Prove that the line segment joining the midpoints of two with with

D A B C Figure 5.20 D A C Figure 5.21 B A C GIVEN: with right ; See Figure 5.20. PROVE: PLAN: Show that in Figure 5.21. Then use CSSTP. NOTE: Although and are both segments of the hypotenuse, is the segment adjacent to Lemma 5.4.3 opens the doors to a proof of the famous Pythagorean Theorem, one of the most frequently applied relationships in geometry. Although the theorem’s title gives credit to the Greek geometer Pythagoras, many other proofs are known, and the ancient Chinese were aware of the relationship before the time of Pythagoras. AC . AD DB AD 䉭ADC 䉭ACB AB AC = AC AD CD ⬜ AB ∠ACB 䉭ABC c C a b a A B Figure 5.22 D A B C b b a y x c The length of each leg of a right triangle is the geometric mean of the length of the hy- potenuse and the length of the segment of the hypotenuse adjacent to that leg. LEMMA 5.4.3 Exs 1, 2 The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the legs. THEOREM 5.4.4 왘 Pythagorean Theorem Thus, where c is the length of the hypotenuse and a and b are the lengths of the legs, . GIVEN: In Figure 5.22a, with right PROVE: PROOF: Draw , as shown in Figure 5.22b. Denote and . By Lemma 5.4.3, Therefore, Using the Addition Property of Equality, we have But . Thus, a 2 + b 2 = cc = c 2 y + x = x + y = AD + DB = AB = c a 2 + b 2 = cy + cx = cy + x b 2 = cx and a 2 = cy c b = b x and c a = a y DB = y AD = x CD ⬜ AB c 2 = a 2 + b 2 ∠C 䉭ABC c 2 = a 2 + b 2 Discover A video entitled “The Rule of Pythagoras” is available through Project Mathematics at Cal Tech University in Pasadena, CA. It is well worth watching EXAMPLE 1 Given with right in Figure 5.23, find: a RT if and b RT if and c RS if and d ST if and Solution With right , the hypotenuse is . Then , , and . ST = b RS = a RT = c RT ∠S RT = 9 RS = 6 ST = 12 RT = 13 ST = 6 RS = 4 ST = 4 RS = 3 ∠S 䉭RST c S T R b a Figure 5.23 GIVEN: [Figure 5.24a] with sides a, b, and c so that PROVE: is a right triangle. PROOF: We are given for which . Construct the right , which has legs of lengths a and b and a hypotenuse of length x. [See Figure 5.24b.] By the Pythagorean Theorem, . By substitution, and . Thus, by SSS. Then opposite the side of length c must be to , the right angle of . Then is a right angle, and is a right triangle. 䉭RST ∠S 䉭ABC ∠C ⬵ ∠S 䉭RTS ⬵ 䉭ABC x = c x 2 = c 2 x 2 = a 2 + b 2 䉭ABC c 2 = a 2 + b 2 䉭RST 䉭RST c 2 = a 2 + b 2 䉭RST c S T R a b a ? Figure 5.24 x C B A a b b a b c d ST = 3 15 L 6.71 b = 145 = 19 5 = 19 15 = 315 b 2 = 45 6 2 + b 2 = 9 2 : 36 + b 2 = 81 a = 5; RS = 5 a 2 = 25 a 2 + 12 2 = 13 2 : a 2 + 144 = 169 RT = 2 213 L 7.21 c = 152 = 14 13 = 14 113 = 2113 c 2 = 52 4 2 + 6 2 = c 2 : 16 + 36 = c 2 c = 5; RT = 5 c 2 = 25 3 2 + 4 2 = c 2 : 9 + 16 = c 2 쮿 Exs. 3, 4 The converse of the Pythagorean Theorem is also true. If a, b, and c are the lengths of the three sides of a triangle, with c the length of the longest side, and if , then the triangle is a right triangle with the right angle opposite the side of length c. c 2 = a 2 + b 2 THEOREM 5.4.5 왘 Converse of Pythagorean Theorem EXAMPLE 2 Do the following represent the lengths of the sides of a right triangle? a , , b , , c , , d , , Solution a Yes. Because that is, , this triangle is a right triangle. b Yes. Because that is, , this triangle is a right triangle. c No. , which is not that is, 100, so this triangle is not a right triangle. d Yes. Because leads to , this triangle is a right triangle. 2 + 3 = 5 12 2 + 13 2 = 15 2 10 2 7 2 + 9 2 = 49 + 81 = 130 225 + 64 = 289 15 2 + 8 2 = 17 2 25 + 144 = 169 5 2 + 12 2 = 13 2 c = 1 5 b = 1 3 a = 1 2 c = 10 b = 9 a = 7 c = 17 b = 8 a = 15 c = 13 b = 12 a = 5 쮿 Exs. 5, 6 b b 10 cm 10 cm 5 cm 5 cm Figure 5.26 Example 5 also uses the Pythagorean Theorem, but it is considerably more compli- cated than Example 4. Indeed, it is one of those situations that may require some insight to solve. Note that the triangle described in Example 5 is not a right triangle because . 4 2 + 5 2 Z 6 2 Discover Construct a triangle with sides of lengths 3 in., 4 in., and 5 in. Measure the angles of the triangle. Is there a right angle? ANSWER Yes, opposite the 5-in. side. EXAMPLE 3 A ladder 12 ft long is leaning against a wall so that its base is 4 ft from the wall at ground level see Figure 5.25. How far up the wall does the ladder reach? 12 4 h Figure 5.25 Solution The desired height is represented by h, so we have The height is exactly , which is approximately 11.31 ft. 쮿 h = 8 12 h = 1128 = 164 2 = 164 12 = 812 h 2 = 128 16 + h 2 = 144 4 2 + h 2 = 12 2 EXAMPLE 4 One diagonal of a rhombus has the same length, 10 cm, as each side see Figure 5.26. How long is the other diagonal? Solution Because the diagonals are perpendicular bisectors of each other, four right 䉭s are formed. For each right 䉭, a side of the rhombus is the hypotenuse. Half of the length of each diagonal is the length of a leg of each right triangle. Therefore, Thus, the length of the whole diagonal is 쮿 10 13 cm L 17.32 cm. b = 175 = 125 3 = 125 13 = 513 b 2 = 75 25 + b 2 = 100 5 2 + b 2 = 10 2 Reminder The diagonals of a rhombus are perpendicular bisectors of each other. EXAMPLE 5 A triangle has sides of lengths 4, 5, and 6, as shown in Figure 5.27. Find the length of the altitude to the side of length 6. 6 x 6 – x 4 5 h Figure 5.27 It is now possible to prove the HL method for proving the congruence of triangles, a method that was introduced in Section 3.2. Solution The altitude to the side of length 6 separates that side into two parts whose lengths are given by x and 6 x. Using the two right triangles formed, we apply the Pythagorean Theorem twice. Subtracting the first equation from the second, we can calculate x. subtraction Now we use to find h. h = 1175 4 = 125 7 4 = 125 17 4 = 5 17 4 L 3.31 h 2 = 175 16 81 16 + h 2 = 256 16 81 16 + h 2 = 16 a 9 4 b 2 + h 2 = 4 2 x 2 + h 2 = 4 2 x = 9 4 x = 27 12 = 9 4 - 12x = - 27 36 - 12x = 9 x 2 + h 2 = 16 36 - 12x + x 2 + h 2 = 25 x 2 + h 2 = 4 2 and 6 - x 2 + h 2 = 5 2 쮿 If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the triangles are congruent HL. THEOREM 5.4.6 GIVEN: Right with right and right with right see Figure 5.28; and PROVE: PROOF: With right , the hypotenuse of is ; similarly, is the hypotenuse of right . Because , we denote the common length by c; that is, . Because , we also have . Then Then so that . Hence, by SSS. 䉭ABC ⬵ 䉭EDF BC ⬵ DF BC = DF BC = 2c 2 - a 2 and DF = 2c 2 - a 2 a 2 + BC 2 = c 2 and a 2 + DF 2 = c 2 which leads to AC = EF = a AC ⬵ EF AB = DE = c AB ⬵ DE 䉭EDF DE AB 䉭ABC ∠C 䉭ABC ⬵ 䉭EDF AC ⬵ EF AB ⬵ DE ∠F 䉭DEF ∠C 䉭ABC Exs. 7, 8 A C B Figure 5.28 D F E 쮿 Our work with the Pythagorean Theorem would be incomplete if we did not address two issues. The first, Pythagorean triples, involves natural or counting numbers as pos- sible choices of a, b, and c. The second leads to the classification of triangles according to the lengths of their sides as found in Theorem 5.4.7 on page 250. PYTHAGOREAN TRIPLES A Pythagorean triple is a set of three natural numbers a, b, c for which . a 2 + b 2 = c 2 DEFINITION Three sets of Pythagorean triples encountered in this section are 3, 4, 5, 5, 12, 13, and 8, 15, 17. These numbers will always fit the sides of a right triangle. Natural-number multiples of any of these triples will also constitute Pythagorean triples. For example, doubling 3, 4, 5 yields 6, 8, 10, which is also a Pythagorean triple. In Figure 5.29, the triangles are similar by SSS . The Pythagorean triple 3, 4, 5 also leads to 9, 12, 15, 12, 16, 20, and 15, 20, 25. The Pythagorean triple 5, 12, 13 leads to triples such as 10, 24, 26 and 15, 36, 39. Basic Pythagorean triples that are used less frequently include 7, 24, 25, 9, 40, 41, and 20, 21, 29. Pythagorean triples can be generated by using select formulas. Where p and q are natural numbers and , one formula uses 2pq for the length of one leg, for the length of other leg, and for the length of the hypotenuse See Figure 5.30.. Table 5.1 lists some Pythagorean triples corresponding to choices for p and q. The triples printed in boldface type are basic triples, also known as primitive triples. In ap- plication, knowledge of the primitive triples and their multiples will save you consid- erable time and effort. In the final column, the resulting triple is provided in the order from a small to c large. p 2 + q 2 p 2 - q 2 p 7 q 5 3 4 Figure 5.29 6 8 10 p 2 + q 2 2 pq p 2 – q 2 Figure 5.30 TABLE 5.1 Pythagorean Triples a or b b or a c p q 2 pq a, b, c 2 1

3 4

5 3, 4, 5

3 1

8 6 10 6, 8, 10 3 2 5 12 13 5, 12, 13

4 1

15 8

17 8, 15, 17 4 3 7 24 25 7, 24, 25

5 1

24 10 26 10, 24, 26 5 2 21 20 29

20, 21, 29

5 3

16 30

34 16, 30, 34

5 4

9 40 41

9, 40, 41

p 2 + q 2 p 2 - q 2 THE CONVERSE OF THE PYTHAGOREAN THEOREM The Converse of the Pythagorean Theorem allows us to recognize a right triangle by knowing the lengths of its sides. A variation on the converse allows us to determine whether a triangle is acute or obtuse. This theorem is stated without proof. Exs. 9–11 Let a, b, and c represent the lengths of the three sides of a triangle, with c the length of the longest side.

1. If , then the triangle is obtuse and the obtuse angle lies

opposite the side of length c.

2. If , then the triangle is acute.

c 2 6 a 2 + b 2 c 2

7 a

2 + b 2 THEOREM 5.4.7 EXAMPLE 6 Determine the type of triangle represented if the lengths of its sides are as follows: a 4, 5, 7 b 6, 7, 8 c 9, 12, 15 d 3, 4, 9 Solution a Choosing , we have , or 25; the triangle is obtuse. b Choosing , we have , or ; the triangle is acute. c Choosing , we have , or ; the triangle is a right triangle. d Because , no triangle is possible. Remember that the sum of the lengths of two sides of a triangle must be greater than the length of the third side. 9 7 3 + 4 225 = 81 + 144 15 2 = 9 2 + 12 2 c = 15 64 6 36 + 49 8 2 6 6 2 + 7 2 c = 8 49 7 16 + 7 2 7 4 2 + 5 2 c = 7 쮿 Exs. 12, 13 Exercises 5.4

1. By naming the vertices in order, state three different

triangles that are similar to each other.

2. Use Theorem 5.4.2 to form a proportion

in which SV is a geometric mean. HINT:

3. Use Lemma 5.4.3 to form a proportion in

which RS is a geometric mean. HINT: 4. Use Lemma 5.4.3 to form a proportion in which TS is a geometric mean. HINT: 5. Use Theorem 5.4.2 to find RV if and . 6. Use Lemma 5.4.3 to find RT if and .

7. Find the length of if:

a and b and

8. Find the length of if:

a and b and EF = 6 13 DF = 12 EF = 5 DF = 13 DE EF = 3 DE = 5 EF = 6 DE = 8 DF VR = 4 RS = 6 VT = 8 SV = 6 䉭 TVS 䉭TSR 䉭 RVS 䉭RST 䉭 SVT 䉭RVS

9. Find EF if:

a and b and

10. Find DF if:

a and b and

11. Determine whether each triple a, b, c is a Pythagorean

triple. a 3, 4, 5 c 5, 12, 13 b 4, 5, 6 d 6, 13, 15

12. Determine whether each triple a, b, c is a Pythagorean

triple. a 8, 15, 17 c 6, 8, 10 b 10, 13, 19 d 11, 17, 20

13. Determine the type of triangle represented if the lengths

of its sides are: a , , and b , , and c , , and d , , and c = 15 b = 8 a = 3 c = 1 7 b = 1 3 a = 2 c = 6 b = 5 a = 4 c = 5 b = 3 a = 4 EF = 6 DE = 12 EF = 5 DE = 12 DE = 8 12 DF = 12 DE = 15 DF = 17 V T S R Exercises 1–6 E D F Exercises 7–10

35. If ,

, and , show that .

36. Given that the line segment shown has

length 1, construct a line segment whose length is .

37. Using the line segment from Exercise 36, construct a line

segment of length 2 and then a second line segment of length . 15 12 c 2 = a 2 + b 2 c = p 2 + q 2 b = 2pq a = p 2 - q 2

14. Determine the type of triangle represented if the lengths

of its sides are: a , , and b , , and c , , and d , , and

15. A guy wire 25 ft long supports an

antenna at a point that is 20 ft above the base of the antenna. How far from the base of the antenna is the guy wire secured?

16. A strong wind holds a kite

30 ft above the earth in a position 40 ft across the ground. How much string does the girl have out to the kite?

17. A boat is 6 m below the

level of a pier and 12 m from the pier as measured across the water. How much rope is needed to reach the boat?

18. A hot-air balloon is held in place by the ground crew at a

point that is 21 ft from a point directly beneath the basket of the balloon. If the rope is of length 29 ft, how far above ground level is the basket? c = 9 b = 7 a = 5 c = 16 b = 12 a = 10 c = 29 b = 21 a = 20 c = 2.5 b = 2 a = 1.5 20 ft 40 ft 12 m 12 m 12 m 8 104

19. A drawbridge that is 104 ft in length is raised at its

midpoint so that the uppermost points are 8 ft apart. How far has each of the midsections been raised?

22. A right triangle has legs of lengths x and and a

hypotenuse of length . What are the lengths of its sides?

23. A rectangle has base length , altitude length

, and diagonals of length 2x each. What are the lengths of its base, altitude, and diagonals?

24. The diagonals of a rhombus measure 6 m and 8 m. How

long are each of the congruent sides?

25. Each side of a rhombus measures 12 in. If one diagonal is

18 in. long, how long is the other diagonal?

26. An isosceles right triangle has a hypotenuse of length

10 cm. How long is each leg?

27. Each leg of an isosceles right triangle has a length of

. What is the length of the hypotenuse?

28. In right with right

, and . Find the length of if M is the midpoint of .

29. In right with right

, and . Find the length of if M and N are the midpoints of and , respectively.

30. Find the length of the altitude to the 10-in. side of a

triangle whose sides are 6, 8, and 10 inches in length.

31. Find the length of the altitude to the 26-in. side of a

triangle whose sides are 10, 24, and 26 inches in length.

32. In quadrilateral ABCD, and

. If , , and , determine DA. DC = 12 BC = 3 AB = 4 DC ⬜ diagonal AC BC ⬜ AB BC AB MN BC = 15 AB = 17 ∠C 䉭ABC AC MB BC = 8 AB = 10 ∠C 䉭ABC 6 12 in x + 1 x + 3 2x + 3 2x + 2

20. A drawbridge that is 136 ft in length is raised at its

midpoint so that the uppermost points are 16 ft apart. How far has each of the midsections been raised? HINT: Consider the drawing for Exercise 19.

21. A rectangle has a width of 16 cm and a diagonal of length

20 cm. How long is the rectangle? D B A C U R S T A C B 1 Exercises 36, 37

33. In quadrilateral RSTU, and .

If , , and , find UT . RU = 15 ST = 8 RS = 6 UT ⬜ diagonal RT RS ⬜ ST

34. Given: is not a right

Prove: a 2 + b 2 Z c 2 䉭 䉭ABC [ NOTE: and ] CB = a . AC = b , AB = c , © Sonya EtchisonShutterstock