 25 : 49 The similarity ratio is 2 : 3 . By the Perimeters and Areas of Similar Figures Theorem,

the ratio of the perimeters is also 2 : 3 . 3 2. The corresponding sides of two similar parallelograms are in the ratio 4 . The area of the larger parallelogram is 96 in.2. Find the area of the smaller parallelogram. Quick Check. 2 2 54 in.2 4. The areas of two similar rectangles are 1875 ft and 135 ft . Find the ratio of their perimeters. 5!5 : 3

3. The similarity ratio of the dimensions of two similar pieces of window glass is 3 : 5. The smaller piece costs $2.50. What should be the cost of the larger piece? $6.94 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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48 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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1 A 2 18(cos 36) 180(sin 36) Substitute for a and p. Lesson 10-5 Trigonometry and Area A 1620 (cos 36) (sin 36) Simplify. NAEP 2005 Strand: Lesson Objectives Measurement A 770.355778 Use a calculator. 1 Find the area of a regular polygon Topic: Measuring Physical Attributes using trigonometry The area is about 770 ft2. 2 Find the area of a triangle using Local Standards: ______trigonometry 2 Surveying A triangular park has two sides that measure 200 ft and 300 ft Key Concepts. and form a 65 angle. Find the area of the park to the nearest hundred square feet. Use Theorem 10-8: The area of a triangle is one half the Theorem 10-8: Area of a Triangle Given SAS 300 ft product of the lengths of two sides and the sine of the included The area of a triangle is one half the product of B angle. 65 the lengths of two sides and a 200 ft c 1 the sine of the included angle . Area 2 side length side length sine of included angle Theorem 9-1 A C 1 1 b Area 2 200 300 sin 65 Substitute. Area of ᭝ABC 2bc(sin A) Area 30,000 sin 65 Substitute.

27189.23361 Use a calculator.

2 Examples. The area of the park is approximately 27,200 ft . 1 Finding Area The radius of a garden in the shape of a regular pentagon is 18 feet. Find the area of the garden. Quick Check. Find the perimeter p and apothem a, and then ﬁnd the area using the C 1. Find the area of a regular octagon with a perimeter of 80 in. Give the area 1 formula A ϭ ap . to the nearest tenth. 2 18 ft 482.8 in.2 360 Because a pentagon has five sides, mЄACB 72 . 5 A MB

CA and CB are radii,so CA CB . Therefore, ᭝ACM ᭝BCM by 1 the HL Theorem , so mЄACM 2mЄACB 36 . 2. Two sides of a triangular building plot are 120 ft and 85 ft long. They include Use the cosine ratio to find a. Use thesine ratio to ﬁnd AM. an angle of 85 . Find the area of the building plot to the nearest square foot. a cos AM 5081 ft2 36 18 Use the ratio. sin 36 18 a 18(cos 36) Solve. AM 18(sin 36)

Use AM to ﬁnd p. Because ᭝ACM ᭝BCM, AB 2 ? AM . Because Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. the pentagon is regular, p 5 ? AB .

So p 5(2 ? AM ) 10 ? AM 10 ? 18(sin 36) 180(sin 36) . 1 Finally, substitute into the area formula A 2ap.

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Lesson 10-6 Circles and Arcs R R R Lesson Objectives NAEP 2005 Strand: Measurement S S S 1 Find the measures of central angles Topic: Measuring Physical Attributes T P T P T P and arcs 2 Find circumference and arc length Local Standards: ______TRS is a semicircle . RS is a minor arc . RTS is a major arc . mTRS 180 mRS mЄRPS mRTS 360 mRS Vocabulary and Key Concepts. A semicircle is half of a A minor arc is smaller than A major arc is greater circle. a semicircle. than a semicircle. Postulate 10-1: Arc Addition Postulate The measure of the arc formed by two adjacent arcs is the sum of the B C Adjacent arcs are arcs of the same circle that have exactly one point in common. measures of the two arcs. 1 0 0 ABC mAB 1 mBC A m Concentric circles are circles that lie in the same plane and have the same center.

Theorem 10-9: Circumference of a Circle a fraction of a circle’s circumference. The circumference of a circle is p times the diameter . Arc length is d r Congruent arcs are arcs that have the same measure and are in the same circle or in C pd or C 2pr O congruent circles. Theorem 10-10: Arc Length C The length of an arc of a circle is the product of the ratio A Example. r 0 1 measure of the arc 1 Finding the Measures of Arcs Find mXY and mDXM in circle C. and the circumference of the circle . O B 360 0 0 0 mXY mXD mDY Arc Addition Postulate 0 M 0 mAB AB ? 2pr 0 0 The measure of a minor arc is the measure length of 360 mXY m/ XCD mDY of its corresponding central angle. 0 Y mXY 56 40 Substitute. 0 40 A circle is the set of all points equidistant from a given point called the center. mXY C W 96 Simplify. D 56 1 0 1 mDXM mDX mXWM Arc Addition Postulate A center of a circle is the point from which all points are equidistant. 1 X mDXM 56 180 Substitute. 1 A radius is a segment that has one endpoint at the center and the other endpoint on the mDXM 236 Simplify. circle. Congruent circles have congruent radii. Quick Check. 1 0 1 A diameter is a segment that contains the center of a circle and has both endpoints on the Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 1. Use the diagram All rights reserved. in Example 1. Find mЄYCD,mYMW ,mMW , and mXMY.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson circle. All rights reserved. 40; 180; 96; 264 A central angle is an angle whose vertex is the center of the circle. Circumference of a circle is the distance around the circle. Pi (π) is the ratio of the circumference of a circle to its diameter.

196 197 Geometry Lesson 10-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-6

49 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Examples. Lesson 10-7 Areas of Circles and Sectors

2 Applying Circumference A circular swimming pool 16 feet in Lesson Objective NAEP 2005 Strand: Measurement diameter will be enclosed in a circular fence 4 ft from the pool. What 1 Find the areas of circles, sectors, and Topic: Measuring Physical Attributes length of fencing material is needed? Round your answer to the next segments of circles POOL whole number. Local Standards: ______The pool and the fence are concentric circles. The diameter of the pool 4 ft16 ft 4 ft is 16 ft, so the diameter of the fence is 16 4 424ft. Use Vocabulary and Key Concepts. the formula for the circumference of a circle to ﬁnd the length of fencing material needed. Theorem 10-11: Area of a Circle The area of a circle is the product of p and the square of the radius . r C pd Formula for the circumference of a circle A pr 2 O C π 24 Substitute. C < 3.14 24 Use 3.14 to approximate π. Theorem 10-12: Area of a Sector of a Circle () measure of the arc < 75.36 The area of a sector of a circle is the product of the ratio A C Simplify. 360 r B About 76 ft of fencing material is needed. and the area of the circle. . O 1 0 B mAB ? pr2 3 Finding Arc Length Find the length of ADB in circle M in terms of π. Area of sector AOB 360 0 Because mAB 150, 1 18 cm A sector of a circle is a region bounded by two radii and their mADB 360 150 210 . Arc Addition Postulate 150 M Segment 1 intercepted arc. 1 mADB of a circle length of ADB ? 2 pr Arc Length Postulate A D 360 A segment of a circle is the part bounded by an arc and the segment 210 joining its endpoints. 2π()18 Substitute. 360 Examples. 21 π Sector 1 1 Applying the Area of a Circle A circular archery target has a 2-ft diameter. of a circle ADB π The length of is 21 cm. It is yellow except for a red bull’s-eye at the center with a 6-in. diameter. Find the area of the yellow region to the nearest whole number. Quick Check. First ﬁnd the areas of the archery target and the red bull’s-eye. 1 2. The diameter of a bicycle wheel is 22 in. To the nearest whole number, how The radius of the archery target is 2 ( 2 ft) 1 ft 12 in. many revolutions does the wheel make when the bicycle travels 100 ft? The area of the archery target is πr2 π()12 2 144p in.2 17 revolutions 1 The radius of the red bull’s-eye region is 2 ( 6 in.) 3 in. The area of the red region is πr2 π()3 2 9p in.2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

π area of area of area of 3. Find the length of a semicircle with radius 1.3 m in terms of . archery target red region yellow region 1.3p m 144 π 9 π 135 π 424.11501 Simplify.

The area of the yellow region is about 424 in.2

198 199 Geometry Lesson 10-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-7

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2 Finding the Area of a Sector of a Circle Find the area of sector ACB. Example. B Leave your answer in terms of π. A 120 C 3 Finding the Area of a Segment of a Circle Find the area of 24 ft mAB π 2 the shaded segment. Round your answer to the nearest tenth. O area of sector ACB r 6 m 360 100 B Step 1 Find the area of sector AOB. 0 mAB 100 2 A area of sector AOB ? πr2 Use the formula for area of a sector . π( 6 ) 360 360 120 2 ? π ( 24 )r Substitute. 5 360 36 π 1 18 ? 576π 192 π Simplify. 3 5 10 π Step 2 Find the area of ᭝AOB. ft The area of sector ACB is 10π m2. 123 B ft 30 You can use a 30 -60 -90 triangle to ﬁnd the 123 height h of ᭝AOB and the length of AB . Quick Check. A 60 24 ft 24 2 h hypotenuse 2 ? shorter leg 60 1. How much more pizza is in a 14-in.-diameter pizza than in a 12-in. pizza? 12 ft O 12 h Divide each side by 2 . 2 about 41 in. AB 2 !3 ? 12 12"3 longer leg "3 ? shorter leg AB 24 "3 Multiply each side by 2 . ᭝AOB has base 12"3 ft 12"3 ft, or 24"3 ft, and height 12 ft. 1 A 5 bh Area of a triangle . 2 1 A 5 2 ()()24"3 12 Substitute 24"3 for b and 12 for h. A 144 "3 Simplify. Step 3 Subtract the area of ᭝AOB from the area of sector AOB to ﬁnd the area of the segment of the circle. area of segment 192 π 144 "3 2. Critical Thinking A circle has a diameter of 20 cm. What is the area of a sector bounded by a 208 major arc? Round your answer to the nearest tenth. area of segment< 353.77047 Use a calculator.

181.5 cm2 To the nearest tenth, the area of the shaded segment is 353.8 ft2.

Quick Check. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson 3. A circle has a radius All rights reserved. of 12 cm. Find the area of the smaller segment of the

13.0 cm2

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50 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 10-8 Geometric Probability Example. 2 Finding Probability Using Area A circle is inscribed in a square target Lesson Objective NAEP 2005 Strand: Data Analysis and Probability 20 cm with 20-cm sides. Find the probability that a dart landing randomly within 1 Use segment and area models to ﬁnd Topic: Probability the probabilities of events the square does not land within the circle. Local Standards: ______Find the area of the square. A s2 202 400 cm2 Vocabulary. Find the area of the circle. Because the square has sides of length 20 cm, Geometric probability is a model in which you let points represent outcomes. the circle’s diameter is 20 cm, so its radius is 10 cm. A πr2 π()10 2 100π cm2 Find the area of the region between the square and the circle. Example. A (400 100π ) cm2 1 Finding Probability Using Segments A gnat lands at random on the Use areas to calculate the probability that a dart landing randomly in the edge of the ruler below. Find the probability that the gnat lands on a point square does not land within the circle. Use a calculator. Round to the between 2 and 10. nearest thousandth. area between square and circle P(between square and circle) area of square

400 100π

1 2 3 4 5 6 7 8 9 10 11 400

π 1 0.215 The length of the segment between 2 and 10 is 10 2 8 . 4

The length of the ruler is 12 . The probability that a dart landing randomly in the square does not land P(landing between 2 and 10) within the circle is about 21.5 %.

length of favorable segment 8 2 3 length of entire segment 12 Quick Check. 2. Use the diagram in Example 2. If you change the radius of the circle as Quick Check. indicated, what then is the probability of hitting outside the circle? a. Divide the radius by 2. 1. A point on AB is selected at random. What is the probability that it is a point on CD ? about 80.4% ACDB

012345678910 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. b. Divide the radius by 5. 2 5 about 96.9%

202 203 Geometry Lesson 10-8 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-8

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Example. Lesson 11-1 Space Figures and Cross Sections 3 Applying Geometric Probability To win a prize, you must toss a quarter Lesson Objective NAEP 2005 Strand: Geometry so that it lands entirely within the outer region of the circle at right. Find the 15 1 Recognize polyhedra and their parts Topic: Dimension and Shape probability that this happens with a quarter of radius in. Assume that the 9 in. 32 2 Visualize cross sections of space quarter is equally likely to land anywhere completely inside the large circle. figures Local Standards: ______15 15 The center of a quarter with a radius of 32 in. must land at least 32 in. 12 in. beyond the boundary of the inner circle in order to lie entirely outside the Vocabulary and Key Concepts. inner circle. Because the inner circle has a radius of 9 in., the quarter must 15 915 land outside the circle whose radius is 9 in. 32 in., or 32 in. Euler’s Formula 15 Find the area of the circle with a radius of 932 in. The numbers of faces (F), vertices (V), and edges (E) of a polyhedron 2 are related by the formula F ϩ V ϭ E ϩ 2 . 15 2 A pr 5 p ° 9 ¢ < 281.66648 in.2 32 15 Similarly, the center of a quarter with a radius of 32 in. must land at least A polyhedron is a three-dimensional figure whose surfaces are polygons. 15 32 in. within the outer circle. Because the outer circle has a radius of 12 in., 15 the quarter must land inside the circle whose radius is 12 in. 32 in., or 1117 32 in. A face is a flat surface of a polyhedron in the shape of a polygon. 17 Rectangular prism Find the area of the circle with a radius of 1132 in. 2 Faces 2 17 A pr 5 p ° 11 ¢ < 417.73672 in.2 Edge 32 An edge is a segment that is formed by the intersection of two Vertex Use the area of the outer region to find the probability that the quarter faces. lands entirely within the outer region of the circle. 417.73672 281.66648 P(outer region) area of outer circle area of large circle 417.73672 A vertex is a point where three or more edges intersect. 136.07024 P(outer region) 0.32573 417.73672 The probability that the quarter lands entirely within the outer region of the circle is about 0.326 or 32.6 %. A cross section is the intersection of a solid and plane.

Quick Check. Examples 3. Critical Thinking Use Example 3. Suppose you toss 100 quarters. Would you

expect to win a prize? Explain. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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51 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Examples. Example.

1 Identifying Vertices, Edges, and Faces How many vertices, edges, B 3 Verify Euler’s Formula for the two-dimensional net of and faces are there in the polyhedron shown? Give a list of each. A the solid in Example 1.

There are four vertices:A, B, C, and D . Count the regions: F 4 AB, BC, CD, DA, AC, BD There are six edges:and . Count the vertices: V 6 There are four faces:kABC, kABD, kACD, and kBCD . C Count the segments: E 9 D 2 Using Euler’s Formula Use Euler’s Formula to ﬁnd the number of 4 6 9 1 edges on a solid with 6 faces and 8 vertices. F V E 2 Euler’s Formula 68E 2 Substitute the number of faces and vertices. Quick Check. 12 E Simplify. 3. The figure at the right is a trapezoidal prism. A solid with 6 faces and 8 vertices has 12 edges. a. verify Euler’s Formula F V E 2 for the prism.

6 8 12 2 Quick Check.

1. List the vertices, edges, and faces of the polyhedron. R

R, S, T, U, V; RS, RU, RT, VS, VU, VT, SU, UT, TS; kRSU, kRUT, kRTS, kVSU, kVUT, kVTS S T b. Draw a net for the prism. U Possible answer:

2. Use Euler’s Formula to find the number of edges on a polyhedron V with eight triangular faces.

12 edges

c. Verify Euler’s Formula F V E 1 for your two-dimensional net.

6 14 19 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

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Surface Areas of Prisms and Cylinders a polyhedron with exactly two congruent, parallel Lesson 11-2 A prism is Pentagonal prism faces. Lesson Objectives NAEP 2005 Strand: Measurement 1 Find the surface area of a prism Topic: Measuring Physical Attributes The bases of a polyhedron are are the parallel faces. 2 Find the surface area of a cylinder Local Standards: ______Bases Lateral faces are the faces on a prism that are not bases. Lateral Vocabulary and Key Concepts. An altitude is a perpendicular segment that joins the planes edges of the bases. Theorem 11-1: Lateral and Surface Area of a Prism

The lateral area of a right prism is the product of the The height is the length of the altitude. Lateral Bases perimeter of the base and the height . faces L.A. ph The lateral area of a prism is the sum of the areas of the lateral The surface area of a right prism is the sum of the lateral area and faces. Triangular prism the areas of the two bases . The surface area of a prism or a cylinder is the sum of the lateral area and the areas of the S. A. L.A. 2B two bases. Perimeter of base c c a b c d d d Base a b h adb c h adb c A right prism is a prism whose lateral faces are rectangular regions h h h and a lateral edge is an altitude. d Base c right prisms

Perimeter Height Lateral Area An oblique prism is a prism whose lateral faces are not h Area of a base Lateral Area ph Surface Area L.A. 2B perpendicular to the bases. oblique prism Theorem 11-2: Lateral and Surface Area of a Cylinder The lateral area of a right cylinder is the product of the A cylinder is a three-dimensional ﬁgure with two congruent circular Bases h circumference of the base and the height of the cylinder . bases that lie in parallel planes. L.A. 2πrh, or L.A. πdh In a cylinder, the bases are parallel circles. The surface area of a right cylinder is the sum of the lateral area h and the areas of the two circular bases . The lateral area of a cylinder is the area of the curved surface. S.A. L.A. 2B, or S.A. 2πrh 2πr2 right cylinders

r r

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A right cylinder is one where the segment joining the centers of the bases is an altitude. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

h Lateral h Surface Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Area Area An oblique cylinder is one in which the segment joining the centers h 2πr r Area of a base of the bases is not perpendicular to the bases. B π r 2 oblique cylinder

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52 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Examples. 2 Finding Surface Area of Cylinders A company sells cornmeal and barley in cylindrical containers. The diameter of the base of the 6-in.-high cornmeal 1 Using Formulas to Find Surface Area Find the surface area of a 10-cm- container is 4 in. The diameter of the base of the 4-in.-high barley container high right prism with triangular bases having 18-cm edges. Round to the is 6 in. Which container has the greater surface area? nearest whole number. d Find the surface area of each container. Remember that r 2. ϭ Use the formula L.A. ph to ﬁnd the lateral area and Cornmeal Container Barley Container ϭ ϩ the formula S.A. L.A. 2B to ﬁnd the surface area of the S.A. 5 L.A. 1 2 B Use the formula for surface area. S.A. 5 L.A. 1 2 B 1 prism. The area B of the base is ap , where a is the apothem and p is the Substitute the formulas for lateral 2 5 2prh 1 p 2 5 p 2pr2 2 r area of a cylinder and area of a circle. 2 r h perimeter . 2π()()2263 2π( 2) Substitute for r and h. 2π()()4 2π( 3 2) The triangle has sides of length 18 cm, so p 3 ? 18 cm, or 54 cm. 5 24 p 1 8 p Simplify. 5 24 p 1 18 p

5 32 p Combine like terms. 5 42 p

18 cm Because 42π in.2 Ͼ 32π in.2, the barley container has the greater

18 cm surface area. Use the diagram of the base. 60 a Use 30°-60°-90° triangles to find theapothem . Quick Check. 30 1. Use formulas to ﬁnd the lateral area and surface area of the prism. 9 cm 9 cm 432 m2; about 619 m2 a ? "3 9 !3 ? shorter leg longer leg 12 m 6 m a 9 Divide both sides by "3 . "3

9 "3 9!3 2. Find the surface area of a cylinder with height 10 cm and radius 10 cm in ? "3 a 3 Rationalize the denominator. π "3 "3 3 terms of . 1 p 2 B 5 2ap 400 cm 1 B 2 ??3 "3 54 Substitute 81!3 for a and 54 for p.

B 81 "3 Simplify.

2 The area of each base of the prism is 81!3 cm . 3. The company in Example 2 wants to make a label to cover the cornmeal container. The label will cover the container all the way around, but will not S.A. 5 L.A.1 2B Use the formula for surface area. cover any part of the top or bottom. What is the area of the label to the ph Substitute ph for L.A. 2B Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson nearest tenth of a All rights reserved. square inch? 54 ! ! ()()10 2()81 3 Substitute 54 for p,10 for h and 81 3 for B. 75.4 in.2 540 162!3 Simplify.

< 820.59223 Use a calculator.

Rounded to the nearest whole number, the surface area is 821 cm2 .

210 211 Geometry Lesson 11-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-2

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a cone in which the altitude is perpendicular to the Lesson 11-3 Surface Areas of Pyramids and Cones A right cone is Slant height plane of the base. Lesson Objectives NAEP 2005 Strand: Measurement Altitude The slant height of a cone is the distance from the vertex to the edge 1 Find the surface area of a pyramid Topic: Measuring Physical Attributes / h 2 Find the surface area of a cone of the base. Local Standards: ______r The lateral area of a cone is the area of the curved lateral surface. Base Vocabulary and Key Concepts. The surface area of a cone the sum of the lateral area and the area Right cone of the base. Theorem 11-3: Lateral and Surface Area of a Regular Pyramid The lateral area of a regular pyramid is half the product of the ᐉ perimeter of the base and the slant height . Examples. 1pᐉ L.A. 2 B 1 Finding Surface Area of a Pyramid Find the surface area of a square pyramid with base edges 7.5 ft and slant height 12 ft. The surface area of a regular pyramid is the sum of the lateral area and the area of the base . 12 ft S.A. L.A. B

Theorem 11-4: Lateral and Surface Area of a Cone ᐉ The lateral area of a right cone is half the product of the

circumference of the baseand the slant height . r 1 B 7.5 ft L.A. 2 ? 2πr ? ᐉ, or L.A. prᐉ The perimeter p of the square base is 4 7.5 ft, or 30 ft. The surface area of a right cone is the sum of the lateral area and the area of the base. You are given ᐍ 12 ft and you found that p 30 ft, so you can S.A. L.A. B ﬁnd the lateral area.

1 a pyramid whose base is a regular polygon and whose lateral faces are A regular pyramid is L.A. 5 p ᐍ Use the formula for lateral area of a pyramid. congruent isosceles triangles. 2

The altitude of a pyramid or a cone is the perpendicular segment from the vertex to the plane 5 1 30 12 L.A. 2()()Substitute. of the base. L.A. 5 180 Simplify. The height of a pyramid or a cone is the length of Lateral Vertex the altitude. edge Lateral face Find the area of the square base. The slant height of a regular pyramid is the length Altitude Because the base is a square with side length 7.5 ft, of the altitude of a lateral face. Base B s2 7.5 2 56.25 . Base edge Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. S.A. 5 L.A. 1 B Use the formula for surface area of a pyramid. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson The lateral area of a pyramid is is the sum All rights reserved. of the Regular Pyramid S.A. 5 180 1 56.25 Substitute. areas of the congruent lateral faces. 5 236.25 The surface area of a pyramid is is the sum of the lateral area and the area of the base. S.A. Simplify.

The surface area of the square pyramid is 236.25 ft2.

212 213 Geometry Lesson 11-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-3

53 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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2 Finding Lateral Area of a Cone Leandre uses paper cones to cover her Lesson 11-4 Volumes of Prisms and Cylinders plants in the early spring. The diameter of each cone is 1 ft, and its height is 1.5 ft. How much paper is in the cone? Round your answer to the nearest Lesson Objectives NAEP 2005 Strand: Measurement tenth. 1 Find the volume of a prism Topic: Measuring Physical Attributes 2 Find the volume of a cylinder Use the formula L.A. prᐍ to ﬁnd the lateral area of the cone. Local Standards: ______

The cone’s diameter is 1 ft, so its radius r is 0.5 ft. Vocabulary and Key Concepts. The altitude of the cone, radius of the base, and slant height form a righttriangle. Use the Pythagorean Theorem Theorem 11-5: Cavalieri’s Principle to ﬁnd the slant height ᐍ. If two space ﬁgures have the same height and the same cross sectional area ᐉ 2 2 0.5 1.5 0.25 2.25 "2.5 at every level, then they have the same volume . Find the lateral area. Theorem 11-6: Volume of a Prism L.A. 5 p r ᐍ Use the formula for lateral area of a cone. The volume of a prism is the product of the area of a base h 5 p "2.5 "2.5 ᐍ and the height of the prism . L.A. ()0.5Substitute 0.5 for r and for . B L.A. < 2.4836471 Use a calculator. V Bh

2 The lateral area of the cone is about 2.5 ft . Theorem 11-7: Volume of a Cylinder The volume of a cylinder is the product of the area of the base h and the height of the cylinder . Quick Check. r p 2 B 1. Find the surface area of a square pyramid with base edges 5 m and slant V Bh, or V r h height 3 m. 2 55 m Volume is the space that a figure occupies. A composite space figure is a three-dimensional figure that is the combination of two or more simpler figures.

Examples. 1 Finding Volume of a Cylinder Find the volume of the cylinder. Leave your answer in terms of π. 2. Find the lateral area of a cone with radius 15 in. and height 20 in. to the The formula for the volume of a cylinder is V pr2h . 9 ft nearest square inch. The diagram shows h and d, but you must ﬁnd r. 2 1178 in. 1 r 5 2d 5 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 16 ft V 5 p r2 h Use the formula for the volume of a cylinder.

5 p ? 82 ? 9 Substitute.

5 576 p Simplify. The volume of the cylinder is 576π ft3.

214 215 Geometry Lesson 11-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-4

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2 Finding Volume of a Triangular Prism Find the volume of the prism. 29 m Quick Check. The prism is a right triangular prism with triangular bases. The 1. Find the volume of the triangular prism. base of the triangular prism is a right triangle where one leg 40 m 150 m3 is the base and the other leg is the altitude . Use the 10 m Pythagorean Theorem to calculate the length of the other leg. 20 m 5 m 6 m 2 2 29 20 841 400 441 21 1 1 The area B of the base is 2bh 2()()20 21 210 . Use the area of the base to ﬁnd the volume of the prism.

V B h Use the formula for the volume of a prism. 2. The cylinder shown is oblique. V 210? 40 Substitute. a. Find its volume in terms of π. V 8400 Simplify. 16 m 256p m3 The volume of the triangular prism is 8400 m3.

8 m

3 Finding Volume of a Composite Figure Find the volume 25 cm of the composite space ﬁgure. 14 cm You can use three rectangular prisms to ﬁnd the volume.

25 b. Find its volume to the nearest tenth of a cubic meter.

cm 10 cm 10 cm 3 6 cm 6 cm 804.2 m 14 cm 4 cm 4 cm cm cm 4 4 6 cm 6 cm 6 cm 6 cm

4 cm 4 cm Each prism’s volume can be found using the formula V Bh .

14 ? ? 1400 Volume of Prism I Bh ( 425) 3. Find the volume of the composite space ﬁgure. 2 in. Volume of Prism II Bh 64? ? 25 600 12 in.3 ( ) 4 in.

6 ? 4 ? 25 600 Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Volume of Prism III Bh ( ) Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 2 in. 1 in. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Sum of the volumes 1400 600 600 2600 4 in.

The volume of the composite space ﬁgure is 2600 cm3.

216 Geometry Lesson 11-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-4 217

54 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 11-5 Volumes of Pyramids and Cones 2 Finding Volume of a Pyramid Find the volume of a square pyramid with base edges 16 m and slant height 17 m. 17 m Lesson Objectives NAEP 2005 Strand: Measurement 1 Find the volume of a pyramid Topic: Measuring Physical Attributes The altitude of a right square pyramid intersects the base h 2 Find the volume of the cone at the center of the square. Because each side of the square Local Standards: ______base is 16 m, the leg of the right triangle along the base is 8 m, as shown. 16 m Key Concepts.

Theorem 11-8: Volume of a Pyramid 16 m 17 m The volume of a pyramid is one third the product of the h area of the base and the height of the pyramid . h 1 V 3Bh 8 m B Step 1 Find the height of the pyramid. Theorem 11-9: Volume of a Cone 2 2 2 17 5 8 1 h Use the Pythagorean Theorem . The volume of a cone is one third the product of the 289 5 64 1 h2 area of the base and the height of the cone . Simplify. 2 1 h 225 5 h Subtract 64 from each side. Bh 1p 2 V 3 , or V 3 r h r h 5 15 Find the square root of each side. B Step 2 Find the volume of the pyramid. 1 V 5 Use the formula for Bh the volume of a pyramid. Examples. 3 1 Finding Volume of a Pyramid Find the volume of a square pyramid with 5 1 ? 3 ( 1616 ) 15 Substitute. base edges 15 cm and height 22 cm. 5 1280 Simplify. Because the base is a square, B 15 ? 15 225 . The volume of the square pyramid is 1280 m3. 1 V 5 Bh Use the formula for volume of a pyramid. 3 3 Finding Volume of an Oblique Cone Find the volume π 1 of the oblique cone in terms of . 3()()225 22 Substitute 225for B and 22 for h. 1 r 5 5 V 5 1650 Simplify. d 3 in. 2 11 in. The volume of the square pyramid is 1650 cm3. 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson V 5 p All rights reserved. r2 h Use the formula for the volume of a cone. 3 5 1 p 2 3 ( 3 )(11 ) Substitute 3 for r and 11 for h. 6 in. 5 33 p Simplify.

The volume of the cone is 33p in.3.

218 219 Geometry Lesson 11-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-5

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Quick Check. Lesson 11-6 Surface Areas and Volumes of Spheres 1. Find the volume of a square pyramid with base edges 24 m and slant height 13 m. Lesson Objective NAEP 2005 Strand: Measurement 3 960 m 1 Find the surface area and volume Topic: Measuring Physical Attributes of a sphere Local Standards: ______

Vocabulary and Key Concepts.

Theorem 11-10: Surface Area of a Sphere 2. Find the volume of each cone in terms of π and also rounded as indicated. The surface area of a sphere is four times the product of p and a. to the nearest cubic meter r the square of the radius of the sphere.

12 m S.A. 4pr2

6 m Theorem 11-11: Volume of a Sphere 144p m3; 452 m3 The volume of a sphere is four thirds the product of p and r the cube of the radius of the sphere.

4p 3 S.A. 3 r

A sphere is the set of all points in space equidistant from a b. to the nearest cubic millimeter center diameter given point. 42 mm The center of a sphere is the given point from which all points 21 mm on the sphere are equidistant. r

a segment that has one endpoint at 6174p mm3; 19,396 mm3 The radius of a sphere is the center and the other endpoint on the sphere. d

The diameter of a sphere is a segment passing through the circumference center with endpoints on the sphere. sphere

3. A small child’s teepee is 6 ft tall and 7 ft in diameter. Find the volume of the A great circle is the intersection of a plane and a sphere teepee to the nearest cubic foot. containing the center of the sphere. Hemispheres 3 A great circle divides a sphere into two congruent 77 ft Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. hemispheres. Great © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. circle The circumference of a sphere is the circumference of its great circle.

220 221 Geometry Lesson 11-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-6

55 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Examples. Example. 1 Finding Surface Area The circumference of a rubber ball is 13 cm. 3 Using Volume to Find Surface Area The volume of a sphere is 1 in.3. Calculate its surface area to the nearest whole number. Find its surface area to the nearest tenth. Step 1 First, find the radius. 4 V 5 p r3 Use the formula for volume of a sphere. C 5 2 p r Use the formula for circumference. 3 13 5 2pr Substitute 13 for C. 4 3 1 3πr Substitute. 13 5 r Solve for r. 3 3 2p 5 r Solve for r3. 4p Step 2 Use the radius to ﬁnd the surface area. S.A. 5 4 p r2 Use the formula for surface area of a sphere. 3 3 r Find the cube root of each side. 13 2 13 4p 5 4p ? Substitute for r. ()2p 2p S. A. 4.83597587 Use r, S.A. 4πr2, and a calculator. 169 To the nearest tenth, the surface area of the sphere is 4.8 in.2. 5 Simplify. p < 53.794371 Use a calculator. Quick Check. To the nearest whole number, the surface area of the rubber ball is 54 cm2. 2. Find the surface area of a spherical melon with circumference 18 in. Round your answer to the nearest ten square inches. 2 Finding Volume Find the volume of the sphere. 100 in.2 Leave your answer in terms of π.

4 30 cm V 5 p r3 Use the formula for volume of a sphere. 3

4 30 5 p ? 15 3 Substitute r 15 . 3 2 3. Find the volume to the nearest cubic inch of a sphere with diameter 60 in.

5 4500 p Simplify. 113,097 in.3

The volume of the sphere is 4500p cm3.

Quick Check.

1. Find the surface area of a sphere with d 14 in. Give your answer in two 3

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson 4. The volume of a All rights reserved. sphere is 4200 ft . Find the surface area to the nearest tenth. ways, in terms of π and rounded to the nearest square inch. 1258.9 ft2 196p in.2, 616 in.2

222 223 Geometry Lesson 11-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-6

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Lesson 11-7 Areas and Volumes of Similar Solids 2 Finding the Similarity Ratio Find the similarity ratio of two similar cylinders with surface areas of 98π ft2 and 2π ft2. Lesson Objective NAEP 2005 Strand: Measurement Use the ratio of the surface areas to ﬁnd the similarity ratios. 1 Find relationships between the ratios Topic: Systems of Measurement 2 98p of the areas and volumes of similar a 5 2 2 solids Local Standards: ______2 The ratio of the surface areas is a :b . b 2p

49 a2 Vocabulary and Key Concepts. 2 5 Simplify. b 1 Theorem 11-12: Areas and Volumes of Similar Solids a 7 If the similarity ratio of two similar solids is a : b, then b 5 Take the square root of each side. 1 (1) the ratio of their corresponding areas is a2 : b2 , and The similarity ratio is 7 :1 . (2) the ratio of their volumes is a3 :b3 .

3 Using a Similarity Ratio Two similar square pyramids have volumes of 3 and 162 cm3. The surface area of the larger pyramid is 135 cm2. Similar solids have the same shape and all of their corresponding parts are proportional. 48 cm Find the surface area of the smaller pyramid. Step 1 Find the similarity ratio. The similarity ratio of two similar solids is the ratio of their corresponding linear dimensions. 3 48 a 3 3 3 5 The ratio of the volumes is a :b . b 162

8 a3 Examples. 3 5 Simplify. b 27 1 Identifying Similar Solids Are the two solids similar? If so, give the similarity ratio. a 2 b 5 Take the cube root of each side. 3

Step 2 Use the similarity ratio to ﬁnd the surface area S1 of the smaller pyramid. 22 S1 5 The ratio of the surface areas is a2 :b2 . S2 26 in. 32 S 4 8 in. 1 5 Simplify. S2 9 9 in. S 4 1 Substitute135 for S2, the surface area of 3 in. 135 9 the larger pyramid. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Both ﬁgures have the same shape. Check that the ratios of the 4 corresponding dimensions are equal. S 5 ? Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 1 135 Solve for S1. 3 8 9 The ratio of the radii is 9 , and the ratio of the heights is 26 . S1 5 60 Simplify. 3 ϶ 8 The cones are not similar because 9 26 . The surface area of the smaller pyramid is 60 cm2.

224 225 Geometry Lesson 11-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-7

56 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Quick Check. Lesson 12-1 Tangent Lines 1. Are the two cylinders similar? If so, give the similarity ratio. Lesson Objectives NAEP 2005 Strand: Geometry yes; 6 : 5 1 Use the relationship between a radius Topic: Relationships Among Geometric Figures and a tangent 12 10 2 Use the relationship between two Local Standards: ______6 5 tangents from one point

Vocabulary and Key Concepts.

Theorem 12-1 If a line is tangent to a circle, then the line is perpendicular to the radius drawn A P to the point of tangency. * ) O B AB ' OP 2. Find the similarity ratio of two similar prisms with surface areas 144 m2 and 324 m2. Theorem 12-2 2 : 3 A If a line in the plane of a circle is perpendicular to a radius at its endpoint P on the circle, then the line is tangent to the circle . * ) O B AB is tangent to ᭪ O.

Theorem 12-3

The two segments tangent to a circle from a point outside the circle A B are congruent . O AB > CB C

A tangent to a circle is a line, segment, or ray in the plane of the circle that intersects the 3. The volumes of two similar solids are 128 m3 and 250 m3. The surface area circle in exactly one point. of the larger solid is 250 m2. What is the surface area of the smaller solid? The point of tangency is the point where a circle and a tangent intersect. 160 m2

A tangent

B point of tangency © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. A triangle is inscribed in a circle if all vertices of the triangle lie on the circle.

A triangle is circumscribed about a circle if each side of the triangle is tangent to the circle.

226 227 Geometry Lesson 11-7 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-1

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Examples. Examples. 1 Finding Angle Measures BA is tangent to ᭪ C at point A. Find the C 3 Finding a Tangent ᭪ O has radius 5. Point P is outside ᭪ ᭪ value of x. x° O such that PO 12, and point A is on O such that PA 13. Is PA tangent to ᭪ O at A? Explain. O 12 Because BA is tangent to ᭪ C, ЄA must be a right angle. A P Use the Triangle Angle-Sum Theorem to ﬁnd x. For PA to be tangent to ᭪ O at A, ЄA must be a 5 13 mЄA mЄB mЄC 180 Triangle Angle-Sum Theorem right angle, ᭝OAP must be a right triangle, and 2 2 2 A 90 22 180 PO PA ϩ OA . x Substitute. 22° 112 x 180 Simplify. B PO2 ՘ PA2 OA2 Is ᭝OAP a right triangle? 68 Solve. x 12 2 ՘ 132 5 2 Substitute.

144 194 Simplify. 2 Applying Tangent Lines A belt ﬁts tightly around Y two circular pulleys, as shown. Find the distance 15 cm Because PO2 PA 2 ϩ OA2 ,PA is not tangent X between the centers of the pulleys. Round your 7 cm to ᭪ O at A. answer to the nearest tenth. 3 cm O P OP OD ZW Draw . Then draw parallel to to form rectangle 4 Circles Inscribed in Polygons ᭪ C is inscribed in quadrilateral XYZW. OZ ᭪ Z D ODWZ. Because is a radius of O, OZ 3 cm. 15 cm Find the perimeter of XYZW. Because opposite sides of a rectangle have the same measure, Y W 8 ft DW 3 cm and OD 15 cm. R S Because ЄODP is the supplement of a right angle, 11 ft Z ЄODP is also a right angle, and ᭝OPD is a right triangle. C 6 ft X Because the radius of ᭪ P is 7 cm, PD 7 Ϫ 3 ϭ 4 cm. T 7 ft OD2 PD2 OP2 Pythagorean Theorem U W 15 2 4 2 OP2 Substitute. 241 OP2 Simplify. UX XR 11 ft YS RY 8 ft By Theorem 12-3, two segments tangent to a circle from OP 15.524175 Use a calculator to ﬁnd the square root. a point outside the circle are . SZ ZT 6 ft congruent The distance between the centers of the pulleys is about 15.5 cm. WU TW 7 ft

Quick Check. p XY YZZW WX Deﬁnition of perimeter p 1. ED is tangent to ᭪ O. Find the value of x. XR RY YS SZZT TWWU UX Segment Addition Postulate D E Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 52 x 11 8 8 6 6 7 711 Substitute. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. O 64 Simplify. 38 The perimeter is 64 ft.

228 229 Geometry Lesson 12-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-1

57 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Quick Check. Lesson 12-2 Chords and Arcs 2. A belt ﬁts tightly around two circular pulleys, as shown. Find the distance Lesson Objectives NAEP 2005 Strand: Geometry between the centers of the pulleys. 1 Use congruent chords, arcs, and Topic: Relationships Among Geometric Figures 35 in. central angles 2 Recognize properties of lines through Local Standards: ______14 in. 8 in. the center of a circle

Vocabulary and Key Concepts.

Theorem 12-4 about 35.5 in. Within a circle or in congruent circles (1) Congruent central angles have congruent chords. (2) Congruent chords have congruent arcs. (3) Congruent arcs have congruent central angles.

Theorem 12-5 Within a circle or in congruent circles (1) Chords equidistant from the center are congruent . 3. In Example 3, if OA 4, AP 7, and OP 8, is PA tangent to ᭪ O at A? (2) Congruent chords are equidistant from the center. Explain.

No; 42 ϩ 72 82 Theorem 12-6 In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs .

Theorem 12-7 In a circle, a diameter that bisects a chord (that is not a diameter) is perpendicular to the chord.

4. ᭪ O is inscribed in ᭝PQR. ᭝PQR has a perimeter of 88 cm. Find QY. Q Theorem 12-8 15 cm X In a circle, the perpendicular bisector of a chord contains the center 12 cm P of the circle. O Y

Z Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. 17 cm R A chord is a segment whose endpoints are on a circle. P Q O

230 231 Geometry Lesson 12-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-2

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Examples. Quick Check. 1 Using Theorem 12-4 In the diagram, radius OX bisects ЄAOB. A 1. If you are given that BC DF in the circles, what can you conclude? B D What can you conclude? 0 0 lO OlP; BC O DF ЄAOX lBOX by the deﬁnition of an angle bisector. O C P F AX BX because congruent central angles have O X ᭪O ᭪P congruent chords. 0 0 AX BX because congruent chords have congruent arcs. B

2 Using Theorem 12-5 Find AB. B 2. Find the value of x in the circle at the right. 18 QS QR RS Segment Addition Postulate 4 A 16 16 QS 77Substitute. C 18 x Q 4 36 QS 14 Simplify. 7 R AB QS Chords that are equidistant from the center of a circle 7 S are congruent.

AB 14 Substitute 14 for QS.

3 Using Diameters and Chords P and Q are points on ᭪ O. 16 in. Q The distance from O to PQ is 15 in., and PQ 16 in. 3. Use the circle at the right. Find the radius of ᭪ O. M a. Find the length of the chord to the nearest unit. 15 in. 6.8 The distance from the center of ᭪ O to PQ is measured along a about 11 4 perpendicular line. P x r O 1 A diameter that is perpendicular PM PQ to a chord bisects the chord. 2

1 PM ()16 8 Substitute. 2

OP2 PM2 OM2 Use the Pythagorean Theorem. b. Find the distance from the midpoint of the chord to the midpoint of its r2 8 2 15 2 Substitute. minor arc. 2 r 289 Simplify. 2.8

r 17 Find the square root of each side. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. The radius of ᭪ O is 17 in.

232 Geometry Lesson 12-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-2 233

58 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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Lesson 12-3 Inscribed Angles Examples. ° 1 Using the Inscribed Angle Theorem Find the values of x and y. E 70 Lesson Objectives NAEP 2005 Strand: Geometry 1 F Find the measure of an inscribed Topic: Relationships Among Geometric Figures 1 1 80° angle x mDEF Inscribed Angle Theorem 2 Find the measure of an angle formed Local Standards: ______2 by a tangent and a chord C 1 0 0 y° x (m DE m EF ) Arc Addition Postulate D Vocabulary and Key Concepts. 2 x° 1 90° Theorem 12-9: Inscribed Angle Theorem x ( 80 70 ) Substitute. G 2 The measure of an inscribed angle is A half the measure of its intercepted arc. B x 75 Simplify. 1 1 0 Є 2mAC C Because EFG is the intercepted arc of ЄD, you need to ﬁnd mFG in order m B 1 to ﬁnd mEFG . The arc measure of a circle is 360, so Theorem 12-10 0 mFG 360 70 80 90 120 . The measure of an angle formed by a tangent and a chord is B D B half the measure of the intercepted arc. 1 1 D mEFG 1 y Inscribed Angle Theorem 1mBDC 2 mЄC 2 C C 1 0 0 y (m EF m FG ) Arc Addition Postulate 2 Corollaries to the Inscribed Angle Theorem 1 1. Two inscribed angles that intercept the same arc are congruent . y ( 70 120 ) Substitute. 2 2. An angle inscribed in a semicircle is a right angle. y 95 Simplify. 3. The opposite angles of a quadrilateral inscribed in a circle are supplementary .

2 Using Corollaries to Find Angle Measures Find the values of a and b.

An inscribed angle has a vertex that is on a A By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a Intercepted arc circle and sides that are chords of the circle. semicircle is a right angle, so a 90 . 32° B C b° O Inscribed angle The sum of the measures of the three angles of the triangle inscribed in ᭪ O is 180 . Therefore, the angle whose intercepted arc has An intercepted arc is an arc with endpoints on a° 90 32 58 the sides of an inscribed angle and its other Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson measure b must All rights reserved. have measure 180 or . points in the interior of the angle. Because the inscribed angle has half the measure of the intercepted arc, the intercepted arc has twice the measure of the inscribed angle, so b 2()58 116 .

234 235 Geometry Lesson 12-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-3

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Angle Measures B 3 *Using) Theorem 12-10 RS and TU are diameters of circle A. T Lesson 12-4 and Segment Lengths RB is tangent to ᭪ A at point R. Find mЄBRT and mЄTRS. R Lesson Objectives NAEP 2005 Strand: Geometry 1 0 mЄBRT mRT Theorem 12-10 1 Find the measures of angles formed Topic: Relationships Among Geometric Figures 2 A by chords, secants, and tangents 0 0 0 126° 2 Find the lengths of segments Local Standards: ______mRT m URT m UR Arc Addition Postulate N associated with circles

1 Substitute 180 for mURT S mЄBRT ( 180 126 ) Vocabulary and Key Concepts. 2 and126 for mUR. U

mЄBRT 27 Simplify. Theorem 12-11 Use the properties of tangents to find mЄTRS. The measure of an angle formed by two lines that (1) intersect inside a circle is half the sum of the measures of the ° y° A tangent is perpendicular to x 1 mЄBRS 90 intercepted arcs. the radius of a circle at its point of tangency. 1 mЄ1 (x 1 y) mЄBRS mЄ BRT mЄ TRS Angle Addition Postulate 2 Є 1 90 27 m TRS Substitute. (2) intersect outside a circle is half the y° 1 y° y° 1 x° x° x° 63 mЄTRS Simplify. difference of the measures of the intercepted arcs. Є 1 x 2 y Quick Check. m 1 2( ) 1. In Example 1, find mЄDEF. 105 Theorem 12-12 For a given point and circle, the product of the lengths of the two segments from 2. For the diagram at the right, find the measure of each numbered angle. ° the point to the circle is constant along any line through the point and circle. 1 60 ml1 5 90, ml2 5 110, ml3 5 90, ml4 5 70 2 4 80° I. II. III. 3 a t c x w b P P P y y d z z a b c d (w x)w (y z)y (y z)y t2

3. In Example 3, describe two ways to ﬁnd mЄNRS using Theorem 12-10.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson A secant is a line All rights reserved. that intersects a circle at two points. mlNRS 5 mlNRU 1 mlURS 5 1mRU 1 27 5 63 1 27 5 90 A secant 0 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson Є 1 1 All rights reserved. B m NRS 2mRUS 2(180) 90 O

236 237 Geometry Lesson 12-3 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-4

59 All-In-One Answers Version A Geometry Geometry: All-In-One Answers Version A (continued)

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Examples. Quick Check. 1 Finding Arc Measures An advertising agency wants a frontal photo of a “flying A 1. Find the value of w. saucer” ride at an amusement park. The photographer stands at the vertex of w° the angle formed by tangents to the “flying saucer.” What is the measure of the 250 ° arc that will be in the photograph? In the diagram, the photographer stands at 110 ) ) 0 1 Y 70° TX TY XY XA Y point 0T. and intercept minor arc and major arc . Let mXY x. X 72° 1 Then mXAY 360 x. T 2. Critical Thinking To photograph a 160 arc of the “flying saucer” in Example 1, 1 1 0 should you move toward or away from the object? What angle should the Є mXAY mXY m T ( ) Theorem 12-11(2) tangents form? 2 away; 20° 1 Ϫ 72 2[(360 x ) x ] Substitute. 1 72 2 ( 360 2x) Simplify. 72 180 x Distributive Property.

x 72 180 Add x to both sides. 3. Find the value of x to the nearest tenth. 20 14 x 108 Solve for x. 13.8 x 16 A 108Њ arc will be in the advertising agency’s photo.

2 Finding Segment Lengths A tram travels from point A to point B along the arc of a circle with a 50 ft radius of 125 ft. Find the shortest distance from AB 4. The basis of a design of a rotor for a Wankel engine is an equilateral Arc point A to point B. xx triangle. Each side of the triangle is a chord to an arc of a circle. The x The perpendicular bisector of the chord AB contains opposite vertex of the triangle is the center of the arc. In the diagram 8 in. 200 ft at the right, each side of the equilateral triangle is 8 in. long. the center of the circle. Because the radius is 125 ft, the a. Use what you know about equilateral triangles and ﬁnd the value of x. diameter is 2 125 250 ft. The length of (8 2 4"3) in. the other segment along the diameter is Center 250 ft 50 ft, or200 ft.

x x 50 200 Theorem 12-12(1) b. Critical Thinking Complete the circle with the given center. Then use x2 10,000 Multiply. Theorem 12-12 to ﬁnd the value of x. Show that your answers to parts

x 100 Solve for x. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson (a) and (b) are All rights reserved. equal. 16 16 The shortest distance from point A to point B is 2x or 200 ft. in.; (8 2 4"3) 5 1.07 8 1 4"3 8 1 4"3

238 239 Geometry Lesson 12-4 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-4

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Circles in the Lesson 12-5 Coordinate Plane 3 Graphing a Circle Given Its Equation Find the center and radius of the circle with equation (x 4)2 (y 1)2 25. Then graph the circle. Lesson Objectives NAEP 2005 Strand: Geometry 2 2 1 Write an equation of a circle Topic: Position and Direction (x 4 ) (y 1 ) 25 2 Relate the equation to the standard form Find the center and radius of a circle x 4 2 y 1 2 5 2 Local Standards: ______( ) ( ) () (x h)2 (y k)2 r2. ccc hkr Key Concepts. y The center is (Ϫ4, 1) and the radius is 5 . Theorem 12-13 (x, y) r y The standard form of an equation of a circle with center (h, k) 6 2 2 2 (h, k) and radius r is (x h) (y k) r . 4 O x 2

x Examples. 8 6 4 2 O 2468 2 1 Writing the Equation of a Circle Write the standard equation of a circle with center (8, 0) and radius "5 . 4

(x hk)2 (y )2 r 2 Standard form 4 Applying the Equation of a Circle A diagram locates a radio tower at [x ()8 ]2 (y 0 )2 ()"5 2 Substitute ()8, 0 for (h, k) and (6, 12) on a coordinate grid where each unit represents 1 mi. The radio "5 for r. signal’s range is 80 mi. Find an equation that describes the position and 2 range of the tower. (x 8 ) y2 5 Simplify. The center of a circular range is at (6, Ϫ12) , and the radius is 80 . (x h)2 (y k)2 r 2 Use standard form. 2 Using the Center and a Point on a Circle Write the standard equation of a circle with center (5, 8) that passes through the point (15, 13). (x 6 )2 [y ()]122 80 2 Substitute. This is an equation for position First ﬁnd the radius. 6 2 12 2 6400 (x ) (y ) and range of the tower. r #(x h )2 (y k )2 Use the Distance Formula to find r. # 15 582 13 2 Substitute 5, 8 for (h, k) and 15, 13 for (x, y). ( ) ( ) () () Quick Check. #( 20 )2 ( 21 )2 Simplify. 1. Write the standard equation of each circle. "2 a. center (3, 5); radius 6 b. center ( 2, 1); radius # 400 441 (x Ϫ 3)2 ϩ (y Ϫ 5)2 ϭ 36 (x ϩ 2)2 ϩ (y ϩ 1)2 ϭ 2 # 841 29 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved.

Then ﬁnd the standard equation of the circle with center (5, 8) and radius29 . Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. (x h )2 (y k )2 r 2 Standard form (x 5 )2 (y 8 )2 ()29 2 Substitute ()5, 8for (h, k) and 29 for r. (x 5 )2 (y 8 )2 841 Simplify.

240 Geometry Lesson 12-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-5 241

60 Geometry All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued)

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2. Write the standard equation of the circle with center (2, 3) that passes Lesson 12-6 Locus: A Set of Points through the point (1, 1). (x Ϫ 2)2 ϩ (y Ϫ 3)2 ϭ 13 Lesson Objective NAEP 2005 Strand: Geometry 1 Draw and describe a locus Topic: Dimension and Shape

Local Standards: ______

3. Find the center and radius of the circle with equation y 2 2 (x 2) (y 3) 100. Then graph the circle. Vocabulary. 16 center: (2, 3); radius: 10 A locus is a set of points, all of which meet a stated condition. 8 (2, 3)

O 4 x 8 Examples. 1 Describing a Locus in a Plane Draw and describe the locus of points in a ᭪ 4. When you make a call on a cellular phone, y plane that are 1.5 cm from C with radius 1.5 cm. a tower receives the call. In the diagram, Use a compass to draw ᭪ C with radius 1.5 cm. the centers of circles O, A, and B are locations of cellular telephone towers. 24 Write an equation that describes the position and range of Tower O. 20 B 2 2 x ϩ y ϭ 144 16

1.5 cm A C 8 1.5 cm 4

8 4 O 4 8 16 20 24 x 4

The locus of points in the interior of ᭪ C that are 1.5 cm from ᭪ C is the center C.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson The locus of points All rights reserved. exterior to ᭪ C that are 1.5 cm from ᭪ C is a circle with center C and radius 3 cm. The locus of points in a plane that are 1.5 cm from a point on ᭪ C with radius 1.5 cm is point C and a circle with radius 3 cm and center C.

242 243 Geometry Lesson 12-5 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-6

Name______Class______Date______Name______Class______Date ______

2 Drawing a Locus for Two Conditions Point P is 10 cm from point Q. Quick Check. Describe the locus of points in a plane that are 6 cm from point P and 8 cm * ) 1. Draw and describe the locus: In a plane, the points 2 cm from line XY . from point Q. * ) * ) Two lines parallel to XY , each 2 cm from XY .

2 cm 6 cm 8 cm

PQ 1010 cm X Y 6 cm 8 cm 2 cm

The set of points in a plane that are 6 cm from point P is a circle with 2. Draw the locus: In a plane, the points equidistant from two points X and Y 6 center P and radius cm. and 2 cm from the midpoint of XY . The set of points in a plane that are 8 cm from point Q is a circle with center Q and radius 8 cm. Because the sum (14 cm) of the radii is greater than 10 cm, the circles 2 cm overlap by 4 cm on PQ . The circles intersect at two points, so the locus } is the two points where the circles intersect . X Y 2 cm 3 Describing a Locus in Space Describe the locus of points in space that are } 4 cm from a plane M. 3. Draw and describe each locus. a. In a plane, the points that are equidistant from two parallel lines. A line midway between the two parallel lines, parallel to and equidistant from each.

44 cm M

44 cm

b. In space, the points that are equidistant from two parallel planes.

ഞ A plane midway between the two parallel planes, parallel to and equidistant from each. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Hall. Prentice Education, Inc., publishing as Pearson © Pearson All rights reserved. Imagine plane M as a horizontal plane. Because distance is measured along © Pearson Education, Inc., publishing as Pearson Prentice Hall. Prentice Education, Inc., publishing as Pearson © Pearson a perpendicular segment from a point to All rights reserved. a plane, the locus of points in space that are 4 cm from a plane M are a plane 4 cm above and parallel to plane M and another plane 4 cm below and parallel to plane M. A B

244 245 Geometry Lesson 12-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-6

61 All-In-One Answers Version A Geometry