DOCSLIB.ORG

Geometrygeometry

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text

Load more

Park Forest Math Team Meet #3 GeometryGeometry Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number Theory: Divisibility rules, factors, primes, composites 4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics 5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities Important Information you need to know about GEOMETRY… Properties of Polygons, Pythagorean Theorem Formulas for Polygons where n means the number of sides: • Exterior Angle Measurement of a Regular Polygon: 360÷n • Sum of Interior Angles: 180(n – 2) • Interior Angle Measurement of a regular polygon: • An interior angle and an exterior angle of a regular polygon always add up to 180° Interior angle Exterior angle Diagonals of a Polygon where n stands for the number of vertices (which is equal to the number of sides): • • A diagonal is a segment that connects one vertex of a polygon to another vertex that is not directly next to it. The dashed lines represent some of the diagonals of this pentagon. Pythagorean Theorem • a2 + b2 = c2 • a and b are the legs of the triangle and c is the hypotenuse (the side opposite the right angle) c a b • Common Right triangles are ones with sides 3, 4, 5, with sides 5, 12, 13, with sides 7, 24, 25, and multiples thereof—Memorize these! Category 2 50th anniversary edition Geometry 26 Y Meet #3 - January, 2014 W 1) How many cm long is segment 6 XY ? All measurements are in centimeters (cm). Z X 8 C 2) Angle ABC is a right X angle. Triangle BCD is E an isosceles triangle D such that DB = BC. 120 150 Find the value of X if it is the measure in degrees of angle 70 EDC and X < 180. B A 3) Moe and Larry race from point B to point S at a rectangular field. Moe runs from B to A to S at an average rate of 5 feet every second. Larry runs diagonally across the field from B to S at an average rate of 10 feet every 3 seconds. If they both leave point B at the same time, then who wins the race? Also, by how many seconds does the winner finish ahead of the runner-up? (You must answer both questions correctly to receive credit.) E S ANSWERS 1) __________ cm 50 feet 2) __________ B 120 feet A 3) __________ winner _________ seconds www.imlem.org Solutions to Category 2 Geometry Meet #3 - January, 2014 Answers 1) Use the Pythagorean Theorem twice - first to find the length of WX and then XY. 1) 24 2) 155 3) Moe Use this result to find XY: 5 2) The measure of angle DBA is 20 degrees, because the sum of the angles of a quadrilateral is 360 degrees. The measure of angle DBC is 70 degrees, because angle ABC is a right angle (90 degrees). Since two sides (DB and BC) of triangle DBC are congruent, the angles opposite those sides are congruent. The vertex angle, DBC, measures 70 degrees, so the base angles are 55 degrees each, including angle BDC. angle X + 150 + 55 = 360, so X = 155. 3) Both answers must be answered correctly in order for students to receive credit. Moe: runs 120 + 50, or 170 feet. At a rate of 5 feet per second, it takes him 170 / 5, or 34 seconds to reach point S. Larry: Use the Pythagorean Theorem to find that he has run 130 feet. At a rate of 10 feet every 3 seconds, it takes him (130 / 10) x 3, or 39 seconds to reach point S. Moe, therefore, reaches point S ahead of Larry by 39 - 34, or by 5 seconds, so Moe wins the race. www.imlem.org Meet #3 January 2012 Category 2 – Geometry (11, 4) 1. Given the coordinates in the diagram, what is the distance between the two points? (-1, -1) 2. How many diagonals are there in a regular polygon with sides (a Hexadecagon)? 3. The sum of interior angles in a regular polygon is times as great as the measure of each of its exterior angles. How many sides does the polygon have? Answers 1. __________ Units 2. _________ Diagonals 3. __________ Sides www.imlem.org Meet #3 January 2012 Solutions to Category 2 – Geometery Answers 1. 1. The horizontal distance is units, and the vertical distance is 2. units, so the total distance is √ units. 3. 2. The formula for the number of diagonals in a polygon with N sides is: so in our case we’ll have diagonals. 3. The exterior angles of a polygon all add up to 360 degrees, so if there are sides to the polygon, then each exterior angle measures degrees. Every interior angle measures degrees, and their sum is therefore degrees. So in our case we’re told that: whice we can rewrite as: . Though this is technically a quadratic equation, we know that is a natural number and can easily find that is a solution (an Octagon). [The other solution, , is clearly not an answer to our problem]. www.imlem.org Category 2 - Geometry Meet #3, January 2010 1. The number of diagonals in a polygon is four times the number of its vertices. How many vertices does it have? (A diagonal is a line segment that connects two non-adjacent vertices of a polygon). 2. The exterior angle to a regular polygon N (with N sides) is half that of a regular polygon M (with M sides). Polygon N has 7 times as many diagonals as polygon M. What is the value of 푀 ∙ 푁? 3. Tom stands exactly 2 miles west of Jerry. At 10:00am Tom starts walking east at 5 mph (miles per hour). At 10:20am Jerry starts heading north at 9 mph. How many miles between them at 12:00pm (noon)? Answers 1. _______________ 2. _______________ Remember: You do not have to specify units. Specifying 3. _______________ the wrong units will disqualify your answer. www.imlem.org Solutions to Category 2 - Geometry Answers Meet #3, January 2010 1. 11 2. 50 3. 17 1. If we call the number of vertices V, then the number of diagonals is given by the 푉×(푉−3) 푉−3 formula: and for this to equal 4 × 푉 we get = 4 or 푉 = 11. 2 2 360 2. The exterior angle is so it should be clear that 푁 = 2푀. 푁푢푚푏푒푟 표푓 푠푑푒푠 푀∙(푀−3) 푁∙(푁−3) 2푀∙(2푀−3) Polygon M will have diagonals, and polygon N will have = 2 2 2 diagonals. So we need to solve: 2푀 ∙ 2 ∙ 푀 − 3 = 7 ∙ 푀 ∙ 푀 − 3 which we can simplify to 4 ∙ 푀 − 6 = 7푀 − 21 and then to 3 ∙ 푀 = 15 and the solution is 푀 = 5, 푁 = 10 and 푀 ∙ 푁 = 50. 3. At 12:00pm, having walked for two hours, Tom is 10 miles to the east of his original location, which means he’s 8 miles east of Jerry’s original location. Jerry, after running at 9 mph for 1 hour and 40 minutes is 15 miles north of his (own) original location. The distance between them is 82 + 152 = 289 = 17 푚푙푒푠. Note that we have to compare their locations to some ‘fixed’ or agreed-upon point. You could have used any other point-of-reference and get the same answer. www.imlem.org Category 2 Geometry Meet #3, January 2008 1. If the exterior angle of a regular polygon is 24 degrees, how many sides does the polygon have? 2. In the figure to the right, four of the sides of a regular polygon are shown. Also drawn are all of the diagonals that use A as one of the endpoints of the diagonal; however the full lengths of the diagonals are not always shown. How many diagonals does this polygon have? A 3. In the figure below, triangle ABC is a right triangle with right angle at A and quadrilateral BCDE is a square. If AB = 8 and AC = 18 , what is the area of the pentagon ABEDC? B C A E Answers 1. _______________ D 2. _______________ 3. _______________ Solutions to Category 2 Geometry Meet #3, January 2008 Answers 1. Since the sum of the exterior angles is always 360 degrees, if you divide 360 degrees by 24 degrees you get 15 which is the 1. 15 number of equal exterior angles and therefore 15 vertices and sides to the polygon. 2. 44 3. 20 2. If there are 8 diagonals and only 2 of the diagonals’ other endpoints are shown there must be 6 vertices that are not shown. Since 5 vertices are shown there are a total of 11 vertices and 11 sides. If there are 11 sides there are ͥͥʚͥͥͯͧʛ diagonals total. ͦ Ɣ ̍̍ 3. Using the Pythagorean Theorem: → ͦ ͦ → ͦ ͦ ͦ ͦ ͦ Since̼̽ŬŬŬŬ Ɣ Ŭ ̻̼ŬŬŬ isƍ the ̻̽ŬŬŬ Ŭarea of the̼̽ŬŬŬŬ square,Ɣ √8 theƍ area√18 of square̼̽ŬŬŬŬ EBCDƔ 8ƍ18 is 26. Ɣ The 26 pentagon ͦ we are̼̽Ŭ ŬlookingŬŬ for though is equal to the area of the square minus the area of the right triangle ABC. The area of triangle ABC = √ͬ√ͥͬ √ͥͨͨ ͥͦ . The area of pentagon ABEDC = 26 – 6 = 20 . ͦ Ɣ ͦ Ɣ ͦ Ɣ 6 B C A E D Category 2 Geometry Meet #3, January 2006 1. How many degrees are in the measure of an interior angle of a regular 15-gon? 1 2.
Recommended publications
  • Section 5.5 Right Triangle Trigonometry 385

    Section 5.5 Right Triangle Trigonometry 385

    Section 5.5 Right Triangle Trigonometry 385 Section 5.5 Right Triangle Trigonometry In section 5.3 we were introduced to the sine and cosine function as ratios of the sides of a triangle drawn inside a circle, and spent the rest of that section discussing the role of those functions in finding points on the circle. In this section, we return to the triangle, and explore the applications of the trigonometric functions to right triangles where circles may not be involved. Recall that we defined sine and cosine as (x, y) y sin( θ ) = r r y x cos( θ ) = θ r x Separating the triangle from the circle, we can make equivalent but more general definitions of the sine, cosine, and tangent on a right triangle. On the right triangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle. Right Triangle Relationships Given a right triangle with an angle of θ opposite sin( θ) = hypotenuse hypotenuse opposite adjacent cos( θ) = hypotenuse θ opposite adjacent tan( θ) = adjacent A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.” 386 Chapter 5 Example 1 Given the triangle shown, find the value for cos( α) . The side adjacent to the angle is 15, and the 17 hypotenuse of the triangle is 17, so 8 adjacent 15 cos( α) = = α hypotenuse 17 15 When working with general right triangles, the same rules apply regardless of the orientation of the triangle.
  • Islamic Geometric Ornaments in Istanbul

    Islamic Geometric Ornaments in Istanbul

    ►SKETCH 2 CONSTRUCTIONS OF REGULAR POLYGONS Regular polygons are the base elements for constructing the majority of Islamic geometric ornaments. Of course, in Islamic art there are geometric ornaments that may have different genesis, but those that can be created from regular polygons are the most frequently seen in Istanbul. We can also notice that many of the Islamic geometric ornaments can be recreated using rectangular grids like the ornament in our first example. Sometimes methods using rectangular grids are much simpler than those based or regular polygons. Therefore, we should not omit these methods. However, because methods for constructing geometric ornaments based on regular polygons are the most popular, we will spend relatively more time explor- ing them. Before, we start developing some concrete constructions it would be worthwhile to look into a few issues of a general nature. As we have no- ticed while developing construction of the ornament from the floor in the Sultan Ahmed Mosque, these constructions are not always simple, and in order to create them we need some knowledge of elementary geometry. On the other hand, computer programs for geometry or for computer graphics can give us a number of simpler ways to develop geometric fig- ures. Some of them may not require any knowledge of geometry. For ex- ample, we can create a regular polygon with any number of sides by rotat- ing a point around another point by using rotations 360/n degrees. This is a very simple task if we use a computer program and the only knowledge of geometry we need here is that the full angle is 360 degrees.
  • Geometry Honors Mid-Year Exam Terms and Definitions Blue Class 1

    Geometry Honors Mid-Year Exam Terms and Definitions Blue Class 1

    Geometry Honors Mid-Year Exam Terms and Definitions Blue Class 1. Acute angle: Angle whose measure is greater than 0° and less than 90°. 2. Adjacent angles: Two angles that have a common side and a common vertex. 3. Alternate interior angles: A pair of angles in the interior of a figure formed by two lines and a transversal, lying on alternate sides of the transversal and having different vertices. 4. Altitude: Perpendicular segment from a vertex of a triangle to the opposite side or the line containing the opposite side. 5. Angle: A figure formed by two rays with a common endpoint. 6. Angle bisector: Ray that divides an angle into two congruent angles and bisects the angle. 7. Base Angles: Two angles not included in the legs of an isosceles triangle. 8. Bisect: To divide a segment or an angle into two congruent parts. 9. Coincide: To lie on top of the other. A line can coincide another line. 10. Collinear: Lying on the same line. 11. Complimentary: Two angle’s whose sum is 90°. 12. Concave Polygon: Polygon in which at least one interior angle measures more than 180° (at least one segment connecting two vertices is outside the polygon). 13. Conclusion: A result of summary of all the work that has been completed. The part of a conditional statement that occurs after the word “then”. 14. Congruent parts: Two or more parts that only have the same measure. In CPCTC, the parts of the congruent triangles are congruent. 15. Congruent triangles: Two triangles are congruent if and only if all of their corresponding parts are congruent.
  • Applying the Polygon Angle

    Applying the Polygon Angle

    POLYGONS 8.1.1 – 8.1.5 After studying triangles and quadrilaterals, students now extend their study to all polygons. A polygon is a closed, two-dimensional figure made of three or more non- intersecting straight line segments connected end-to-end. Using the fact that the sum of the measures of the angles in a triangle is 180°, students learn a method to determine the sum of the measures of the interior angles of any polygon. Next they explore the sum of the measures of the exterior angles of a polygon. Finally they use the information about the angles of polygons along with their Triangle Toolkits to find the areas of regular polygons. See the Math Notes boxes in Lessons 8.1.1, 8.1.5, and 8.3.1. Example 1 4x + 7 3x + 1 x + 1 The figure at right is a hexagon. What is the sum of the measures of the interior angles of a hexagon? Explain how you know. Then write an equation and solve for x. 2x 3x – 5 5x – 4 One way to find the sum of the interior angles of the 9 hexagon is to divide the figure into triangles. There are 11 several different ways to do this, but keep in mind that we 8 are trying to add the interior angles at the vertices. One 6 12 way to divide the hexagon into triangles is to draw in all of 10 the diagonals from a single vertex, as shown at right. 7 Doing this forms four triangles, each with angle measures 5 4 3 1 summing to 180°.
  • Polygon Review and Puzzlers in the Above, Those Are Names to the Polygons: Fill in the Blank Parts. Names: Number of Sides

    Polygon Review and Puzzlers in the Above, Those Are Names to the Polygons: Fill in the Blank Parts. Names: Number of Sides

    Polygon review and puzzlers ÆReview to the classification of polygons: Is it a Polygon? Polygons are 2-dimensional shapes. They are made of straight lines, and the shape is "closed" (all the lines connect up). Polygon Not a Polygon Not a Polygon (straight sides) (has a curve) (open, not closed) Regular polygons have equal length sides and equal interior angles. Polygons are named according to their number of sides. Name of Degree of Degree of triangle total angles regular angles Triangle 180 60 In the above, those are names to the polygons: Quadrilateral 360 90 fill in the blank parts. Pentagon Hexagon Heptagon 900 129 Names: number of sides: Octagon Nonagon hendecagon, 11 dodecagon, _____________ Decagon 1440 144 tetradecagon, 13 hexadecagon, 15 Do you see a pattern in the calculation of the heptadecagon, _____________ total degree of angles of the polygon? octadecagon, _____________ --- (n -2) x 180° enneadecagon, _____________ icosagon 20 pentadecagon, _____________ These summation of angles rules, also apply to the irregular polygons, try it out yourself !!! A point where two or more straight lines meet. Corner. Example: a corner of a polygon (2D) or of a polyhedron (3D) as shown. The plural of vertex is "vertices” Test them out yourself, by drawing diagonals on the polygons. Here are some fun polygon riddles; could you come up with the answer? Geometry polygon riddles I: My first is in shape and also in space; My second is in line and also in place; My third is in point and also in line; My fourth in operation but not in sign; My fifth is in angle but not in degree; My sixth is in glide but not symmetry; Geometry polygon riddles II: I am a polygon all my angles have the same measure all my five sides have the same measure, what general shape am I? Geometry polygon riddles III: I am a polygon.
  • Squaring the Circle a Case Study in the History of Mathematics the Problem

    Squaring the Circle a Case Study in the History of Mathematics the Problem

    Squaring the Circle A Case Study in the History of Mathematics The Problem Using only a compass and straightedge, construct for any given circle, a square with the same area as the circle. The general problem of constructing a square with the same area as a given figure is known as the Quadrature of that figure. So, we seek a quadrature of the circle. The Answer It has been known since 1822 that the quadrature of a circle with straightedge and compass is impossible. Notes: First of all we are not saying that a square of equal area does not exist. If the circle has area A, then a square with side √A clearly has the same area. Secondly, we are not saying that a quadrature of a circle is impossible, since it is possible, but not under the restriction of using only a straightedge and compass. Precursors It has been written, in many places, that the quadrature problem appears in one of the earliest extant mathematical sources, the Rhind Papyrus (~ 1650 B.C.). This is not really an accurate statement. If one means by the “quadrature of the circle” simply a quadrature by any means, then one is just asking for the determination of the area of a circle. This problem does appear in the Rhind Papyrus, but I consider it as just a precursor to the construction problem we are examining. The Rhind Papyrus The papyrus was found in Thebes (Luxor) in the ruins of a small building near the Ramesseum.1 It was purchased in 1858 in Egypt by the Scottish Egyptologist A.
  • Interior and Exterior Angles of Polygons 2A

    Interior and Exterior Angles of Polygons 2A

    Regents Exam Questions Name: ________________________ G.CO.C.11: Interior and Exterior Angles of Polygons 2a www.jmap.org G.CO.C.11: Interior and Exterior Angles of Polygons 2a 1 Which type of figure is shown in the accompanying 5 In the diagram below of regular pentagon ABCDE, diagram? EB is drawn. 1) hexagon 2) octagon 3) pentagon 4) quadrilateral What is the measure of ∠AEB? 2 What is the measure of each interior angle of a 1) 36º regular hexagon? 2) 54º 1) 60° 3) 72º 2) 120° 4) 108º 3) 135° 4) 270° 6 What is the measure, in degrees, of each exterior angle of a regular hexagon? 3 Determine, in degrees, the measure of each interior 1) 45 angle of a regular octagon. 2) 60 3) 120 4) 135 4 Determine and state the measure, in degrees, of an interior angle of a regular decagon. 7 A stop sign in the shape of a regular octagon is resting on a brick wall, as shown in the accompanying diagram. What is the measure of angle x? 1) 45° 2) 60° 3) 120° 4) 135° 1 Regents Exam Questions Name: ________________________ G.CO.C.11: Interior and Exterior Angles of Polygons 2a www.jmap.org 8 One piece of the birdhouse that Natalie is building 12 The measure of an interior angle of a regular is shaped like a regular pentagon, as shown in the polygon is 120°. How many sides does the polygon accompanying diagram. have? 1) 5 2) 6 3) 3 4) 4 13 Melissa is walking around the outside of a building that is in the shape of a regular polygon.
  • The Construction, by Euclid, of the Regular Pentagon

    The Construction, by Euclid, of the Regular Pentagon

    THE CONSTRUCTION, BY EUCLID, OF THE REGULAR PENTAGON Jo˜ao Bosco Pitombeira de CARVALHO Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria, Ilha do Fund˜ao, Rio de Janeiro, Brazil. e-mail: [email protected] ABSTRACT We present a modern account of Ptolemy’s construction of the regular pentagon, as found in a well-known book on the history of ancient mathematics (Aaboe [1]), and discuss how anachronistic it is from a historical point of view. We then carefully present Euclid’s original construction of the regular pentagon, which shows the power of the method of equivalence of areas. We also propose how to use the ideas of this paper in several contexts. Key-words: Regular pentagon, regular constructible polygons, history of Greek mathe- matics, equivalence of areas in Greek mathematics. 1 Introduction This paper presents Euclid’s construction of the regular pentagon, a highlight of the Elements, comparing it with the widely known construction of Ptolemy, as presented by Aaboe [1]. This gives rise to a discussion on how to view Greek mathematics and shows the care on must have when adopting adapting ancient mathematics to modern styles of presentation, in order to preserve not only content but the very way ancient mathematicians thought and viewed mathematics. 1 The material here presented can be used for several purposes. First of all, in courses for prospective teachers interested in using historical sources in their classrooms. In several places, for example Brazil, the history of mathematics is becoming commonplace in the curricula of courses for prospective teachers, and so one needs materials that will awaken awareness of the need to approach ancient mathematics as much as possible in its own terms, and not in some pasteurized downgraded versions.
  • Formulas Involving Polygons - Lesson 7-3

    Formulas Involving Polygons - Lesson 7-3

    you are here > Class Notes – Chapter 7 – Lesson 7-3 Formulas Involving Polygons - Lesson 7-3 Here’s today’s warmup…don’t forget to “phone home!” B Given: BD bisects ∠PBQ PD ⊥ PB QD ⊥ QB M Prove: BD is ⊥ bis. of PQ P Q D Statements Reasons Honors Geometry Notes Today, we started by learning how polygons are classified by their number of sides...you should already know a lot of these - just make sure to memorize the ones you don't know!! Sides Name 3 Triangle 4 Quadrilateral 5 Pentagon 6 Hexagon 7 Heptagon 8 Octagon 9 Nonagon 10 Decagon 11 Undecagon 12 Dodecagon 13 Tridecagon 14 Tetradecagon 15 Pentadecagon 16 Hexadecagon 17 Heptadecagon 18 Octadecagon 19 Enneadecagon 20 Icosagon n n-gon Baroody Page 2 of 6 Honors Geometry Notes Next, let’s look at the diagonals of polygons with different numbers of sides. By drawing as many diagonals as we could from one diagonal, you should be able to see a pattern...we can make n-2 triangles in a n-sided polygon. Given this information and the fact that the sum of the interior angles of a polygon is 180°, we can come up with a theorem that helps us to figure out the sum of the measures of the interior angles of any n-sided polygon! Baroody Page 3 of 6 Honors Geometry Notes Next, let’s look at exterior angles in a polygon. First, consider the exterior angles of a pentagon as shown below: Note that the sum of the exterior angles is 360°.
  • A Congruence Problem for Polyhedra

    A Congruence Problem for Polyhedra

    A congruence problem for polyhedra Alexander Borisov, Mark Dickinson, Stuart Hastings April 18, 2007 Abstract It is well known that to determine a triangle up to congruence requires 3 measurements: three sides, two sides and the included angle, or one side and two angles. We consider various generalizations of this fact to two and three dimensions. In particular we consider the following question: given a convex polyhedron P , how many measurements are required to determine P up to congruence? We show that in general the answer is that the number of measurements required is equal to the number of edges of the polyhedron. However, for many polyhedra fewer measurements suffice; in the case of the cube we show that nine carefully chosen measurements are enough. We also prove a number of analogous results for planar polygons. In particular we describe a variety of quadrilaterals, including all rhombi and all rectangles, that can be determined up to congruence with only four measurements, and we prove the existence of n-gons requiring only n measurements. Finally, we show that one cannot do better: for any ordered set of n distinct points in the plane one needs at least n measurements to determine this set up to congruence. An appendix by David Allwright shows that the set of twelve face-diagonals of the cube fails to determine the cube up to conjugacy. Allwright gives a classification of all hexahedra with all face- diagonals of equal length. 1 Introduction We discuss a class of problems about the congruence or similarity of three dimensional polyhedra.
  • Graphing a Circle and a Polar Equation 13 Graphing Calculator Lab (Use with Lessons 91, 96)

    Graphing a Circle and a Polar Equation 13 Graphing Calculator Lab (Use with Lessons 91, 96)

    SSM_A2_NLB_SBK_Lab13.inddM_A2_NLB_SBK_Lab13.indd PagePage 638638 6/12/086/12/08 2:42:452:42:45 AMAM useruser //Volumes/ju110/HCAC061/SM_A2_SBK_indd%0/SM_A2_NL_SBK_LAB/SM_A2_Lab_13Volumes/ju110/HCAC061/SM_A2_SBK_indd%0/SM_A2_NL_SBK_LAB/SM_A2_Lab_13 LAB Graphing a Circle and a Polar Equation 13 Graphing Calculator Lab (Use with Lessons 91, 96) Graphing a Circle in Function Mode 2 2 Graphing 1. Enter the equation of the circle (x - 3) + (y - 5) = 21 as the two Calculator functions, y = ± √21 - (x - 3)2 + 5. Refer to Calculator Lab 1 a. Press to access the Y= equation on page 19 for graphing a function. editor screen. 2 b. Type √21 - (x - 3) + 5 into Y1 2 and - √21 - (x - 3) + 5 into Y2 c. Press to graph the functions. d. Use the or button to adjust the window. The graph will have a more circular shape using 5 rather than 6. Graphing a Polar Equation 2. Change the mode of the calculator to polar mode. a. Press .Press two times to highlight RADIAN, and then press enter. b. Press to FUNC, and then two times to highlight POL, and then press . c. Press two more times to SEQUENTIAL, press to highlight SIMUL, and then press . d. To exit the MODE menu, press . 3. Enter the equation of the circle (x - 3)2 + (y - 5)2 = 34 as the polar equation r = 6 cos θ - 10 sin θ. a. Press to access the r1 = equation editor screen. b. Type in 6 cos θ - 10 sin θ into r1 by pressing Online Connection www.SaxonMathResources.com . 638 Saxon Algebra 2 SSM_A2_NLB_SBK_Lab13.inddM_A2_NLB_SBK_Lab13.indd PagePage 639639 6/13/086/13/08 3:45:473:45:47 PMPM User-17User-17 //Volumes/ju110/HCAC061/SM_A2_SBK_indd%0/SM_A2_NL_SBK_LAB/SM_A2_Lab_13Volumes/ju110/HCAC061/SM_A2_SBK_indd%0/SM_A2_NL_SBK_LAB/SM_A2_Lab_13 Graphing 4.
  • Right Triangles and the Pythagorean Theorem Related?

    Right Triangles and the Pythagorean Theorem Related?

    Activity Assess 9-6 EXPLORE & REASON Right Triangles and Consider △​ ABC​ with altitude ​​CD‾ ​​ as shown. the Pythagorean B Theorem D PearsonRealize.com A 45 C 5√2 I CAN… prove the Pythagorean Theorem using A. What is the area of △​ ​ABC? ​Of △​ACD? Explain your answers. similarity and establish the relationships in special right B. Find the lengths of ​​AD‾ ​​ and ​​AB‾ ​​. triangles. C. Look for Relationships Divide the length of the hypotenuse of △​ ABC​ VOCABULARY by the length of one of its sides. Divide the length of the hypotenuse of ​ △ACD​ by the length of one of its sides. Make a conjecture that explains • Pythagorean triple the results. ESSENTIAL QUESTION How are similarity in right triangles and the Pythagorean Theorem related? Remember that the Pythagorean Theorem and its converse describe how the side lengths of right triangles are related. THEOREM 9-8 Pythagorean Theorem If a triangle is a right triangle, If... ​△ABC​ is a right triangle. then the sum of the squares of the B lengths of the legs is equal to the square of the length of the hypotenuse. c a A C b 2 2 2 PROOF: SEE EXAMPLE 1. Then... ​​a​​ ​ + ​b​​ ​ = ​c​​ ​ THEOREM 9-9 Converse of the Pythagorean Theorem 2 2 2 If the sum of the squares of the If... ​​a​​ ​ + ​b​​ ​ = ​c​​ ​ lengths of two sides of a triangle is B equal to the square of the length of the third side, then the triangle is a right triangle. c a A C b PROOF: SEE EXERCISE 17. Then... ​△ABC​ is a right triangle.