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CHAPTER The Pythagorean 9 CHAPTER 9 OBJECTIVES

● Understand the Pythagorean Theorem more deeply ● Discover the Converse of the Pythagorean Theorem ● Practice working with radical expressions ● Discover relationships among the of the sides of a 45°-45°-90° and among the lengths of the sides of a 30°-60°-90° triangle ● Apply the Pythagorean Theorem and its converse ● Discover and apply the Pythagorean relationship on a coordinate (the formula) ● Derive the of a But serving up an action, suggesting OBJECTIVES from the distance the dynamic in the static, has become a In this chapter you will formula hobby of mine . . . .The “flowing” on that ● discover the Pythagorean motionless plane holds my attention to such a ● Practice using Theorem, one of the most degree that my preference is to try and make it into important concepts in tools a cycle. ● ● Develop reading M. C. ESCHER use the Pythagorean Theorem to calculate the comprehension, Waterfall, M. C. Escher, 1961 distance between any problem-solving skills, ©2002 Cordon Art B. V.–Baarn–Holland. All rights reserved. two points and cooperative behavior ● use conjectures related to the Pythagorean Theorem ● Learn new vocabulary to solve problems

Escher has cleverly used right to form his [Ask] “What impossible things do you see?” artwork known as Water fall. The picture contains [Water seems to be traveling up an incline, yet it is three uses of the impossible tribar created by running a mill wheel.] “Which surfaces appear to British mathematician Roger Penrose (b 1931) in be horizontal? Vertical? Sloped? There are three 1954. In 1934 Swedish artist Oscar Reutersvard impossible tribars in the picture; where are they?” (b 1915), “father of impossible figures,”had created [They all have flowing water along two sides; twice an impossible tribar that consisted of a triangular one of the bars is replaced by the waterfall, and Penrose tribar arrangement of . once one bar is replaced by a group of four columns.] The shapes topping the towers in Escher’s work are, on the left, a compound of three cubes and, on the right, a stellation of the rhombic dodecahedron.

CHAPTER 9 The Pythagorean Theorem 461 LESSON LESSON The Theorem of 9.1 9.1 In a , the side opposite the right is called the In a right triangle, the side opposite the , here with c. PLANNING is called the hypotenuse. I am not young enough to The other two sides are called legs. In c a the figure at right, a and b represent the The other two sides know everything. are legs, here with LESSON OUTLINE lengths of the legs, and c represents the b OSCAR WILDE lengths a and b. One day: length of the hypotenuse. 15 min Investigation 5 min Sharing 10 min Examples 15 min Closing and Exercises

MATERIALS

construction tools FUNKY WINKERBEAN by Batiuk. Reprinted with special permission of North America Syndicate. scissors There is a special relationship between the lengths of the legs and the length of the Pythagorean Theorem (W) for One step hypotenuse. This relationship is known today as the Pythagorean Theorem. Dissection of (W), optional Sketchpad demonstration Three , optional Investigation The Three Sides of a Right Triangle TEACHING You will need The puzzle in this investigation is intended to help you ● scissors recall the Pythagorean Theorem. It uses a dissection, ● a compass which means you will cut apart one or more Many students may already know ● a straightedge geometric figures and make the pieces fit into O the Pythagorean Theorem as ● patty paper another figure. 2 2 2 k a b c . In this lesson they Step 1 Construct a scalene right triangle in review what the letters stand for b the middle of your paper. Label the a and discover proofs showing why hypotenuse c and the legs a and b. the relationship holds for all Construct a on each side of c right triangles. the triangle. j Step 2 To locate the center of the square on the INTRODUCTION longer leg, draw its . Label the Direct students’ attention to center O. Improving Your Visual Thinking Step 3 Through O, construct j Skills on page 454. Ask what they to the hypotenuse and line k can conclude about right trian- perpendicular to line j. Line k is to the hypotenuse. Lines j and k divide the square on the longer leg into four parts. gles, and help them state the Pythagorean Theorem using Step 4 Cut out the square on the shorter leg and the four parts of the square on the of squares and the terms longer leg. Arrange them to exactly cover the square on the hypotenuse. hypotenuse and legs.

Guiding the Investigation trying a special case first—here, an isosceles right with compass and straightedge.It is also easy to triangle. As needed, point out that good pieces might create several examples using geometry software. be formed if they draw lines through the smaller One step Hand out a copy of the Step 2 As needed, remind students that the legs are squares parallel to edges of the largest square. Pythagorean Theorem worksheet the sides other than the hypotenuse, so the “longer to each group. Challenge students [Language] A dissection is the result of separating leg” is not the hypotenuse. Suggest that students to cut up one or both of the something into pieces. minimize clutter by making these diagonals very smaller squares and assemble the light or by drawing only the portion near the center Step 1 Using the Dissection of Squares worksheets or pieces on top of the largest of the square. square. As you circulate, you the Sketchpad demonstration will speed the investiga- might remind students of the tion, but the use of many different triangles drawn by Step 4 Ask students to take care in drawing and problem-solving technique of the students strengthens the inductive conclusion. cutting out pieces so they will fit together well. The constructions are quicker with patty paper than Students may want to tape the pieces together. 462 CHAPTER 9 The Pythagorean Theorem Step 5 State the Pythagorean Theorem. SHARING IDEAS You might make a transparency C-82 The Pythagorean Theorem of the Dissection of Squares In a right triangle, the sum of the squares of the lengths of the legs equals the worksheets for students to use square of the length of the hypotenuse. If a and b are the lengths of the legs, in presenting their ideas. and c is the length of the hypotenuse, then ? . a2 b2 c2 Ask about in the dissected square on the hypot- enuse. The method of the Inves- tigation gives 4-fold rotational History symmetry. [Ask] “What if the triangle isn’t Pythagoras of Samos (ca. 569–475 B.C.E.), a right triangle? Do you think depicted in this statue, is often described as “the first pure mathematician.”Samos was a there’s still a relationship among principal commercial center of Greece and is the lengths of the sides?” You need located on the island of Samos in the Aegean not answer this question now; Sea. The ancient town of Samos now lies in it’s addressed later. [Link] The ruins, as shown in the photo at right. Pythagorean Theorem is a special Mysteriously, none of Pythagoras’s writings still exist, and we know very little about his life. He case of the founded a mathematical society in Croton, in c2 a2 b2 2ab cos C, what is now Italy, whose members discovered irrational numbers and the five where C is the angle opposite regular solids. They proved what is now called the Pythagorean Theorem, side c;when mC 90°, we although it was discovered and used 1000 years earlier by the Chinese and Babylonians. Some math historians believe that the ancient Egyptians also have cos C 0. used a special case of this property to construct right angles. [Ask] “What is the longest side of a right triangle?”“Is it the same as A theorem is a conjecture that has been proved. Demonstrations like the one in the longest leg?” [The hypotenuse, the investigation are the first step toward proving the Pythagorean Theorem. not the longest leg, is the longest Believe it or not, there are more than 200 proofs of the Pythagorean Theorem. side.] If you ask why the longest Elisha Scott Loomis’s Pythagorean Proposition, first published in 1927, contains side is always the hypotenuse and original proofs by Pythagoras, , and even Leonardo da Vinci and U. S. what can be said about the longer President James Garfield. One well-known proof of the Pythagorean Theorem of the two legs, you can review is included below. You will complete another proof as an exercise. the Conjecture and the Side-Angle Inequality ba Paragraph Proof: The Pythagorean Theorem b Conjecture. c 2 2 2 a c You need to show that a b equals c for the right triangles in the figure at left. 2 2 2 [Ask] “What is a theorem?” [It’s a c The of the entire square is a b or a 2ab b . The area of any triangle 1 c a is ab, so the sum of the areas of the four triangles is 2ab. The area of the conjecture that has been proved b 2 in the center is a2 2ab b2 2ab, or a2 b2. deductively within a deductive a b system.] So far in this course no If the quadrilateral in the center is a square then its area also equals c2.You now axiom system has been devel- need to show that it is a square. You know that all the sides have length c, but you also need to show that the angles are right angles. The two acute angles in oped, so there are no real theo- the right triangle, along with any angle of the quadrilateral, add up to 180°. The rems; this conjecture is called a acute angles in a right triangle add up to 90°. Therefore the quadrilateral angle theorem in this book because it measures 90° and the quadrilateral is a square. If it is a square with side length c, has been proved within an axiom 2 2 2 2 then its area is c .So,a b c , which proves the Pythagorean Theorem. system and it’s so well known by that name. If students are not very familiar NCTM STANDARDS LESSON OBJECTIVES with the Pythagorean Theorem, CONTENT PROCESS Understand the Pythagorean Theorem more deeply ask how this theorem about areas of squares might be used to Number Problem Solving Practice using geometry tools calculate lengths. Direct students’ Algebra Reasoning Learn new vocabulary attention to the two examples. Geometry Communication Measurement Connections /Probability Representation

LESSON 9.1 The Theorem of Pythagoras 463 EXAMPLE A The Pythagorean Theorem works for right triangles, but does it work for all triangles? A quick check demonstrates that it doesn’t hold for other triangles. After working through the example, ask students what Acute triangle square roots are and how to find 2 2 2 2 2 2 them using their calculators. 7 6 a b 45 a b 45 a b 45 c 2 25 c 2 45 c 2 57.2 Remind them that many of the 8 square roots they’ll find with 62 72 82 their calculators are approxima- tions. [Ask] “What are some numbers whose square roots are Obtuse triangle whole numbers?” [4, 9, 16, 25, 36, 17 25 a 2 b 2 c 2 a 2 b 2 c 2 a 2 b 2 c 2 and so on] Students will begin to recognize more examples of 38 For an interactive version of this sketch, visit www.keymath.com/DG . perfect squares as they work 172 252 382 through the chapter. Let’s look at a few examples to see how you can use the Pythagorean Theorem to find the distance between two points. The equation h2 375 is not a perfect model for this problem EXAMPLE A How high up on the wall will a 20-foot ladder touch if the because it has two solutions, one foot of the ladder is placed 5 feet from the wall? positive and one negative. The negative solution is ignored Solution The ladder is the hypotenuse of a right triangle, so 20 ft h because all in geometry a2 b2 c2. are positive. 52 h2 202 Substitute. 5 ft EXAMPLE B 25 h2 400 Multiply. 2 In Example A, the book mentions h 375 Subtract 25 from both sides. that 19.4 is an approximation of h 375 19.4 Take the of each side. 375. Point out that in Example The top of the ladder will touch the wall about 19.4 feet up from the ground. B the calculation is exact. See whether students recognize 256 Notice that the exact answer in Example A is 375.However,this is a practical as a perfect square. Again, the application, so you need to calculate the approximate answer. negative square root is being EXAMPLE B ignored. Find the area of the rectangular rug if the width 12 ft is 12 feet and the measures 20 feet. Process text (the reason for each algebraic step) is included with Solution Use the Pythagorean Theorem to find the length. each step in Example A. In 2 2 2 Example B,it has been left up to a b c 2 2 2 students to figure out what alge- 12 L 20 2 L 20 ft braic steps are being carried out. 144 L 400 [Ask] “Why does [this step] L2 256 follow from the previous step?” L 256 L 16 Assessing Progress The length is 16 feet. The area of the is You can assess students’ under- и standing of right triangle, square, 12 16, or 192 square feet. diagonal, and perpendicular. Also watch for their ability to follow instructions and to work together in groups. See how well Closing the Lesson they realize when exact answers The Pythagorean Theorem is a claim about areas are appropriate and when they of squares built on the sides of a right triangle: need to use approximations. The area of the square on the hypotenuse is the sum of the areas of the squares on the legs. Its primary applications are in finding the length of one side of a right triangle in which the lengths of the other two sides are known. The theorem can be proved by dissection.

464 CHAPTER 9 The Pythagorean Theorem EXERCISES BUILDING UNDERSTANDING ᮣ In Exercises 1–11, find each missing length. All measurements are in centimeters. Give approximate answers accurate to the nearest tenth of a centimeter. After students have solved some 1. a ? 12 cm 2. c ? 19.2 cm 3. a ? 5.3 cm exercises, you may want to have several groups report on how 13 a 5 15 they solved them. 12 6

a c 8 ASSIGNING HOMEWORK 1–16 4. d ? 10 cm 5. s ? 26 cm 6. c ? 8.5 cm Essential 8 Portfolio 17, 18 s s c Journal 17, 18 d 10 6 6 Group 18 24 Review 19–22

7. ? 8. ? 9. b 24 cm x 3.6 cm The base is a circle. | x ? 40 cm ᮣ Helping with the Exercises 8 [Alert] As needed, remind 3.9 7 x 41 x students that the hypotenuse is 25 always the longest side and is b 1.5 9 opposite the right angle. 1.5 Exercise 4 Students might wonder why the upper triangle is a right 10. s ? 3.5 cm 11. r ? 13 cm y triangle. [Ask] “What kind of (5, 12) figure is the quadrilateral? Do you see congruent triangles?” 5 r x Exercise 7 [Alert] (0, 0) Some students may want to use the measure 8, s which is not needed in the calculation. 12. A baseball infield is a square, each side measuring 90 feet. To the Second base nearest foot, what is the distance from home plate to second base? Exercise 10 If students are having 127 ft difficulty, ask what shape the 13. The diagonal of a square measures 32 meters. What is the area of quadrilateral is. the square? 512 m2 Exercise 11 This is good prepara- 14. What is the length of the diagonal of a square whose area is 64 cm2? 11.3 cm tion for work with the 15. The lengths of the three sides of a right triangle are consecutive in . . Find them. 3, 4, 5 Home plate Exercises 12–16 Students may find 16. A rectangular garden 6 meters wide has a diagonal measuring it helpful to draw and label 10 meters. Find the of the garden. 28 m pictures. Exercise 12 As an extension, you might have students measure the lengths of the distances between the bases on a local baseball field.

LESSON 9.1 The Theorem of Pythagoras 465 17. The area of the large square 17. One very famous proof of the Pythagorean Theorem is by the is 4 area of triangle area of Hindu mathematician Bhaskara. It is often called the “Behold” proof small square. because, as the story goes, Bhaskara drew the diagram at right and offered no verbal argument other than to exclaim, “Behold.” Use 2 1 2 c 4 2 ab (b a) algebra to fill in the steps, explaining why this diagram proves the c2 2ab b2 2ab a2 Pythagorean Theorem. 2 2 2 c a b History

b Bhaskara (1114–1185, India) was one of the first mathematicians to gain a a thorough understanding of number systems and how to solve , several centuries before European mathematicians. He wrote six books on b a mathematics and astronomy, and led the astronomical observatory at Ujjain.

c 18. Is ABC XYZ? Explain your reasoning. C Z Sample answer: Yes, ABC XYZ by SSS. Both triangles are right triangles, so 4 cm 4 cm you can use the Pythagorean Theorem to find that CB ZY 3 cm. Exercise 18 This problem illustrates AB5 cm XY5 cm a special case in which SSA is a shortcut. That is, when the non-included angle is a right ᮣ Review angle, there is only one triangle 19. 20. that can be formed by SSA. This 8.1 The two are squares. 7.4 Give the arrangement for the Find x. x 21 cm 2-uniform tessellation. 36 32.4.3.4 shortcut is sometimes called HL / (Hypotenuse-Leg). Students can prove HL using the thinking process used in the solution to this 225 cm2 exercise. Given two right triangles 36 cm2 with congruent, corresponding with length c and x corresponding legs with length a, use the Pythagorean Theorem to 4.1 21. Explain why m n 120°. 6.3 22. Calculate each lettered angle, measure, show that the other two corre- or arc. EF is a ; 1 and 2 are sponding legs are congruent; the tangents. triangles are congruent by SSS. 1 Students will prove the HL u F Theorem in Lesson 13.7 t ° m 120° 58 21. Mark the unnamed angles v a n n r as shown in the figure below. g By the Linear Pair Conjecture, h p 120° 180°. p 60°. By s f e d c b AIA, m q. By the Triangle E ° 106 2 Sum Conjecture, q p n 180°. Substitute m q and p 60° to get m 60° n 180°. m n 120°.

22. a 122°, b 74°, c 106°, d 16°, e 90°, f 74°, g 74°, h 74°, n 74°, r 32°, m 120° s 74°, t 74°, u 32°, v 74° p n q

Or use the Exterior Angle Conjecture q n 120°. By AIA q m. Substituting, m n 120°.

466 CHAPTER 9 The Pythagorean Theorem [Context] Eadweard Muybridge, the man who created The Horse in Motion, was born in England CREATING A GEOMETRY FLIP BOOK but moved to the United States Have you ever fanned the pages of a flip book and as a boy.As part of scientific watched the pictures seem to move? Each page shows research to improve techniques a picture slightly different from the previous one. Flip of horseracing, he devised an books are basic to animation technique. For more ingenious method of setting up information about flip books, see www.keymath.com/DG . a dozen or more cameras to go off in rapid sequence. He was a leader in the early days of photography, improving on existing methods, including the art of film developing. He These five frames start off the photo series titled The Horse in Motion, by photographer, invented a precursor to the innovator, and motion picture pioneer Eadweard Muybridge (1830–1904). movie projector based on the Here are two dissections that you can animate to demonstrate the Pythagorean Theorem. idea of a zoetrope, a slitted drum (You used another dissection in the Investigation The Three Sides of a Right Triangle.) with pictures on the inside that is spun to show the illusion of motion when the pictures are watched through the slits.

EXTENSIONS A. Have students research and try one of numerous other dissections that demonstrate the b Pythagorean Theorem. a a B. Use Take Another Look activity 1, 2, or 3 on You could also animate these drawings to demonstrate area formulas. pages 501–502.

Choose one of the animations mentioned above and create a flip book that demonstrates it. Be ready to explain how your flip book demonstrates the formula you chose. Here are some practical tips. ᮣ Draw your figures in the same position on each page so they don’t jump around when the pages are flipped. Use graph paper or tracing paper to help. ᮣ The smaller the change from picture to picture, and the more pictures there are, the smoother the motion will be. ᮣ Label each picture so that it’s clear how the process works.

Supporting the OUTCOMES ᮣ Movement in the flip book is smooth. Suggest that students use a small graph paper ᮣ The student can explain the dissection used. tablet to help keep the nonmoving features in ⅷ The student adds other animation, for the same position from page to page. They can example, drawing a hand flipping the pages of glue tracing papers of the flip book onto cards a flip book, thus creating a flip book of a flip for a firmer flip. book! ⅷ An animation is created using geometry software.

LESSON 9.1 The Theorem of Pythagoras 467 LESSON LESSON The Converse of the 9.2 9.2 Pythagorean Theorem In Lesson 9.1, you saw that if a triangle is a right PLANNING triangle, then the square of the length Any time you see someone of its hypotenuse is equal to the sum more successful than you are, of the squares of the lengths of LESSON OUTLINE the two legs. What about the they are doing something One day: converse? If x, y, and z you aren’t. are the lengths of the 20 min Investigation MALCOLM X three sides of a triangle 10 min Sharing and they satisfy the Pythagorean equation, 5 min Closing a2 b2 c2,must the 10 min Exercises triangle be a right triangle? Let’s find out. MATERIALS

ⅷ rulers

ⅷ string (two 1-meter strings per group) Investigation ⅷ paper clips (six per group) Is the Converse True? ⅷ protractors You will need Three positive integers that work in the Pythagorean equation are called ⅷ patty paper ● string Pythagorean triples. For example, 8-15-17 is a because 2 2 2 ● a ruler 8 15 17 . Here are nine sets of Pythagorean triples. TEACHING ● paper clips 3-4-5 5-12-13 7-24-25 8-15-17 ● a piece of patty paper 6-8-10 10-24-26 16-30-34 The Converse of the Pythagorean 9-12-15 Theorem can also be proved. 12-16-20 Step 1 Select one set of Pythagorean triples from the list above. Mark off four points, Guiding the Investigation A, B, C, and D, on a string to create three consecutive lengths from your set of triples. One step Pose this problem: “What can you say about triangles in 8 cm 15 cm 17 cm which the side lengths satisfy the AB C D 2 2 2 equation a b c ?” As you 1234 6789 11 12 13 14 16 17 0 5 10 15 circulate, encourage groups to try various triples from the list in the student book or to make up their Step 2 Loop three paper clips onto the string. Tie the ends together so that points A and D meet. own lengths, perhaps not integers. Step 3 Three group members should each pull a paper clip at point A, B, or C to As needed, remind students that stretch the string tight. the converse of a true statement might be false. The Pythagorean triples are arranged in families. [Ask] “How are the triples in a column You might let the class do this investigation outside, LESSON OBJECTIVES with the students themselves acting as “rope related?” [They are multiples of ⅷ Discover the Converse of the Pythagorean Theorem the first triple in that column.] stretchers” of long ropes. ⅷ Learn new vocabulary “What would be the next triple Step 1 Suggest that leaving excess string at both ⅷ in the second column?” [15, 36, ends will make it easier to tie the ends together in Develop reading comprehension, problem-solving skills, and cooperative behavior 39] “Why is 3-4-5 or 5-12-13 Step 2. called a primitive Pythagorean triple?” [The numbers in the triple have no common factor.]

468 CHAPTER 9 The Pythagorean Theorem Step 4 right triangle Step 4 With your paper, check the largest angle. What type of triangle is formed? Step 4 Students can also use a Step 5 Select another set of triples from the list. Repeat Steps 1–4 with your corner of a piece of paper (or new lengths. of some other object such as a Step 6 Compare results in your group. State your results as your next conjecture. carpenter’s square) to verify a right angle. C-83 Converse of the Pythagorean Theorem SHARING IDEAS If the lengths of the three sides of a triangle satisfy the Pythagorean equation, Have students show their work then the triangle ? . is a right triangle with a variety of lengths. Let the class discuss measurement errors. Elicit the idea that a slight error in the measurement of a length This ancient Babylonian can result in a measurably tablet, called Plimpton 322, dates sometime between different angle. Keep asking 1900 and 1600 B.C.E.It students whether they think the suggests several advanced converse is always true. Agree Pythagorean triples, such as 1679-2400-2929. on a statement of the conjecture without resolving the question of truth. [Ask] “Will the conjecture hold if you use different measurement units so that the triples change, possibly to non-integers?” [Ask] “Suppose one triangle has sides whose lengths are a Pythagorean triple and another triangle has sides whose lengths History are a multiple of that Pythagorean triple. How are the triangles Some historians believe Egyptian “rope stretchers” used the Converse of the related?” [They are similar.] Pythagorean Theorem to help reestablish land boundaries after the yearly flooding of the Nile and to help construct the pyramids. Some ancient tombs Students can try drawing such show workers carrying ropes tied with equally spaced knots. For example, triangles and looking for patterns, 13 equally spaced knots would divide the rope into 12 equal lengths. If one but you need not answer the person held knots 1 and 13 together, and two others held the rope at knots question yet. It foreshadows the 4 and 8 and stretched it tight, they could have created a 3-4-5 right triangle. ideas of in Chapter 11. Also ask whether students who believe the conjecture is true can prove it deductively. Then have the class read the proof outline in the student book.

MAKING THE CONNECTION Pythagoras himself is believed to have studied in Egypt, and he may have learned the triangle relationship there. Although NCTM STANDARDS some historians discount the CONTENT PROCESS rope stretchers tale, there’s no doubt that Egyptian mathemati- Number Problem Solving cians knew the relationship. Algebra Reasoning Geometry Communication Measurement Connections Data/Probability Representation

LESSON 9.2 The Converse of the Pythagorean Theorem 469 Proof The proof of the Converse of the Pythagorean Theorem is very interesting because Ask students to critique this it is one of the few instances where the original theorem is used to prove the outline in order to deepen their converse. Let’s take a look. One proof is started for you below. You will finish it as an exercise. understanding of it. As they discuss it, monitor their facial Proof: Converse of the Pythagorean Theorem expressions and try to include Conjecture: If the lengths of the three sides of a triangle C students who seem to have ideas work in the Pythagorean equation, then the triangle is but aren’t speaking up. You may a right triangle. ab want to let the discussion lead to Given: a, b, c are the lengths of the sides of ABC and filling in details as a class and a2 b2 c2 BAc then writing up a good model Show: ABC is a right triangle proof. Or, if your students are Plan: Begin by constructing a second triangle, right F fairly comfortable with proof, triangle DEF (with F a right angle), with legs of you can challenge them to write lengths a and b and hypotenuse of length x. The ab up the details as homework. plan is to show that x c, so that the triangles are congruent. Then show that C and F are congruent. EDx Assessing Progress Once you show that C is a right angle, then ABC is You can check how well students a right triangle and the proof is complete. can measure lengths and angles, experiment systematically, and follow a deductive proof. EXERCISES

Closing the Lesson ᮣ In Exercises 1–6, use the Converse of the Pythagorean Theorem to determine whether each triangle is a right triangle. The Converse of the Pythagorean Theorem is true and can be used 1.yes 2. yes 3. no to determine right angles. A 12 36 17 130 common proof actually uses the 8 50 Pythagorean Theorem itself. 35 15 120

4.no 5.no 6. no BUILDING 22 10 20 1.73 1.41 UNDERSTANDING 12

18 24 2.23 These exercises help students practice both the Pythagorean In Exercises 7 and 8, use the Converse of the Pythagorean Theorem and its converse. Theorem to solve each problem. 7. Is a triangle with sides measuring 9 feet, 12 feet, and 18 feet a ASSIGNING HOMEWORK right triangle? no Essential 1–7, 9–15, 19, 20 8. A window frame that seems rectangular has height 408 cm, length 306 cm, and one diagonal with length 525 cm. Is the Performance window frame really rectangular? Explain. 8 assessment No, the given lengths are not a Pythagorean triple. Portfolio 16 Journal 20 Group 17, 18 Review 21–25 Exercise 7 Students’ justifications might cite Exercise 8 After students decide that the angle isn’t Pythagorean multiples. The numbers 9, 12, and 18 right, you might ask whether students could have | ᮣ Helping with the Exercises have a common factor of 3, so a primitive would be known the window frame was rectangular if the 3-4-6. This is not a Pythagorean triple; the familiar angle had turned out to be right. Having one right Exercise 2 This is a good place to 3-4-5 triple says that if the two legs have lengths 3 angle is not a sufficient condition for a quadrilateral point out the power of recog- and 4, the hypotenuse must have length 5 to give a to be a rectangle. nizing Pythagorean multiples. right triangle. The triple 50-120-130 is based on the Pythagorean primitive 5-12-13, so sides of those lengths form a right triangle.

470 CHAPTER 9 The Pythagorean Theorem In Exercises 9–11, find y. Exercise 9 Here is another situa- y 24 units 9. Both quadrilaterals 10.y 11. y 17.3 m tion in which Pythagorean triples are squares. y 25 cm (–7, y) can be applied. For the triangle with y as the hypotenuse, the legs 15 cm have lengths 15 cm and 20 cm. 25 Each length has a common x 25 m y y 25 cm2 factor of 5 cm. Because 3-4-5 is a 18 m common Pythagorean primitive, the hypotenuse must have length 5(5 cm) 25 cm.

12. The lengths of the three sides of a right 13. Find the area of a right triangle with Exercise 10 This exercise provides triangle are consecutive even integers. hypotenuse length 17 cm and one leg groundwork for Chapter 12 Find them.6, 8, 10 length 15 cm. 60 cm2 work with the unit circle in trigonometry. 14. How high on a building will a 15-foot ladder 15. The congruent sides of an touch if the foot of the ladder is 5 feet from measure 6 cm, and the base measures 8 cm. Exercises 12–16 Encourage 2 the building?14.1 ft Find the area. 17.9 cm students to draw pictures. 16. Find the amount of fencing in linear feet needed for the perimeter of a Exercise 12 Students might miss rectangular lot with a diagonal length 39 m and a side length 36 m. 102 m the condition that the integers 17. A rectangular piece of cardboard fits snugly on a diagonal in are even. If appropriate, [Ask] “If this box. x is the first even , how 2 a. What is the area of the cardboard rectangle? 1442 cm 20 cm would the second integer be b. What is the length of the diagonal of the described?” [x 2] “Is there cardboard rectangle? 74.8 cm only one triple of consecutive 60 cm even integers?”[Halfofany such 18. Look back at the start of the proof of the Converse of the 40 cm Pythagorean Theorem. Copy the conjecture, the given, the triple is a triple of consecutive show, the plan, and the two diagrams. Use the plan to integers, say, x 1, x, and x 1. complete the proof. The algebraic equation (x 1)2 x2 (x 1)2 has only two solu- 19. What’s wrong with this picture? 20. Explain why ABC is a right triangle. tions: x 0 and x 4.] Sample answer: The B Sample answer: 2 2 2 2 cm numbers given satisfy BD 6 3 27; Exercise 15 [Ask] “In an isosceles 3.75 cm 2 2 2 the Pythagorean 6 m BC BD 9 108; triangle, what does an 4.25 cm Theorem, so the trian- then AB2 BC2 gle is a right triangle; AC 2 (36 108 144), from the vertex angle to the base but the right angle AD3 m 9 m Cso ABC is a right tri- do to the base?” [It is the perpen- should be inscribed in angle by the Converse dicular bisector of the base; it is an arc of 180°. Thus of the Pythagorean the same as the median.] the triangle is not a Theorem. right triangle. Exercise 16 [Language] Linear feet refers to the length of the fence, ᮣ Review whereas square feet measures area.

3.8 21. Identify the point of concurrency from the Exercise 17 [Alert] Some students construction marks. centroid will have difficulty visualizing the box. They might create a three-dimensional model. Students are being gradually introduced to the Pythagorean Theorem in three . 18. Because DEF is a right triangle, a2 b2 x2. As in Exercises 11 and 13–16, By substitution, c2 x2 and c x. Therefore, exactness of answers will vary. EFD BCA by SSS and C F by CPCTC. The answers given assume that Hence, C is a right angle and BCA is a right measurements given in the exer- triangle. cises are exact and thus answers to the nearest square cm or nearest tenth of a cm are reason- able. In real life the exactness of an answer depends on how it will be used and further knowl- edge of given measurements.

LESSON 9.2 The Converse of the Pythagorean Theorem 471 22. Line CF is tangent to circle D at C. 23. What is the probability of randomly selecting Exercise 22 This exercise can be 6.3 approached through finding that The arc measure of CE is a. Explain three points that form an isosceles triangle 1 3 a why x a. from the 10 points in this isometric grid? the complement of x is 90 2, 2 10 through using the properties of E an isosceles right triangle, or by considering the limit as a secant a D line through C rotates to become x the tangent line (and the inter- F cepted arc becomes the arc with measure a). C Exercise 23 If students are having difficulty, ask how they might count triangles systematically. 2.3 24. If the pattern of blocks continues, what will be the surface area of the 50th solid in One approach is to consider the pattern? 790 square units where the vertex angle might go. Symmetry helps. Exercise 24 As needed, focus students’ attention on the number of square faces being added at each step. 1.8 25. Sketch the solid shown, but with the EXTENSIONS two blue cubes removed and the A. Pose this problem: If the sum red moved to cover the visible face of the green cube. of the squares of the lengths of the two shorter sides of a triangle is less than the square of the length of the longest side, what can you conjecture about the angle opposite the longest side? The outlines of stacked If the sum of the squares of the cubes create a visual lengths of the two shorter sides impact in this untitled module unit sculpture of a triangle is greater than the by conceptual artist square of the length of the longest Sol Lewitt. side, what can you conjecture about the angle opposite the longest side? [If the sum is less, the angle is obtuse. If the sum is IMPROVING YOUR ALGEBRA SKILLS greater, the angle is acute.] Algebraic Sequences III B. Have students use geometry software to draw triangles. Have Find the next three terms of this algebraic sequence. them label and measure angles x9,9x8y,36x7y2,84x6y3, 126x5y4, 126x4y5,84x3y6, ? , ? , ? and sides, calculate squares of the lengths of the sides or construct squares on the sides and measure their area, and drag vertices until the sum of the squares of the lengths of the two smallest sides equals the square of the length of IMPROVING ALGEBRA SKILLS the largest side. Students should Students might think of number combinations, the term that includes x6 is the third term, so its find a right angle. Pascal’s triangle, or symmetry. Or they might 7(36) coefficient is 3 84. The next three terms see this pattern: After the first term, each coeffi- in the sequence are 36x2y7,9xy8,and y9. cient can be determined by multiplying the previous coefficient by the exponent on x in the previous term and then dividing by the number of the term (counting from zero). For example, See page 775 for answers to Exercises 22 and 25.

472 CHAPTER 9 The Pythagorean Theorem LGEBRA SKILLSUSING 1 ● USINGYOUR YOUR ALGEBRA ALGEBRA SSKILLSKILLS 1 ●8USING YOUR ALGEBRA SKILLS 8 ● USING YO USING YOUR ALGEBRA SKILLS 8 Radical Expressions

When you work with the Pythagorean Theorem, you often get radical expressions, PLANNING such as 50. Until now you may have left these expressions as radicals, or you may have found a decimal approximation using a calculator. Some radical expressions can be simplified. To simplify a square root means to take the square root of any LESSON OUTLINE perfect-square factors of the number under the radical sign. Let’s look at an One day or partial day: example. 15 min Examples 20 min Exercises EXAMPLE A Simplify 50. MATERIALS ᮣ Solution One way to simplify a square root is to look for perfect-square factors. none The largest perfect-square factor of 50 is 25. TEACHING 50 25 2 25 25 2 Being able to work with radical 25 is a perfect square, so expressions will help students you can take its square root. later see patterns in special right triangles. Another approach is to factor the number as far as possible with prime factors. One step Draw a right triangle with legs labeled with lengths Write 50 as a set Look for any square factors 4 and 8. Ask what multiple of of prime factors. (factors that appear twice). 4 the length of the hypotenuse is. Students might divide the 2 2 50 5 5 2 5 2 5 252 hypotenuse into four pieces and Squaring and taking the square look at squares of each piece. root are inverse operations— Or they might rewrite 80 as they undo each other. a product of 4 and a square So, 5 2 equals 5. root. During Sharing, ask about working backward, as in Example B. You might argue that 52 doesn’t look any simpler than 50.However,in the days before calculators with square root buttons, mathematicians used paper-and- pencil algorithms to find approximate values of square roots. Working with the INTRODUCTION smallest possible number under the radical made the algorithms easier to use. [Language] The symbol for the Giving an exact answer to a problem involving a square root is important in a nonnegative square root, ,is number of situations. Some patterns are easier to discover with simplified square an example of a radical. Roots roots than with decimal approximations. Standardized tests often express answers to other powers are also called in simplified form. And when you multiply radical expressions, you often have to radicals. simplify the answer. ᮣ EXAMPLE A You might give a definition such NCTM STANDARDS LESSON OBJECTIVES as “Prime factors are factors that can’t be broken down into CONTENT PROCESS ⅷ Learn to simplify square roots smaller factors.” ⅷ Number Problem Solving Learn to multiply radical expressions Emphasize that 52 equals 50 Algebra Reasoning exactly; it is not an approximation. Geometry Communication Measurement Connections Data/Probability Representation

USING YOUR ALGEBRA SKILLS 8 Radical Expressions 473 ᮣ EXAMPLE B ALGEBRA SKILLS 8 ● USING YOUR ALGEBRA SKILLS 8 ● USING YOUR ALGEBRA SKILLS 8 ● USING Y You may want to mention that the commutative property of multiplication allows you to EXAMPLE B Multiply 36 by 52. rewrite 3 6 5 2 as 3 5 6 2. ᮣ Solution To multiply radical expressions, associate and multiply the quantities outside the radical sign, and associate and multiply the quantities inside the radical sign. SHARING IDEAS 3652 3 5 6 2 15 12 15 4 3 15 23 303 You might ask the class to critique this shortcut for Example A as a shortcut for all cases: 5 5 2 EXERCISES immediately becomes 52 because “you can move any ᮣ In Exercises 1–5, express each product in its simplest form. factor that appears twice inside 1. 32 6 2. 52 5 3. 3623 4. 732 147 5. 222 8 the radical to one appearance 182 outside the radical.” Keep asking In Exercises 6–20, express each square root in its simplest form. for justification, using the general 6. 18 32 7. 40 210 8. 75 53 9. 85 85 10. 96 46 rules about square roots of prod- ucts and of squares. 11. 576 24 12. 720 125 13. 722 192 14. 784 28 15. 828 623 Ask whether the relationship 16. 2952 682 17. 5248 882 18. 8200 1082 19. 11808 1282 20. 16072 1482 a a b a b can be repre- 21. What is the next term in the pattern? 2952, 5248, 8200, 11808, 16072,... sented by lengths, remembering 20992 1682 that square roots are often repre- sented by sides of squares. This question foreshadows work with area in Lesson 11.5. IMPROVING YOUR VISUAL THINKING SKILLS Assessing Progress Folding Cubes II You can assess students’ previous Each cube has designs on three faces. When unfolded, which figure at right could understanding of radical expres- it become? sions as well as their ability to 1. see and generalize patterns. A. B. C. D.

Closing the Lesson Two rules are useful in rewriting square roots and in multiplying 2. A. B. C. D. radical expressions: The (posi- tive) square root of a product is the product of the square roots, and the square root of the square of a number is the ( of) the number.

BUILDING UNDERSTANDING | ᮣ Helping with the Exercises Though students are taught to IMPROVING VISUAL THINKING SKILLS simplify radical expressions, they Exercise 2 Squaring and taking the square root are are not taught to rationalize the opposite operations. If students are having difficulty, point out denominators. You might decide that a good problem-solving technique is to Exercise 9 It is possible to factor 85 as 5 17, but to teach that also. eliminate as many choices as possible. For neither factor is a perfect square. example, 1A can be eliminated because the ASSIGNING HOMEWORK Exercise 21 If students are mystified, refer them to base of the green face is touching the red Exercises 16–20. face, unlike its position in the original cube. Essential 1–21 odds 1. D 2. C Group 2–20 evens

474 CHAPTER 9 The Pythagorean Theorem LESSON LESSON Two Special Right Triangles 9.3 In this lesson you will use the Pythagorean Theorem to discover some relationships 9.3 between the sides of two special right triangles. One of these special triangles is an isosceles right triangle, also called a 45°-45°-90° triangle. Each isosceles right triangle is half a square, so they show up often in PLANNING In an isosceles triangle, mathematics and engineering. In the next investigation, you will look for a shortcut the sum of the square roots for finding the length of an unknown side in a 45°-45°-90° triangle. LESSON OUTLINE of the two equal sides is equal to the square root of One day: 45° the third side. 25 min Investigation THE SCARECROW IN THE 1939 10 min Sharing FILM THE WIZARD OF OZ 5 min Closing 45° 5 min Exercises An isosceles right triangle MATERIALS

ⅷ square and isometric dot paper

TEACHING Investigation 1 Isosceles Right Triangles Two special right triangles occur so often in real life (and on Step 1 Sketch an isosceles right triangle. Label the college entrance exams and legs l and the hypotenuse h. achievement tests) that it’s good h Step 2 Pick any integer for l, the length of the legs. l to know the ratios of their side Use the Pythagorean Theorem to find h. lengths. Simplify the square root. l Step 3 Repeat Step 2 with several different One step Ask students to draw on dot paper right triangles in values for l. Share results with 1 your group. Do you see any which the two legs are congruent 1 ? 2 2 pattern in the relationship and right triangles in which the between l and h? 10 10 ? hypotenuse has length twice that 3 Step 4 State your next 3 of one of the legs. They should conjecture in terms of look for the length of the third ? length l. ? sides of these triangles and measure the angles. Isosceles Right Triangle Conjecture C-84 In an isosceles right triangle, if the legs have length l, then the hypotenuse Guiding Investigation 1 has length ? . l2 Step 3 If students are having difficulty seeing a pattern, suggest that they systematically try consecutive integers and make a table of the results. NCTM STANDARDS LESSON OBJECTIVES

CONTENT PROCESS ⅷ Practice simplifying square roots Number Problem Solving ⅷ Discover relationships among the lengths of the sides of a 45°-45°-90° triangle and a 30°-60°-90° triangle Algebra Reasoning ⅷ Develop problem-solving skills and cooperative behavior Geometry Communication Measurement Connections Data/Probability Representation

LESSON 9.3 Two Special Right Triangles 475 You can also demonstrate this property Guiding Investigation 2 2 on a geoboard or graph paper, as 1 2 2 2 shown at right. 1 3 2 Steps 1–3 Students can use patty 3 paper to answer some of the 2 questions in Steps 2 and 3 The other is 3 without doing the measurements 60° a 30°-60°-90° triangle. If you fold an in Step 1. They might also recall along one of its altitudes, the triangles you get are 30°-60°-90° triangles. A 30°-60°-90° triangle is that every altitude is a median. 30° half an equilateral triangle, so it also shows up often in mathematics and engineering. Let’s see if there is a shortcut for finding the lengths of its sides. SHARING IDEAS A 30°-60°-90° triangle After the class reaches consensus about what conjectures to record in their notebooks, ask how to restate the conjectures using Investigation 2 ratios. [For a 45°-45°-90° triangle, 30º-60º-90º Triangles the side lengths have the Let’s start by using a little deductive thinking to find the C 1:1: 2; for a 30°-60°-90° triangle, relationships in 30°-60°-90° triangles. Triangle ABC is they have the ratio 1:3:2.] equilateral, and CD is an altitude. Ask whether these ratios can Step 1 60°; 30°; 90° Step 1 What are mA and mB? What are mACD and be represented geometrically, mBCD? What are mADC and mBDC? remembering that square roots Step 2 yes, SAS,ASA, Step 2 Is ADC BDC? Why? A D B are often shown as sides of or SAA Step 3 Is AD BD? Why? How do AC and AD compare? squares. Students can draw Step 3 Yes, CPCTC. In a 30°-60°-90° triangle, will this relationship between the hypotenuse and the the triangles on dot paper (or AC 2AD; yes all shorter leg always hold true? Explain. isometric dot paper) and calculate 30°-60°-90° triangles Step 4 Sketch a 30°-60°-90° triangle. Choose any the areas of appropriate squares. are similar. integer for the length of the shorter leg. ? [Ask] “How do you know which Use the relationship from Step 3 and the angle is the 30° angle in a Pythagorean Theorem to find the length of the other leg. Simplify the square root. 14 30°-60°-90° triangle?” [By the ? Step 5 7 4 Side-Angle Inequality Conjec- Repeat Step 4 with several different values 2 ture, it’s the angle opposite the for the length of the shorter leg. Share ? 2 results with your group. What is the 1 shortest side.] relationship between the lengths of the [Ask] “How can these special two legs? You should notice a pattern in your answers. triangles be constructed with Step 6 State your next conjecture in terms of the length of the shorter leg, a. straightedge and compass?” [Students can construct perpen- 30°-60°-90° Triangle Conjecture C-85 dicular lines and then lay out equal segments from the inter- In a 30°-60°-90° triangle, if the shorter leg has length a, then the longer leg has length ? and the hypotenuse has length ? . 2a section point to get a 45°-45°-90° a3 triangle. Now that they know the 30°-60°-90° Triangle Conjecture, they can construct one leg and a hypotenuse to determine the third side.] Wonder aloud w hether the 30°-60°-90° Triangle Conjecture Assessing Progress Closing the Lesson can be proved for all triangles, Assess students’ ability to generate isosceles right even if none of the lengths are triangles and equilateral triangles, to simplify In a 45°-45°-90° triangle, if the legs have length l, integers. After students have square roots, to measure angles, to find patterns, then the hypotenuse has length l2. In a 30°-60°- made suggestions, direct their and to follow a deductive proof. Also check their 90° triangle, if the leg opposite the 30° angle has attention to the proof in the understanding of altitudes of isosceles triangles, length a, then the hypotenuse has length 2a and the student book, asking them to SAS, and the Pythagorean Theorem. other leg has length a3. critique and rewrite the reasoning in order to understand it better.

476 CHAPTER 9 The Pythagorean Theorem You can use algebra to verify that the conjecture will hold true for any 30°-60°-90° triangle. BUILDING UNDERSTANDING Proof: 30°-60°-90° Triangle Conjecture

2a2 a2 b2 Start with the Pythagorean Theorem. The exercises give practice with 30° 2 2 2 the two special right triangles. 4a a b Square 2a. 2a b 3a2 b2 Subtract a2 from both sides. ASSIGNING HOMEWORK a3 b Take the square root of both sides. a Essential 1–11 Although you investigated only integer values, the proof shows that any number, even a non-integer, can be used for a. You can also demonstrate this property for Performance integer values on isometric dot paper. assessment 18 Portfolio 17 3 Group 12–16 3 Review 19–23 2 4 6 3 3 3 2 3 3

3 | 1 ᮣ 3 Helping with the Exercises 2 Exercise 3 If students aren’t sure 3 what to do, suggest that in a 30°-60°-90° triangle it often helps to locate the shortest side first. [Ask] “Where is the 30-degree EXERCISES You will need angle?” Construction tools ᮣ In Exercises 1–8, use your new conjectures to find the unknown lengths. for Exercises 19 and 20 All measurements are in centimeters.

1. a ? 722 cm 2. b ? 13 cm 3. a ? , b ? 10 cm, 53 cm

60° a 13 2 a 5

b 72 b

4. c ? , d ? 5. e ? , f ? 6. What is the perimeter of 103 cm, 10 cm 34 cm, 17 cm square SQRE? 72 cm E R ° 20 60 17 3 f 18 2 d ° c 30 e S Q

LESSON 9.3 Two Special Right Triangles 477 Exercise 7 This is another example 7. The solid is a cube. 8. g ? , h ? 9. What is the area of 2 of using the Pythagorean d ? 123 cm 50 cm, 100 cm the triangle? 16 cm Theorem in three dimensions. H [Ask] “How would you generalize G 120 g E d ° h 30 the Pythagorean Theorem to F 8 three dimensions?” [In a right D 130 C rectangular prism, the space diag- A onal (d) can be found from the 12 cm B three dimensions of the prism (a, b, c): d2 a2 b2 c2. 10. Find the coordinates of P. 11. What’s wrong with this picture? Exercise 10 In situations in which y 1 , 1 A 30°-60°-90° triangle must have you want to find the coordinates 2 2 sides whose lengths are multiples of 1, 2, and 3. The triangle shown of a point, it’s often useful to draw P (?, ?) 8 15 does not reflect this rule. segments whose lengths are those 45° x ° ° coordinates. [Link] Students will (1, 0) 60 30 work with the unit circle in 17 trigonometry. (0, –1) 12. possible answer: 12. Sketch and label a figure to demonstrate that 27 is equivalent to 33. (Use isometric dot paper to aid your sketch.) 2 3 6 13. Sketch and label a figure to demonstrate that 32 is equivalent to 42. 1 (Use square dot paper or graph paper.) 2 3 3 3 14. In equilateral triangle ABC, AE,BF, and CD are all angle bisectors, C 1 medians, and altitudes simultaneously. These three segments divide 2 the equilateral triangle into six overlapping 30°-60°-90° triangles 3 and six smaller, non-overlapping 30°-60°-90° triangles. F E 1 a. One of the overlapping triangles is CDB. Name the other five 3 triangles that are congruent to it. CDA, AEC, AEB, BFA, BFC M 13. possible answer: b. One of the non-overlapping triangles is MDA. Name the other A D B five triangles congruent to it. MDB, MEB, MEC, MFC, MFA

4 2 15. Use algebra and deductive reasoning to show that the Isosceles 4 Right Triangle Conjecture holds true for any isosceles right triangle. c Use the figure at right. x 4 16. Find the area of an equilateral triangle whose sides measure 26 meters. 1693 m2 x Exercise 16 Ask students how to find a special triangle in a 17. An equilateral triangle has an altitude that measures 26 meters. picture of this situation. Find the area of the triangle to the nearest square meter. 390 m2 Exercise 18 Students using geom- 18. Sketch the largest 45°-45°-90° triangle that fits in a 30°-60°-90° triangle. What is the etry software might discover that ratio of the area of the 30°-60°-90° triangle to the area of the 45°-45°-90° triangle? letting the hypotenuses (not the right angles) coincide produces a slightly larger triangle. 18. 3 30° 1 15. c2 x2 x2 Start with the Pythagorean Theorem. c2 2x2 Combine like terms. x 3 c x2 Take the square root of both sides. 2x 45° x 2

x 15°

45° x

478 CHAPTER 9 The Pythagorean Theorem 19. Construct an isosceles right ᮣ Review triangle with legs of length a, construct a 30°-60°-90° triangle Construction In Exercises 19 and 20, choose either patty paper or a compass and with legs of lengths a and a3, straightedge and perform the constructions. and construct a right triangle 19. Given the segment with length a below, construct segments with lengths a2, a3, with legs of lengths a2 and and a5. a3.

a a 5 20. Mini-Investigation Draw a right triangle with sides of lengths 6 cm, 8 cm, and 10 cm. a 3 8.5 a 2 a 2a Locate the midpoint of each side. Construct a on each side with the a midpoints of the sides as centers. Find the area of each semicircle. What a a 3 a 2 relationship do you notice among the three areas? Exercise 20 This mini-investiga- 9.1 21. The Jiuzhang suanshu is an ancient tion foreshadows area ratios in text of 246 problems. Some solutions use the gou gu, the Lesson 11.5. Chinese name for what we call the Pythagorean Theorem. The gou gu reads gou2 gu2 (xian)2. 20. Areas: 4.5 cm2,8 cm2, Here is a gou gu problem translated from the ninth 12.5 cm2. 4.5 8 12.5, chapter of Jiuzhang. that is, the sum of the areas of A rope hangs from the top of a pole with three chih of it the on the two legs lying on the ground. When it is tightly stretched so that its end just touches the ground, it is eight chih from the is equal to the area of the semi- 73 circle on the hypotenuse. base of the pole. How long is the rope? 6 12.2 chih 2.622. Explain why m1 m2 90°. 8.7 23. The lateral surface area of the below is 22. Extend the rays that unwrapped into a sector. What is the angle at form the right angle. the vertex of the sector? 80° m4 m5 180° by the 1 Linear Pair Conjecture, and ? it’s given that m 5 90°. l 2 l m 4 90°. m2 m3 m4 r l 27 cm, r 6 cm m2 m3 90° 180°. m2 m3 90°. m3 m1 by AIA. m1 m2 90°. IMPROVING YOUR VISUAL THINKING SKILLS Mudville Monsters 1 5 The 11 starting members of the Mudville Monsters football team and their coach, 4 Osgood Gipper, have been invited to compete in the Smallville Punt, Pass, and Kick 3 2 Competition. To get there, they must cross the deep Smallville River. The only way across is with a small boat owned by two very small Smallville football players. The boat holds just one Monster visitor or the two Smallville players. The Smallville players agree to help EXTENSION the Mudville players across if the visitors agree to pay $5 each time the boat crosses the river. If the Monsters have a total of $100 among them, do they have enough money Use Take Another Look to get all players and the coach to the other side of the river? activity 4 or 5 on page 502.

IMPROVING VISUAL THINKING SKILLS Each Monster visitor must go across alone. It 20 one-way trips. Even without seeing how to makes no sense for one to return, so before arrange the Smallville players to return the boat each Monster crosses, the two Smallville players at least 11 times, the Monsters don’t have must cross and leave one on the other side to enough money. bring the boat back on the next trip. This will require 23 round trips for the 11 players and the coach. The team has enough money for

LESSON 9.3 Two Special Right Triangles 479 EXPLORATION

PLANNING

LESSON OUTLINE A Pythagorean One day: If you wanted to draw a picture to state the 45 min Activity Pythagorean Theorem without words, you’d probably draw a right triangle with squares on MATERIALS each of the three sides. This is the way you first explored the Pythagorean Theorem in Lesson 9.1. ⅷ The Geometer’s Sketchpad Another picture of the theorem is even simpler: ⅷ The Right Triangle Fractal (W), a right triangle divided into two right triangles. optional Here, a right triangle with hypotenuse c is divided into two smaller triangles, the smaller with ⅷ Sketchpad demonstration A Right Triangle Fractal, optional hypotenuse a and the larger with hypotenuse b. Clearly, their areas add up to the area of the whole triangle. What’s surprising is that all three triangles have the same angle measures. Why? Though different in size, the three triangles all have the same shape. Figures that have the same shape but not necessarily the same size are called TEACHING similar figures. Yo u’ll use these similar triangles to prove the Pythagorean Theorem in a later chapter. Students construct a fractal based on the Pythagorean Theorem. a b The student book includes a a b very brief mention of similarity. c You need not elaborate; the topic will be formally addressed in A beautifully complex fractal combines Chapter 11. Also, a proof of the both of these pictorial representations Pythagorean Theorem based on of the Pythagorean Theorem. The similar triangles is an exercise in fractal starts with a right triangle Lesson 13.7. with squares on each side. Then similar triangles are built onto the squares. Then squares are built uiding the Activity onto the new triangles, and so on. G In this exploration, you’ll create this fractal. Sketchpad provides many tools for constructions like this. For example, half of this fractal can be made easily by constructing a square, then constructing a right triangle on one side, and then using the iteration tool. The full fractal can be made with custom tools. If students are not proficient enough using Sketchpad to create the fractal on their own, use The LESSON OBJECTIVES NCTM STANDARDS Right Triangle Fractal worksheet ⅷ with detailed steps for creating the Explore a right triangle fractal CONTENT PROCESS fractal in Sketchpad or use the ⅷ Look for and describe patterns in the fractal Number Problem Solving Sketchpad demonstration. Algebra Reasoning Step 2 Changing the ratio of the lengths of the legs of the original Geometry Communication triangle affects the shape of the Measurement Connections fractal. See page 56 for a quilt design that uses isosceles right Data/Probability Representation triangles.

480 CHAPTER 9 The Pythagorean Theorem Activity Step 3 At each stage, the sum of the areas of the squares added The Right Triangle Fractal to each branch equals the area You will need The Geometer’s Sketchpad software uses custom of the original square on that ● the worksheet tools to save the steps of repeated constructions. branch. The Right Triangle They are very helpful for like this one. Fractal (optional) 1. Use the diameter of a circle Step 4a Step 1 Use The Geometer’s Sketchpad to create the and an to fractal on page 480. Follow the Procedure Note. make a triangle that always remains a right triangle. Notice that each square has two congruent 2. It is important to construct triangles on two opposite sides. Use a reflection the altitude to the to guarantee that the triangles are congruent. hypotenuse in each triangle in order to divide it into similar triangles. 3. Create custom tools to make At Stage 1, the area of the squares and similar triangles After you successfully make the Pythagorean triangle on the hypotenuse fractal, you’re ready to investigate its fascinating repeatedly. patterns. branch equals the area of the original triangle, and the sum of Step 2 First, try dragging a vertex of the original triangle. the areas of the triangles on the Step 3 Does the Pythagorean Theorem still apply to the branches of this figure? That is, other two branches also equals Step 4c At every new stage, does the sum of the areas of the branches on the legs equal the area of the the area of the original triangle. the area added equals the branch on the hypotenuse? See if you can answer without actually measuring all The sum of the two added the areas. yes area of the Stage 0 figure squares equals the area of one plus the area of the Step 4 Consider your original sketch to be a single right triangle with a square built on of the original squares by the original triangle. each side. Call this sketch Stage 0 of your fractal. Explore these questions. Step 4d Its area would be Pythagorean Theorem. Thus a. At Stage 1, you add three triangles and six squares to your construction. On infinite. the total added area is twice a piece of paper, draw a rough sketch of Stage 1. How much area do you add the area of the original triangle Step 5 Sample observa- to this fractal between Stage 0 and Stage 1? (Don’t measure any areas to tions: Each square on the answer this.) plus the areas of all the original hypotenuse branch has a squares, or the area of the b. Draw a rough sketch of Stage 2. How much area do you add between Stage 1 mirror image on one of the Stage 0 figure plus the area leg branches. The squares and Stage 2? of equal area on the leg c. How much area is added at any new stage? of the original triangle. branches all face the same d. A true fractal exists only after an infinite number of stages. If you could build Step 4b direction; that is, corre- a true fractal based on the construction in this activity, what would be its sponding sides in all the total area? squares are parallel. The number of squares Step 5 Give the same color and shade to sets of squares that are congruent. What do of a particular area on the you notice about these sets of squares other than their equal area? Describe any leg branches is one more patterns you find in sets of congruent squares. than the number of squares of the next Step 6 Describe any other patterns you can find in the Pythagorean fractal. largest area on those branches.

The total added area is, again, the area of the Stage 0 figure plus the area of the original triangle.

Step 4 At each stage, triangles whose total area is Step 5 Students can see many patterns. For example, twice that of the original are added, and squares there is reflectional symmetry over a midsegment of whose total area is twice that of the square on the the square on the original hypotenuse. original hypotenuse are added. If the original Step 6 Students’ responses will vary. Check them for triangle has the usual lengths a, b, and c, then an validity and justification. area equal to ab 2c2 is added at each stage. This number becomes larger without bound as the fractal design expands.

EXPLORATION A Pythagorean Fractal 481 LESSON 9.4 LESSON Story Problems You have learned that drawing a diagram will help you to solve difficult problems. 9.4 By now you know to look for many special relationships in your diagrams, such as congruent , parallel lines, and right triangles. PLANNING You may be disappointed if LESSON OUTLINE you fail, but you are doomed if you don’t try. One day: BEVERLY SILLS 25 min Example and Exercises 15 min Sharing

5 min Closing FUNKY WINKERBEAN by Batiuk. Reprinted with special permission of North America Syndicate.

MATERIALS EXAMPLE What is the longest stick that will fit inside a 24-by-30-by-18-inch box? ⅷ calculators ᮣ Solution Draw a diagram.

You can lay a stick with length d diagonally at the 18 in. x TEACHING bottom of the box. But you can position an even d longer stick with length x along the diagonal of 24 in. 30 in. The Pythagorean Theorem has the box, as shown. How long is this stick? many applications. This lesson Both d and x are the hypotenuses of right triangles, but consists primarily of exercises. finding d 2 will help you find x. One step Pose the problem from 302 242 d2 d2 182 x2 the Example and have students 900 576 d2 1476 182 x2 discuss it without opening their d2 1476 1476 324 x2 books. 1800 x2 x 42.4 ᮣ EXAMPLE The longest possible stick is about 42.4 in. The distance x is called the space diagonal. To find its length, you need d2,but you don’t need to find d. EXERCISES

Assessing Progress ᮣ 1. A giant California redwood tree 36 meters tall cracked Through their work on the in a violent storm and fell as if hinged. The tip of the exercises, you can assess once beautiful tree hit the ground 24 meters from the students’ understanding of base. Researcher Red Woods wishes to investigate the x the Pythagorean Theorem. crack. How many meters up from the base of the tree does he have to climb? 10 m 24 m

SHARING IDEAS 2. Amir’s sister is away at college, and he wants to mail her a 34 in. baseball bat. The Have several groups prepare packing service sells only one kind of box, which measures 24 in. by 2 in. by 18 in. some of their solutions to the Will the box be big enough? No. The space diagonal of the box is 30.1 in. exercises on transparencies and share them with the class. Keep asking whether the results seem LESSON OBJECTIVES NCTM STANDARDS reasonable. ⅷ Apply the Pythagorean Theorem and its converse CONTENT PROCESS Closing the Lesson ⅷ Develop reading comprehension and problem-solving skills Number Problem Solving Emphasize that a good first step Algebra Reasoning in solving application problems is to draw a picture. Then students Geometry Communication should examine the picture for Measurement Connections common geometric shapes, such as right triangles, and apply what Data/Probability Representation they know about those shapes.

482 CHAPTER 9 The Pythagorean Theorem 3. Meteorologist Paul Windward and geologist Rhaina Stone are rushing to a paleontology conference in Pecos Gulch. BUILDING Paul lifts off in his balloon at noon from Lost Wages, UNDERSTANDING heading east for Pecos Gulch Conference Center. With the wind blowing west to east, he averages a land speed of You may want to divide up the 30 km/hr. This will allow him to arrive in 4 hours, just as exercises and have different groups the conference begins. Meanwhile, Rhaina is 160 km north of Lost Wages. At the moment of Paul’s lift off, Rhaina work on different exercises. hops into an off-roading vehicle and heads directly for the conference center. At what average speed must she travel to ASSIGNING HOMEWORK arrive at the same time Paul does? 50 km/hr Essential 1–5 Career Performance assessment 6 Meteorologists study the weather and the atmosphere. They also look at air Portfolio 9 quality, oceanic influence on weather, changes in climate over time, and even other planetary climates. They make forecasts using satellite photographs, Group 7–10 weather balloons, contour maps, and mathematics to calculate wind speed or 11–20 the arrival of a storm. Review

| ᮣ 4. A 25-foot ladder is placed against a building. The bottom of Helping with the Exercises the ladder is 7 feet from the building. If the top of the ladder slips down 4 feet, how many feet will the bottom Exercise 1 If students are having slide out? (It is not 4 feet.) 8 ft difficulty, [Ask] “If the tree was originally 36 meters tall and 5. The front and back walls of an A-frame cabin are isosceles it cracked with x meters left triangles, each with a base measuring 10 m and legs standing, how much has fallen?” measuring 13 m. The entire front wall is made of glass 1 cm thick that cost $120/m2. What did the glass for the [36 x] You might also use front wall cost? area: 60 m2; cost: $7200 specific numbers until students see the general case. Exercise 5 As needed, suggest that 6. A regular hexagonal prism fits students draw altitudes to create perfectly inside a cylindrical box with diameter 6 cm and height 10 cm. What right triangles. is the surface area of the prism? What Exercise 6 [Alert] Some students is the surface area of the ? may have forgotten that a hexagon 2 surface area of prism 273 180 cm inscribed in a circle has sides the 226.8 cm2; surface area of cylinder 78 cm2 245.0 cm2 same length as the circle’s radius. 7. Find the perimeter of an equilateral triangle whose median Exercise 7 After students have measures 6 cm. 36 cm 3 drawn pictures, you might 8. APPLICATION According to the Americans with Disabilities [Ask] “Which side of the Act, the slope of a wheelchair ramp must be no greater than 30°-60°-90° triangle do we 1 12. What is the length of ramp needed to gain a height of know?” [the longer leg] 4 feet? Read the Science Connection on the top of page 484 and then figure out how much force is required to go up the ramp if a person and a wheelchair together weigh 200 pounds. 48.2 ft; 16.6 lb

LESSON 9.4 Story Problems 483 11. 2 2 2 2 Science 2 2 2 2 2 2 It takes less effort to roll objects up an inclined plane, or ramp, than to lift 4 2 them straight up. Work is a measure of force applied over distance, and you 2 2 calculate it as a product of force and distance. For example, a force of 2 2 4 2 100 pounds is required to hold up a 100-pound object. The work required to lift it 2 feet is 200 foot-pounds. But if you use a 4-foot-long ramp to 4 2 roll it up, you’ll do the 200 foot-pounds of work over a 4-foot distance. So you need to apply only 50 pounds of force at any given moment.

For Exercises 9 and 10, refer to the above Science Connection about inclined planes.

9. Compare what it would take to lift an object these three different ways. a. How much work, in foot-pounds, is necessary to lift 80 pounds straight up 2 feet? 160 ft-lb b. If a ramp 4 feet long is used to raise the 80 pounds up 2 feet, how much force, in pounds, will it take? 40 lb c. If a ramp 8 feet long is used to raise the 80 pounds up 2 feet, how much force, in pounds, will it take? 20 lb

10. If you can exert only 70 pounds of force and you need to lift a 160-pound steel drum up 2 feet, what is the minimum length of ramp you should set up? 4.6 ft

ᮣ Review

Making the Connection Recreation [Language] Qi qiao is – – . pronounced [che cheau]. This set of enameled porcelain qi qiao bowls can be arranged to form a 37-by-37 cm square (as shown) or other shapes, or used separately. Each bowl is 10 cm deep. Dishes of this type are usually used to serve candies, nuts, dried fruits, and other snacks on special occasions. The qi qiao, or tangram puzzle, originated in and consists of seven pieces—five isosceles right triangles, a square, and a . The puzzle involves rearranging the pieces into a square, or hundreds of other shapes (a few are shown below).

Private collection, Berkeley, California. Photo by Cheryl Fenton.

Swan

Cat Rabbit Horse with Rider

9.1 11. If the area of the red square piece is 4 cm2, what are the dimensions of the other six pieces?

484 CHAPTER 9 The Pythagorean Theorem 12. Make a set of your own seven tangram pieces and create the Cat, Rabbit, Swan, and 12. Horse with Rider as shown on page 484. 9.3 13. Find the radius of circle Q. 9.1 14. Find the length of AC. 6.215. The two rays are tangent to 12 units 182 cm the circle. What’s wrong with y this picture? A 15. Draw radii CD and CB. 54° (?, 6) C B ABC ADC 90°. For 150° 36 cm quadrilateral ABCD 54° r x Q 45° 30° 90° m C 90° 360°, so AB C mC 126°. BD 126° but D 126° 226° 360°. 226°

7.1 16. In the figure below, point A4.5 17. Which congruence shortcut 3.718. Identify the point of Exercises 19, 20, 18 [Language] The is the image of point A after can you use to show that concurrency in QUO figure names in these exercises a reflection over OT. What ABP DCP?SAA from the construction form a legal phrase: quid pro quo. are the coordinates of A ? marks. orthocenter Quid pro quo means “something y 4, 43 U for something.” It’s used to mean A BD the consideration for a contract, T that is, what each party gets out P Q of it (such as money or advan- ° A AC 30 x O tage). A similar colloquial O (8, 0) expression is “tit for tat.”

5.5 19. In parallelogram QUID, mQ 2x 5° and mI 4x 55°.What is mU? 115° EXTENSION 4.3 20. In PRO, mP 70° and mR 45°. Which side of the triangle is the shortest? PO Have students show algebraically that 3-4-5 is the only Pythagorean triple of consecutive positive integers. IMPROVING YOUR VISUAL THINKING SKILLS Fold, Punch, and Snip A square sheet of paper is folded vertically, a hole is punched out of the center, and then one of the corners is snipped off. When the paper is unfolded it will look like the figure at right. Sketch what a square sheet of paper will look like when it is unfolded after the following sequence of folds, punches, and snips.

Fold once. Fold twice. Snip double-fold corner. Punch opposite corner.

IMPROVING VISUAL THINKING SKILLS

LESSON 9.4 Story Problems 485 LESSON LESSON Distance in Coordinate 9.5 9.5 Geometry Viki is standing on the corner of Seventh Street and 8th Avenue, and her brother PLANNING Scott is on the corner of Second Street and 3rd Avenue. To find her shortest sidewalk We talk too much; we should route to Scott, Viki can simply count blocks. But if Viki wants to know her diagonal distance to Scott, she would need the Pythagorean Theorem to measure across blocks. LESSON OUTLINE talk less and draw more. JOHANN WOLFGANG VON GOETHE 8th Ave One day: V 7th Ave 25 min Investigation and Examples 6th Ave 5 min Sharing 5th Ave 4th Ave 5 min Closing 3rd Ave S 10 min Exercises 2nd Ave 1st Ave MATERIALS First St Fifth St Sixth St

ⅷ Third St Eighth St Fourth St

graph paper Second St Seventh St ⅷ The Distance Formula (T), optional

You can think of a coordinate plane as a grid of streets with two sets of parallel TEACHING lines running perpendicular to each other. Every segment in the plane that is not in the x- or y-direction is the hypotenuse of a right triangle whose legs are in the If your curriculum or your x- and y-directions. So you can use the Pythagorean Theorem to find the distance students’ background requires between any two points on a coordinate plane. that you emphasize both the y distance formula and the equa- tion of a circle, you may want to plan two days for this lesson. One step Pose this problem: “Viki is standing on a street corner and is trying to talk with Scott by walkie-talkie. He says he’s x also at a corner, but static keeps Viki from understanding which corner. If each block is one-tenth Investigation 1 mile long, what are the possible The Distance Formula straight-line distances Scott You will need In Steps 1 and 2, find the length of each segment by using the segment as the could be from Vicki and still be ● graph paper hypotenuse of a right triangle. Simply count the squares on the horizontal and within a half mile?” As needed, vertical legs, then use the Pythagorean Theorem to find the length of the encourage students to draw hypotenuse. right triangles to find diagonal Step 1 Copy graphs a–d from the next page onto your own graph paper. Use each distances. During Sharing, ask segment as the hypotenuse of a right triangle. Draw the legs along the grid lines. for an equation describing Scott’s Find the length of each segment. possible locations if he’s exactly 1 2 mile away but not necessarily at a corner. LESSON OBJECTIVES NCTM STANDARDS ⅷ Discover the Pythagorean relationship on a coordinate plane CONTENT PROCESS (the distance formula) Number Problem Solving ⅷ Derive the equation of a circle from the distance formula Algebra Reasoning ⅷ Use the distance formula to solve problems ⅷ Develop problem-solving skills and cooperative behavior Geometry Communication Measurement Connections Data/Probability Representation

486 CHAPTER 9 The Pythagorean Theorem y y a. b. Guiding Investigation 1 5 5 Step 1 Students used triangles like this earlier in finding slopes of lines. You might provide The x x Distance Formula worksheet so 5 5 20 25 4.5 40 210 6.3 students aren’t tempted to write c.y d. y in their books. You might use pair share within groups or 5 jigsaw between groups to x –5 save time. Step 2 As needed, encourage x –4 students to use right triangles so 5 they can see the distance formula 68 217 8.2 100 10 Step 2 Graph each pair of points, then find the distances between them. as a special application of the Pythagorean Theorem. Pair share a. (1, 2), (11, 7) 13 b. (9,6), (3, 10) 20 works well for this step also. Step 3 Students may want to What if the points are so far apart that it’s not discuss how to handle negative distances. Some students may practical to plot them? For example, what is the B (42, 70) distance between the points A(15, 34) and want to take the absolute value B(42, 70)? A formula that uses the coordinates of so that all distances are positive. ? the given points would be helpful. To find this Others may be comfortable with formula, you first need to find the lengths of the A (15, 34) ? negative values as a measure legs in terms of the x- and y-coordinates. From of directed distance. Still other your work with slope triangles, you know how to calculate horizontal and vertical distances. students may advocate always subtracting the smaller Step 3 Write an expression for the length of the horizontal leg using the x-coordinates. x-coordinate from the larger Step 4 Write a similar expression for the length of the vertical leg using the so the distance is always positive. y-coordinates. Step 3 Subtract one x-coordinate Step 5 Use your expressions from Steps 3 and 4, and the Pythagorean Theorem, to find from the other. the distance between points A(15, 34) and B(42, 70). Step 4 Subtract one y-coordinate Step 6 Generalize what you have learned about the distance between two points in a coordinate plane. Copy and complete the conjecture below. from the other. Step 5 Distance Formula C-86 AB2 (42 15)2 (70 34)2; AB 729 1296 45 The distance between points A x1, y1 and B x2, y2 is given by Step 6 AB2 ? 2 ? 2 or AB ? 2 ? 2 Students may remember the distance formula from algebra 2 2 2 2 x2 x1 y2 y1 x2 x1 y2 y1 I, perhaps using d instead of AB.

Let’s look at an example to see how you can apply the distance formula.

LESSON 9.5 Distance in Coordinate Geometry 487 ᮣ EXAMPLE A EXAMPLE A Find the distance between A(8, 15) and B(7, 23). Because the distance ᮣ 2 2 2 formula is derived directly Solution AB x2 x1 y2 y1 The distance formula. 2 2 from the Pythagorean Theorem, 7 8 23 15 Substitute 8 for x1, 15 for y1, 7 for x2, and 23 for y2. the book begins with the 152 82 Subtract. 2 2 2 a b c version of the AB2 289 Square 15 and 8 and add. distance formula. You could also AB 17 Take the square root of both sides. show that you can start with 2 2 The distance formula is also used to write the equation of a circle. AB x2 x1 y2 y1 . [Alert] If students are using calculators, they can easily get EXAMPLE B Write an equation for the circle with center (5, 4) and radius 7 units. incorrect answers if parentheses y are omitted or not used correctly. ᮣ Solution Let (x, y) represent any point on the circle. The distance from (x, y) to the circle’s center, (5, 4), is 7. Substitute (x, y) ᮣ EXAMPLE B this information in the distance formula. (x 5)2 (y 4)2 72 (5, 4) Students may benefit from being x reminded that an equation for a geometric figure is satisfied by the Substitute Substitute Substitute 7 coordinates of a point if and only (x, y) for (x2, y2). (5, 4) for (x1, y1). as the distance. if the point lies on the figure. So, the equation in standard form is x 52 y 42 72. Guiding Investigation 2 Step 1 [Alert] Students may inat- Investigation 2 tentively think of diameter The Equation of a Circle instead of radius. You will need Find equations for a few more and then generalize the equation for any Step 1a y ● graph paper circle with radius r and center (h, k). Step 1 Given its center and radius, graph each circle on graph paper. x (1, –2) a. Center (1, 2), r 8 b. Center (0, 2), r 6 c. Center (3, 4), r 10 (x, y) Step 2a Step 2 Select any point on each circle; label it (x, y). Use the distance formula to write (x 1)2 (y 2)2 64 an equation expressing the distance between the center of each circle and (x, y). Step 1b y Step 2b Step 3 Copy and complete the conjecture for the equation of a circle. (x, y) x2 (y 2)2 36 (0, 2) Step 2c y C-87 x (x 3)2 (y 4)2 100 Equation of a Circle (x, y) The equation of a circle with radius r and r center (h, k) is (h, k) Step 1c y ? 2 ? 2 ? 2 x y x hkr

x (–3, –4)

(x, y)

SHARING IDEAS When discussing the equation of a circle, ask whether you could get an equivalent formula by Step 2 As needed, encourage Students may have come up with a variety of distance taking the square root, as with the distance formula. students to use the model in formulas. For example, they may have either x2 x1 Students may say r (x h)2 ( y k)2 . Example B. or y2 y1 first, or they might have x1 x2 in place [Ask] “Should the equation include negative values of x2 x1 or y1 y2 in place of y2 y1.Through 2 2 After doing this investigation, student discussion, elicit the idea that these are all of r, r (x h)( yk) ? [No; r is a radius, you may want to work with the same formula after squaring and adding. so it cannot be negative.] The equation with r2 is another example. [Ask] “What called the standard form. is the equation for a circle with [Ask] “In Example A, how would you know to substitute the coordinates (8, 15) for x , y instead Ask how students might use the equation to graph center (2, 3) and radius 4?” 1 1 2 2 a circle with center at (1, 2) and radius 8. They [(x 2) ( y 3) 16] of for x2, y2 ?” Student experimentation can lead to the realization that it doesn’t matter. need to solve the equation for y in terms of x. They’ll get ( y 2)2 64 (x 1)2,from 488 CHAPTER 9 The Pythagorean Theorem Let’s look at an example that uses the equation of a circle in reverse. ᮣ EXAMPLE C Point out that (x 2)2 can be 2 2 EXAMPLE C Find the center and radius of the circle x 2 y 5 36. rewritten as (x (2))2. Assessing Progress ᮣ Solution Rewrite the equation of the circle in the standard form. You can assess students’ famil- x h2 y k2 r 2 iarity with coordinates of points, x (2)2 y 52 62 circles and their radii, right triangles and their hypotenuses, Identify the values of h, k, and r. The center is (2, 5) and the radius is 6. and the Pythagorean Theorem. Also check students’ ability to work separately with the hori- EXERCISES zontal and vertical distances.

ᮣ In Exercises 1–3, find the distance between each pair of points. Closing the Lesson

1. (10, 20), (13, 16) 5 units 2. (15, 37), (42, 73) 45 units 3. (19, 16), (3, 14) 34 units The distance formula, AB x x 2 y y 2, 4. Look back at the diagram of Viki’s and Scott’s locations on page 486. Assume each 2 1 2 1 block is approximately 50 meters long. What is the shortest distance from Viki to and the related equation of a 2 2 2 Scott to the nearest meter? 354 m circle, (x h) ( y k) r , are both derived from the 5. Find the perimeter of ABC with vertices A(2, 4), B(8, 12), and C(24, 0). 52.4 units Pythagorean Theorem. 6. Determine whether DEF with vertices D(6, 6), E(39, 12), and F(24, 18) is scalene, isosceles, or equilateral. isosceles

For Exercises 7 and 8, find the equation of the circle. BUILDING UNDERSTANDING 7. Center (0, 0), r 4 x2 y2 16 8. Center (2, 0), r 5 (x 2)2 y2 25

For Exercises 9 and 10, find the radius and center of the circle. These exercises motivate the need for a formula to find the 2 2 2 2 2 9. x 2 y 5 6 center is (2, 5), r 6 10. x y 1 81 center is (0, 1), r 9 distance between points when 11. The center of a circle is (3, 1). One point on the circle is (6, 2). Find the equation plotting the points is impractical. of the circle. (x 3)2 (y 1)2 18 ASSIGNING HOMEWORK 12. Mini-Investigation How would you find the This point is the graph z distance between two points in a Essential 1–10 three-dimensional coordinate system? of the ordered triple 5 Investigate and make a conjecture. (2, 1, 3). Performance 12 a. What is the distance from the origin (0, 0, 0) –4 assessment to (2, 1, 3)? 14 units Portfolio 6 b. What is the distance between P(1, 2, 3) and y Journal 17 Q(5, 6, 15)? 176 411 units –5 5 c. Complete this conjecture: Group 11 4 If A x1, y1, z1 and B x2, y2, z2 are two Review 13–17 points in a three-dimensional coordinate x –5 system, then the distance AB is | ᮣ ? 2 ? 2 ? 2 . Helping with the Exercises Exercise 3 [Alert] Students may not be careful enough in substi- Sharing Ideas (continued) 11c. 2 2 2 AB x1 x2 y1 y2 z1 z2 tuting the negative numbers into which they may want to take square roots the distance formula. Exercise 12 The radius is determined to be 18 or to get y 2 64 (x 1)2 and hence 32. The equation is then (x 3)2 ( y 1)2 Exercise 11 As needed, encourage y 2 64 (x 1)2 .When they graph 322 or (x 3)2 ( y 1)2 182,or students to break apart the prob- this on a calculator, they’ll get only a semi- simply (x 3)2 ( y 1)2 18. This is a good lem and consider a diagonal of circle, because in taking the square root they elimi- example of a case in which the “unsimplified” form a horizontal or vertical rectangle nated negative values of y 2. To graph the other 18 is more useful than the “simplified” form, as an intermediate step. Or ask half of the circle, they’ll need to graph the second because it’s easier to square. whether the distance could be a 2 function y 2 64 (x 1) . space diagonal of an imaginary Their graphs still might not look very circular. box. Suggest that they use a friendly window. LESSON 9.5 Distance in Coordinate Geometry 489 Exercise 15 As needed, ask whether the triangle is a special ᮣ Review kind. If a student rationalizes the denominator, the solutions are 9.3 13. Find the coordinates of A. 9.114. k ? , m ? 9.1 15. The large triangle is y 3 1 k 2, m 6 equilateral. Find x and y. x 2 3 and y 4 3. , 2 2 x 6 , y 12 Exercise 17 If students are having 60° 3 3 A (?, ?) difficulty, ask if they know any k 12 150° m properties of the center of a y x x rotation. (1, 0) 3

(0, –1)

16. Antonio is a biologist studying life in a pond. He needs to know how deep the water is. He notices a water lily sticking straight up from the water, whose blossom is 8 cm above the water’s surface. Antonio pulls the lily to one side, keeping the stem straight, until the blossom touches the water at a spot 40 cm from where the stem first broke the water’s surface. How is Antonio able to calculate the depth of the water? What is the depth? 96 cm

R 7.1 17. CURT is the image of CURT under a R' rotation transformation. Copy the C' and its image onto patty paper. Find the T center of rotation and the measure of the T' U angle of rotation. Explain your method. The angle of rotation is approximately 77°. Connect two pairs of corresponding points. Construct the perpendi- C cular bisector of each segment. The point where the U' perpendicular bisectors meet is the center of rotation.

IMPROVING YOUR VISUAL THINKING SKILLS The Spider and the Fly The Spider and the Fly Here is another path shorter (attributed to the British puzzlist Henry E. Dudeney, 1857–1930) than 42 feet. It is not the In a rectangular room, measuring 30 by 12 by 12 feet, a spider 30 ft shortest. is at point A on the middle of one of the end walls, 1 foot from Side 40.7 the ceiling. A fly is at point B on the center of the opposite wall Ceiling 17 wall, 1 foot from the floor. What is the shortest distance 12 ft 6 ft 37 that the spider must crawl to reach the fly, which remains Side stationary? The spider never drops or uses its web, but 11 ft 12 ft Back wall wall crawls fairly. 1 ft 30 ft 6 ft

IMPROVING VISUAL THINKING SKILLS [Context] Henry E. Dudeney (1857–1930, two dimensions. They might draw several 30 ft Side 40 England) was a puzzle inventor whose different nets of the room and draw wall A Ceiling 24 6 ft puzzles continue to challenge people today. straight lines between the points on the 32 net. [Ask] “Which net will give the shortest Back His spider-and-fly problem first appeared 12 ft 40 ft in an English newspaper. distance? What does that mean for the wall original room?” 6 ft Floor B If students are stuck, wonder aloud whether Side wall it would help to think of the problem in The shortest path measures 40 feet. 1 ft30 ft 1 ft

490 CHAPTER 9 The Pythagorean Theorem EXPLORATION

PLANNING

Ladder Climb LESSON OUTLINE One day: Suppose a house painter rests a 20-foot ladder against a building, 25 min Activity then decides the ladder needs to 20 min Sharing and Closing rest 1 foot higher against the building. Will moving the ladder 1 foot toward the building do the MATERIALS job? If it needs to be 2 feet lower, ⅷ rulers will moving the ladder 2 feet away from the building do the trick? ⅷ graphing calculators Let’s investigate. TEACHING Activity Climbing the Wall [ESL] Do the trick means “accom- plish the task.” You will need Sketch a ladder leaning against a ● a graphing calculator vertical wall, with the foot of the ladder resting on horizontal ground. Label the sketch using y for height reached by G uiding the Activity the ladder and x for the distance from the base of the wall to the foot of the ladder. Step 1 Step 1 Write an equation relating x, y, and the length of the ladder and solve it for y. Yo u Step 2 Ask students to conjecture y 400x2 now have a function for the height reached by the ladder in terms of the distance whether the amount of change Step 2 0 x 20; from the wall to the foot of the ladder. Enter this equation into your calculator. will always be the same and, 0 y 20 Step 2 Before you graph the equation, think about the settings you’ll want for the graph if not, what it depends on. As Step 3 a quarter-circle window. What are the greatest and least values possible for x and y? Enter needed, help students see where reasonable settings, then graph the equation. Step 4 (3, 19.8); (6, 19.1); the maximum value of each (9, 17.9); (12, 16) Step 3 Describe the shape of the graph. variable occurs (on an axis). No.At first the height reached by the ladder Step 4 Trace along the graph, starting at x 0. Record values (rounded to the nearest Step 3 Unless students happened decreases slowly, but the 0.1 unit) for the height reached by the ladder when x 3, 6, 9, and 12. If you to choose a friendly window in farther the ladder is pulled move the foot of the ladder away from the wall 3 feet at a time, will each move Step 2, they may say that the out, the faster the height result in the same change in the height reached by the ladder? Explain. reached by it decreases. graph is part of a Step 5 Find the value for x that gives a y-value approximately equal to x. How is this rather than an arc of a circle. Step 5 When x y value related to the length of the ladder? Sketch the ladder in this position. What at x 14, the length of the angle does the ladder make with the ground? Step 5 Suggest that students ladder divided by 2,the study a table of these data. ladder forms a 45° Step 6 Should you lean a ladder against a wall in such a way that x is greater than y? angle with the floor and Explain. How does your graph support your explanation? the wall. SHARING IDEAS As students present their ideas about Steps 4–6, [Ask] “At what point will a one-foot change in x NCTM STANDARDS LESSON OBJECTIVES result in a one-foot change in y?” [at the instant y x] CONTENT PROCESS ⅷ Create an algebraic model for the ladder problem Number Problem Solving ⅷ Review graphing an equation Closing the Lesson

ⅷ Algebra Reasoning Apply the Pythagorean Theorem to understand how rates of Rates of change of a function change vary will vary if the graph of the Geometry Communication function is curved. Measurement Connections Data/Probability Representation See page 775 for answer to Step 6.

EXPLORATION Ladder Climb 491 LESSON LESSON Circles and the 9.6 9.6 Pythagorean Theorem In Chapter 6, you discovered a number of properties that involved right angles in PLANNING and around circles. In this lesson you will use the conjectures you made, along with You must do things you think the Pythagorean Theorem, to solve some challenging problems. Let’s review two of the most useful conjectures. LESSON OUTLINE you cannot do. ELEANOR ROOSEVELT Tangent Conjecture: A tangent to a circle is One day: perpendicular to the radius drawn to the point 10 min Examples of tangency. 30 min Exercises Angles Inscribed in a Semicircle Conjecture: Angles inscribed in a semicircle are right angles. 5 min Closing Here are two examples that use these conjectures along MATERIALS with the Pythagorean Theorem.

ⅷ graph paper EXAMPLE A TA is tangent to circle N at A. TA 123 cm. Find the area of the ⅷ Exercise 8 (T) for One step shaded region. T A 30° TEACHING G N The Pythagorean Theorem can be powerful when combined with earlier conjectures about circles. You may either start with the ᮣ Solution TA is tangent at A, so TAN is a right angle and TAN is a 30°-60°-90° one-step investigation; or, for triangle. The longer leg is 123 cm, so the shorter leg (also the radius of the more structure, begin with the circle) is 12 cm. The area of the entire circle is 144 cm2. The area of the shaded 360 60 5 examples. The lesson consists region is 360 ,or 6, of the area of the circle. Therefore the shaded area is 5 2 primarily of exercises. 6(144 ), or 120 cm .

ᮣ EXAMPLE A EXAMPLE B AB 6 cm and BC 8 cm. Find the area of the circle. Some students might miss the C fact that the angle at vertex N has measure 60°. They may also be confused by the proportional reasoning. B A

ᮣ EXAMPLE B ᮣ Solution Inscribed angle ABC is a right angle, so ABC is a semicircle and AC is a Ask students how they know diameter. By the Pythagorean Theorem, if AB 6 cm and BC 8 cm, then that AC is a diameter. [By the AC 10 cm. Therefore the radius of the circle is 5 cm and the area of the circle Inscribed Angle Conjecture, a is 25 cm2. right angle inscribed in a circle intercepts a semicircle.] One step Pose this problem, as you display the transparency for Exercise 8: “In repairing an old from the center of the smaller wheel. What length Closing the Lesson belt should you locate?” As needed, encourage machine, you must find a belt Much of the power of geometry comes from that will make one wheel turn students to review earlier conjectures about circles and special right triangles. combining different conjectures. For example, using another. The wheels’ what we know about tangent segments, arc lengths, have lengths 36 cm and 24 cm, Assessing Progress the Pythagorean Theorem, and special right trian- and their centers are 60 cm As students discuss the examples and work through gles can help us solve a variety of problems. apart. Because the wheels turn in the exercises, you can assess their understanding of opposite directions, the belt has tangent segments, arc lengths, the Pythagorean to cross itself; marks indicate Theorem, and special right triangles. that the crossing point is 24 cm

492 CHAPTER 9 The Pythagorean Theorem EXERCISES You will need BUILDING Construction tools UNDERSTANDING ᮣ In Exercises 1–4, find the area of the shaded region in each figure. Assume for Exercise 20 lines that appear tangent are tangent at the labeled points. You might have several groups 1. OD 24 cm 2. HT 83 cm 3. HA 83 cm 4. HO 83 cm present their solutions to a few 2 2 456 cm 32 323 cm 64 163 cm2 of the exercises. Distribute H 3 transparencies and pens to T A B 105° help students prepare their presentations. 60° 120° O Y H R R ° T 120 D O H ASSIGNING HOMEWORK 64 2 Essential 1–9 64 3 3 cm 5. A 3-meter-wide circular track is shown at right. The radius of the inner circle Performance is 12 meters. What is the longest straight path that stays on the track? (In assessment 13 other words, find AB.) 18 m O A Portfolio 14 T 6. An has a 36 cm of the outer circle that is also tangent to the Group 10–17 inner concentric circle. Find the area of the annulus. 324 cm2 B Review 18–23 7. In her latest expedition, Ertha Diggs has uncovered a portion of circular, terra-cotta pipe that she | believes is part of an early water drainage system. ᮣ To find the diameter of the original pipe, she lays a Helping with the Exercises meterstick across the portion and measures the Exercise 3 If students are having length of the chord at 48 cm. The depth of the difficulty, wonder aloud whether portion from the midpoint of the chord is 6 cm. What was the pipe’s original diameter? 102 cm there’s a way to draw auxiliary lines to create right triangles, perhaps special right triangles. 3. 64 2 64 3 3 cm Exercise 4 As needed, remind students that the area of a 8. APPLICATION A machinery belt needs to be replaced. segment is the difference The belt runs around two wheels, crossing between 24 between the area of a sector them so that the larger wheel turns the smaller and the area of a triangle. wheel in the opposite direction. The diameter of the 36 cm 24 cm larger wheel is 36 cm, and the diameter of the Exercise 7 Some students may smaller is 24 cm. The distance between the centers stop after finding the radius 60 cm of the two wheels is 60 cm. The belt crosses 24 cm instead of finding the diameter. from the center of the smaller wheel. What is the length of the belt? 230 cm Exercise 8 The answer in terms of is 40 603 cm. 9. A circle of radius 6 has chord AB of length 6. If point C is selected randomly on the circle, what is the probability that ABC is obtuse? 5 6 Exercise 9 As needed, [Ask] “Does a chord that’s congruent to a radius remind you of anything?” [Mark the radius six times around a circle to form a hexagon.] “What NCTM STANDARDS LESSON OBJECTIVE happens if point C falls in the arcs intercepted by the various sides of CONTENT PROCESS ⅷ Apply the Pythagorean relationship to problems involving the inscribed regular hexagon?” circles Number Problem Solving [The arc on the hexagon side opposite side AB would be the Algebra Reasoning only one where the third vertex Geometry Communication would create an acute triangle.] Measurement Connections Data/Probability Representation

LESSON 9.6 Circles and the Pythagorean Theorem 493 Exercise 12 As in Exercise 8, this In Exercises 10 and 11, each triangle is equilateral. Find the area of the inscribed circle exercise represents a physical and the area of the circumscribed circle. How many times greater is the area of the situation, so an answer in terms circumscribed circle than the area of the inscribed circle? 6400 2 of , 3 1600 3 cm , 10. AB 6 cm 11. DE 2 3 cm may not be as appropriate as a C Inscribed circle: 3 cm2. F Inscribed circle: cm2. rounded answer. Circumscribed circle: Circumscribed circle: 12 cm2. The area of the 4 cm2.The area ofthe Exercise 13 If students are stuck, circumscribed circle is circumscribed circle is ask where the marked lengths four times as great as the four times as great as the transfer to on the sides of the A B area of the inscribed circle. D E area of the inscribed circle. rectangle and what they know about the rest of the height. 12. The Gothic arch is based on the equilateral 13. Each of three circles of radius 6 cm is tangent triangle. If the base of the arch measures to the other two, and they are inscribed in a 80 cm, what is the area of the shaded region? rectangle, as shown. What is the height of 3931 cm2 the rectangle? 12 63 cm

6

80 cm

6 80 cm

Exercise 14 You might encourage 14. Sector ARC has a radius of 9 cm and an angle 15. APPLICATION Will plans to use a circular that measures 80°. When sector ARC is cut cross section of wood to make a square table. students to make a model for this exercise. out and AR and RC are taped together, they The cross section has a of form a cone. The length of AC becomes the 336 cm. To the nearest centimeter, what is the circumference of the base of the cone. What is side length of the largest square that he can the height of the cone? 77 cm cut from it? 76 cm

C

9 cm h x A 80° R

Exercise 16 Remind students as 16. Find the coordinates of point M. 17. Find the coordinates of point K. y needed that they haven’t worked y 3 1 1 , 1 , much with obtuse angles, so they 2 2 M 2 2 might benefit from drawing an 135° ° acute triangle with point M at 210 x x one vertex. (1, 0) (1, 0) K Exercises 16, 17 [Link] This kind of reasoning is used in trigonometry. (0, –1) (0, –1)

494 CHAPTER 9 The Pythagorean Theorem Exercise 19 This exercise requires ᮣ Review that students be able to recognize the square of the binomial 9.5 18. Find the equation of a circle with center (3, 3) and radius 6. (x 3)2 (y 3)2 36 (x 1)2. 9.5 19. Find the radius and center of a circle with the equation 20. The diameter is the 2 2 x y 2x 1 100. center (1, 0), r 10 transversal, and the chords are 6.3 20. Construction Construct a circle and a chord in a circle. With compass and parallel by the Converse of the straightedge, construct a second chord parallel and congruent to the first chord. Parallel Lines Conjecture. The Explain your method. chords are congruent because they are the same distance from 2.6 21. Explain why the opposite sides of a regular hexagon are parallel. the center. Sample construction: 2.3 22. Find the rule for this number pattern: 1 3 3 4 0 2 4 3 5 1 3 5 3 6 2 4 6 3 7 3 5 7 3 8 4

… 21. Any long diagonal of a n ? ? ? ? regular hexagon divides it into n (n 2) 3 (n 3) (n 1) two congruent quadrilaterals. 6.2 23. APPLICATION Felice wants to determine the Each angle of a regular hexagon diameter of a large heating duct. She places (6 2) 180° a carpenter’s square up to the surface of the is 120°, and 6 cylinder, and the length of each tangent the diagonal divides two of the segment is 10 inches. 120° angles into 60° angles. a. What is the diameter? Explain your reasoning. Look at the diagonal as a b. Describe another way she can find the transversal. diameter of the duct. Possible answer: Measure the circumference The alternate interior angles with string and divide by . are congruent, thus the opposite sides of a regular hexagon are parallel.

IMPROVING YOUR REASONING SKILLS 120° 60° 60° Reasonable ’rithmetic I 120° ° Each letter in these problems represents a different digit. 120 1. What is the value of B? 2. What is the value of J? 60° 60° 120° 372 EF6 384 D 7 23a. Because a carpenter’s 9 B 4 DDF D square has a right angle and both radii are perpendicular to C 7 CA J ED the tangents, a square is formed. HGE D The radius is 10 in., therefore the diameter is 20 in.

10 in.

10 in. 10 EXTENSION in.

IMPROVING REASONING SKILLS Ask students to show that the shaded area in this 10 in. figure is equal to the area of the triangle. See d 20 in. If students are having difficulty, suggest that Leonardo’s Dessert—No by Herbert Wills for they make a table of the letters, listing what further study of problems like this. they can say about each letter. 1. A 0 and C 1, so B 5. 2. D 2; 7F 4 ends in F, so F is 1 or 6. Trying each possibility leads to J 6. a b

LESSON 9.6 Circles and the Pythagorean Theorem 495 CHAPTER REVIEW 9 ● CHAPTERCHAPTER 11 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPT If 50 years from now you’ve forgotten everything else you learned 9 in geometry, you’ll probably still remember the Pythagorean REVIEW Theorem. (Though let’s hope you don’t really forget everything PLANNING else!) That’s because it has practical applications in the mathematics and science that you encounter throughout your LESSON OUTLINE education. First day: It’s one thing to remember the equation a2 b2 c2.It’s another to know what it means and to be able to apply it. Review your 10 min Reviewing work from this chapter to be sure you understand how to use 35 min Exercises special triangle shortcuts and how to find the distance between two points in a coordinate plane. Second day: 30 min Exercises 15 min Student self-assessment EXERCISES You will need Construction tools ᮣ For Exercises 1–4, measurements are given in centimeters. for Exercise 30 REVIEWING 1. x ? 20 cm 2. AB ? 10 cm 3. Is ABC an acute, obtuse, Remind students of the equation of C or right triangle? obtuse 2 2 2 a circle, (x h) ( y k) r , C 240 and ask what that means. Try to 25 70 elicit the idea that it’s the set of all 15 13 AB260 points (x, y) whose coordinates 12 x satisfy the equation. Ask why those points form a circle. Bring A B out the idea of the distance 4. 5. 6. formula. Finally, ask where the The solid is a rectangular Find the coordinates of Find the coordinates of prism. AB ? 26 cm point U. 3 1 point V. 1 1 , , distance formula comes from. See y 2 2 y 2 2 that students understand it comes (0, 1) (0, 1) from the Pythagorean Theorem. B U (?, ?) Wonder aloud how the dist ance A ° ° 225 formula can be derived from 6 24 30 x x (1, 0) (1, 0) the Pythagorean Theorem given 8 that there are no a’s, b’s, or c’s V (?, ?) in the distance formula but the Pythagorean Theorem is 7. What is the area of the 8. The area of this square is 9. What is the area of 2 2 2 a b c . Elicit the idea that triangle?2003 cm2 144 cm2. Find d. d 122 cm ABCD? 246 cm2 the Pythagorean Theorem includes D C a condition: a2 b2 c2 if a 40 cm 15 cm and b are the lengths of the legs d 12 cm 20 cm of a right triangle and c is the 30° length of its hypotenuse. The A B letters a, b, and c are variables and can be replaced by any other representations of those lengths.

ASSIGNING HOMEWORK

It’s probably best to assign every student to work on every problem from Exercises 1–29. The odds 15–29 could be done in groups. Use the mixed review to prepare for a unit exam.

496 CHAPTER 9 The Pythagorean Theorem ● ● ● ● | CHAPTER 9 REVIEW CHAPTER 9 REVIEW CHAPTER 9 REVIEW CHAPTER 9 REVIEW CHAPTER 9 ᮣ Helping with the Exercises Exercise 11 If students are having 10. The arc is a semicircle. 11. Rays TA and TB are tangent to 12. The quadrilateral is a difficulty, suggest that they create What is the area of the circle O at A and B respectively, square, and QE 22 cm. a right triangle and look for shaded region? and BT 63 cm. What is the What is the area of the special triangles. 72 in.2 area of the shaded region?24 cm2 shaded region? (2 4) cm2 Exercise 14 From the way the T E R B problem is stated, students need to do only two calculations to ° find sides of two different lengths; 7 in. 120 O then they can conclude that the A 25 in. SQ triangle is an isosceles right triangle because it’s not equilat- eral. Encourage them, though, to 13. The area of circle Q is 350 cm2. Find the area of square ABCD to the BC verify the congruence and nearest 0.1 cm2. 222.8 cm2 perpendicularity of two sides. Q Slope can help with the latter. 14. Determine whether ABC with vertices A(3, 5), B(11, 3), and C(8, 8) is an equilateral, isosceles, or isosceles right triangle. AD 15. No. The closest she can isosceles right come to camp is 10 km. 15. Sagebrush Sally leaves camp on her dirt bike traveling east at 60 km/hr with a full tank of gas. After 2 hours, Exercise 17 After finding the she stops and does a little prospecting—with no luck. distance apart to be 2 km, So she heads north for 2 hours at 45 km/hr. She stops students might realize that the again, and this time hits pay dirt. Sally knows that she second question can be answered can travel at most 350 km on one tank of gas. Does easily by using closing speed: she have enough fuel to get back to camp? If not, how close can she get? Paul and Peter’s distance apart is closing at a rate of 10 miles 16. A parallelogram has sides measuring 8.5 cm and per hour. 12 cm, and a diagonal measuring 15 cm. Is the parallelogram a rectangle? If not, is the 15 cm Exercise 18 Students have worked diagonal the longer or shorter diagonal? enough with space diagonals No. 15 cm is the longer diagonal. that they may be ready to gener- 17. After an argument, Peter and Paul walk away from each other on separate paths at a right angle alize and find that the space to each other. Peter is walking 2 km/hr, and Paul is diagonal of this box has length walking 3 km/hr. After 20 min, Paul sits down to think. After 30 min, Peter stops. 122 162 142 . Both decide to apologize. How far apart are they? How long will it take them to reach each other if they both start running straight toward each other at 5 km/hr? 1 1.4 km; 82 min 18. Flora is away at camp and wants to mail her flute back home. The flute is 24 inches long. Will it fit diagonally within a box whose inside dimensions are 12 by 16 by 14 inches? yes

19. To the nearest foot, find the original height of a fallen flagpole that cracked and fell as if hinged, forming an angle of 45 degrees with the ground. The tip of the pole hit the ground 12 feet from its base. 29 ft

CHAPTER 9 REVIEW 497 EW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPT

20. You are standing 12 feet from a cylindrical corn-syrup storage tank. The distance from you to a point of tangency on the tank is 35 feet. What is the r 35 feet radius of the tank? 45 ft r

Technology 12 feet

Radio and TV stations broadcast from high towers. Their signals are picked up by radios and TVs in homes within a certain radius. Because Earth is spherical, these signals don’t get picked up beyond the point of tangency.

21. APPLICATION Read the Technology Connection above. What is the maximum broadcasting radius from a radio tower 1800 feet tall (approximately 0.34 mile)? The radius of Earth is approximately 3960 miles, and you can assume the ground around the tower is nearly . Round your answer to the nearest 10 miles. 50 mi

22. A diver hooked to a 25-meter line is searching for the remains of a Spanish galleon in the Caribbean Sea. The sea is 20 meters deep and the bottom is flat. What is the area of circular region that 2 the diver can explore? 707 m 20 m 25 m 23. What are the lengths of the two legs of a 30°-60°-90° triangle if the length of the hypotenuse is 123? 63 and 18

24. Find the side length of an equilateral triangle with an area of 363 m2. 12 m

25. Find the perimeter of an equilateral triangle with a height of 73. 42 26. No. If you reflect one of the 26. Al baked brownies for himself and his right triangles into the center two sisters. He divided the square pan of piece, you’ll see that the area of brownies into three parts. He measured the is almost half again as three 30° angles at one of the corners so that two pieces formed right triangles large as the area of each of the and the middle piece formed a kite. Did other triangles. he divide the pan of brownies equally? Draw a sketch and explain your Extra reasoning. 27. A circle has a central angle AOB that measures 80°.Ifpoint C is selected randomly on the circle, what is the probability that ABC is obtuse? 7 30° 9 30°

30°

Or students might compare areas by assuming the short leg of the 30°-60°-90° triangle is 1. The area of each triangle is then 3 2 and the area of the kite is 3 3.

498 CHAPTER 9 The Pythagorean Theorem CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 28. The quarter-circle gives the maximum area. Triangle: 28. One of the sketches below shows the greatest area that you can enclose in a right- angled corner with a rope of length s. Which one? Explain your reasoning.

45° _1s ___s _ s 2 s 2 s _1s 45° 2 ___s _ 2 A triangle A square A quarter-circle 1 s s s2 A 29. A wire is attached to a block of wood at point A. 1.4 m 2 2 2 4 The wire is pulled over a pulley as shown. How far 3.9 m Square: will the block move if the wire is pulled 1.4 meters 1.5 m A in the direction of the arrow? 1.6 m _1s ? 2

_1s MIXED REVIEW 2

ᮣ 30. Construction Construct an isosceles triangle that has a base length equal to half the 3.1 length of one leg. 1 1 s2 A s s 2.3 31. In a regular octagon inscribed in a circle, how many diagonals pass through the 2 2 4 center of the circle? In a regular nonagon? a regular 20-gon? What is the general rule? Quarter-circle: 6.7 32. A bug clings to a point two inches from the center of a spinning fan blade. The blade spins around once per second. How fast does the bug travel in inches per second? 4, or approximately 12.6 in./sec s In Exercises 33–40, identify the statement as true or false. For each false statement, explain why it is false or sketch a counterexample. 8.1 33. The area of a rectangle and the area of a parallelogram are both given by the __2s_ formula A bh, where A is the area, b is the length of the base, and h is the height. true 7.1 34. When a figure is reflected over a line, the line of reflection is perpendicular to every 1 segment joining a point on the original figure with its image. true s 4 2 r 9.3 35. In an isosceles right triangle, if the legs have length x, then the hypotenuse has 2s r length x3. False. The hypotenuse is of length x2. 2 2 8.2 36. The area of a kite or a can be found by using the formula A (0.5)d d , 1 2s s 1 2 A 4 where A is the area and d1 and d2 are the lengths of the diagonals. true 37. s2 s2 9.5 If the coordinates of points A and B are x1, y1 and x2, y2 , respectively, then 2 2 4 AB x y x y . 2 2 1 1 2 2 false; AB x2 x1 y2 y1 7.3 38. A glide reflection is a combination of a translation and a rotation. 30. False.A glide reflection is a combination of a translation and a reflection.

Exercise 31 Encourage students to make a table and look for a pattern. 31. n 4; 0; 10. The rule is 2 if n is even, but 0 if n is odd.

CHAPTER 9 REVIEW 499 EW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPT

7.4 39. Equilateral triangles, squares, and regular octagons can be used to create monohedral tessellations. False. Equilateral triangles, squares, and regular hexagons can be used to create monohedral tessellations. 9.3 40. In a 30°-60°-90° triangle, if the shorter leg has length x, then the longer leg has length x3 and the hypotenuse has length 2x. true

In Exercises 41–46, select the correct answer.

9.1 41. The hypotenuse of a right triangle is always ? . D A. opposite the smallest angle and is the shortest side. B. opposite the largest angle and is the shortest side. C. opposite the smallest angle and is the longest side. D. opposite the largest angle and is the longest side.

8.2 42. The area of a triangle is given by the formula ? ,where A is the area, b is the length of the base, and h is the height. B 1 A. A bh B. A 2bh C. A 2bh D. A b2h 9.2 43. If the lengths of the three sides of a triangle satisfy the Pythagorean equation, then the triangle must be a(n) ? triangle. A A. right B. acute C. obtuse D. scalene

7.2 44. The ordered pair rule (x, y) → (y, x) is a ? . C A. reflection over the x-axis B. reflection over the y-axis C. reflection over the line y x D. rotation 90° about the origin

7.3 45. The composition of two reflections over two intersecting lines is equivalent to ? . C A. a single reflection B. a translation C. a rotation D. no transformation 8.7 46. The total surface area of a cone is equal to ? ,where r is the radius of the circular base and l is the slant height. D A. r2 2r B. rl C. rl 2r D. rl r2 5.7 47. Create a flowchart proof to show that the diagonal of a rectangle divides the rectangle into two congruent triangles. AB

DC

47. 1 ABCD is 4 D B a rectangle Definition of Given rectangle

2 3 5 7 ABCD is DA CB DAC BCA ABC CDA a parallelogram Definition of AIA Conjecture SAA Congruence Definition of parallelogram Conjecture rectangle 6 AC AC Same segment

500 CHAPTER 9 The Pythagorean Theorem CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9

1.2 48. Copy the ball positions onto patty paper. N a. At what point on the S cushion should a player aim so that the cue ball bounces off and strikes the 8-ball? Mark the point with the letter A. WE b. At what point on the W cushion should a player aim so that the cue ball bounces off and strikes the 8-ball? Mark the point with the letter B. S

8.2 49. Find the area and the perimeter of 8.6 50. Find the area of the shaded region. the trapezoid. 34 cm2; 22 42 27.7 cm 40 2 3 cm 12 cm 3 cm

4 cm 5 cm 120° 5 cm 4 cm

9.1 51. An Olympic swimming pool has length 8.4 52. The side length of a regular is 6 cm, 51. about 55.9 m2 50 meters and width 25 meters. What is and the apothem measures about 4.1 cm. 2 the diagonal distance across the pool? What is the area of the pentagon? 52. about 61.5 cm

9.1 53. The box below has dimensions 25 cm, 8.7 54. The cylindrical container below has an open 36 cm, and x cm. The diagonal shown top. Find the surface area of the container has length 65 cm. Find the value of x. 48 cm (inside and out) to the nearest square foot. 322 ft2 65 cm

25 cm 9 ft x cm 36 cm 5 ft

TAKE ANOTHER LOOK | A ᮣ ᮣ 1. Use geometry software to demonstrate the Take Another Look Pythagorean Theorem. Does your demonstration still 1. Demonstrations should include work if you use a shape other than a square—for shapes other than a square. (Any example, an equilateral triangle or a semicircle? regular ploygon can be used. In 2. Find Elisha Scott Loomis’s Pythagorean Proposition C B fact, any three similar figures will and demonstrate one of the proofs of the Pythagorean work. Students will study similar Theorem from the book. figures in Chapter 11.) 2. Demonstrations will vary.

48. N

8-ball

B W E Cue ball

A

S

CHAPTER 9 REVIEW 501 3. The small square in the center EW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPTER 9 REVIEW ● CHAPT has sides b a, the slanted square has area c2,and the ab triangles each have area 2 .The 3. The Zhoubi Suanjing, one of the oldest sources of Chinese 2 2 ab equation c (b a) 4 2 mathematics and astronomy, contains the diagram at right simplifies to c2 a2 b2. demonstrating the Pythagorean Theorem (called gou gu in China). Find out how the Chinese used and proved the b gou gu, and present your findings.

4. Use the SSS Congruence Conjecture to verify the converse of a the 30°-60°-90° Triangle Conjecture. That is, show that if a c triangle has sides with lengths x, x3,and 2x, then it is a 30°- 60°-90° triangle.

b a

5. Starting with an isosceles right triangle, use geometry software or b a a compass and straightedge to start a right triangle like the one 4 shown. Continue constructing right triangles on the hypotenuse 1 4. One possible proof: Given of the previous triangle at least five more times. Calculate the ABC with BC x, AC length of each hypotenuse and leave them in radical form. 3 x3,and AB 2x, construct 1 2 30°-60°-90° right triangle DEF 1 with right angle F, 30° angle D, 1 and EF x. DF x3 and DE 2x, by the 30°-60°-90° Triangle Conjecture. Assessing What You’ve Learned ABC DEF by SSS. C F and is a right angle, UPDATE YOUR PORTFOLIO Choose a challenging project, Take Another Look A D and is a 30° angle, activity, or exercise you did in this chapter and add it to your portfolio. Explain the strategies you used. and B E and is a 60° angle.

5. 1 1 ORGANIZE YOUR NOTEBOOK Review your notebook and your conjecture list to be 1 sure they are complete. Write a one-page chapter summary. 1 8 7 WRITE IN YOUR JOURNAL Why do you think the Pythagorean Theorem is 9 6 1 considered one of the most important in mathematics? 5 WRITE TEST ITEMS Work with group members to write test items for this chapter. 4 1 Try to demonstrate more than one way to solve each problem.

3 1 GIVE A PRESENTATION Create a visual aid and give a presentation about the 2 1 Pythagorean Theorem. 1

ASSESSING

As part of a final grade, you might ask students to present FACILITATING SELF-ASSESSMENT visual demonstrations or anima- To help students complete the portfolio described tions of various proofs of the in Assessing What You’ve Learned, suggest that Pythagorean Theorem. (See Take they consider for evaluation their work on Another Look activities 1–3.) If Lesson 9.1, Exercises 17, 18; Lesson 9.2, this presentation replaces the Exercise 16; Lesson 9.3, Exercise 17; Lesson 9.4, chapter test, you can give a unit Exercise 9; and Lesson 9.5, Exercise 6. exam after working through the mixed review.

502 CHAPTER 9 The Pythagorean Theorem