In the right triangle, find the length of the altitude drawn

A C D B Figure 8.17 12 ft ? ? 12 ft 20 ft 18 ft Figure 8.18 EXAMPLE 1 Find the perimeter of shown in Figure 8.17 if: a AB ⫽ 5 in., AC ⫽ 6 in., and BC ⫽ 7 in. b Altitude AD ⫽ 8 cm, BC ⫽ 6 cm, and Solution a = 18 in. = 5 + 6 + 7 P ABC = AB + AC + BC AB ⬵ AC 䉭ABC 쮿 Exs. 1–4 HERON’S FORMULA If the lengths of the sides of a triangle are known, the formula generally used to calcu- late the area is Heron’s Formula named in honor of Heron of Alexandria, circa A . D . 75. One of the numbers found in this formula is the semiperimeter of a triangle, which is defined as one-half the perimeter. For the triangle that has sides of lengths a, b, and c, the semiperimeter is . We apply Heron’s Formula in Example 3. The proof of Heron’s Formula can be found at our website. s = 1 2 a + b + c We apply the perimeter concept in a more general manner in Example 2. b With , is isosceles. Then is the bisector of . If BC 6, it follows that DC 3. Using the Pythagorean Theorem, we have Now . NOTE: Because x ⫹ x ⫽ 2x, we have . 173 + 173 = 2173 P ABC = 6 + 173 + 173 = 6 + 2173 L 23.09 cm AC = 173 64 + 9 = AC 2 8 2 + 3 2 = AC 2 AD 2 + DC 2 = AC 2 = = BC ⬜ AD 䉭ABC AB ⬵ AC EXAMPLE 2 While remodeling, the Gibsons have decided to replace the old woodwork with Colonial-style oak woodwork. a Using the floor plan provided in Figure 8.18, find the amount of baseboard in lin- ear feet needed for the room. Do not make any allowances for doors b Find the cost of the baseboard if the price is 1.32 per linear foot. Solution a Dimensions not shown measure 20 12 or 8 ft and 18 12 or 6 ft. The perimeter, or “distance around,” the room is b The cost is . 76 1.32 = 100.32 12 + 6 + 8 + 12 + 20 + 18 = 76 linear ft 쮿 When the lengths of the sides of a quadrilateral are known, we can apply Heron’s Formula to find the area if the length of a diagonal is also known. In quadrilateral ABCD in Figure 8.20, Heron’s Formula can be used to show that the area of is 60 and the area of is 84. Thus, the area of quadrilateral ABCD is 144 units 2 . The following theorem is named in honor of Brahmagupta, a Hindu mathematician born in A . D . 598. We include the theorem without its rather lengthy proof. As it happens, Heron’s Formula for the area of any triangle is actually a special case of Brahmagupta’s Formula, which is used to determine the area of a cyclic quadrilateral. In Brahmagupta’s Formula, as in Heron’s Formula, the letter s represents the numerical value of the semi- perimeter. The formula is applied in essentially the same manner as Heron’s Formula. See Exercises 11, 12, 41, and 42 of this section. 䉭BCD 䉭ABD If the three sides of a triangle have lengths a, b, and c, then the area A of the triangle is given by , where the semiperimeter of the triangle is s = 1 2 a + b + c A = 1s s - as - bs - c THEOREM 8.2.1 왘 Heron’s Formula EXAMPLE 3 Find the area of a triangle which has sides of lengths 4, 13, and 15. See Figure 8.19. Solution If we designate the sides as a ⫽ 4, b ⫽ 13, and c ⫽ 15, the semiperimeter of the triangle is given by . Therefore, = 1161231 = 1576 = 24 units 2 = 11616 - 416 - 1316 - 15 A = 1ss - as - bs - c s = 1 2 4 + 13 + 15 = 1 2 32 = 16 쮿 4 13 15 Figure 8.19 A D C B 15 13 14 17 8 Figure 8.20 For a cyclic quadrilateral with sides of lengths a, b, c, and d, the area is given by where s = 1 2 a + b + c + d A = 1 s - as - bs - cs - d, THEOREM 8.2.2 왘 Brahmagupta’s Formula a b d c Brahmagupta’s Formula becomes Heron’s Formula when the length d of the fourth side shrinks the length d approaches 0 so that the quadrilateral becomes a triangle with sides of lengths a, b, and c. The remaining theorems of this section contain numerical subscripts. In practice, subscripts enable us to distinguish quantities. For instance, the lengths of the two unequal bases of a trapezoid are written b 1 read “b sub 1” and b 2 . In particular, b 1 represents the numerical length of the first base, and b 2 represents the length of the second base. The following chart illustrates the use of numerical subscripts. Exs. 5–8 GIVEN: Trapezoid ABCD with ; AB ⫽ b 1 and DC ⫽ b 2 . PROVE: PROOF: Draw as shown in Figure 8.22a. Now has an altitude of length h and a base of length b 2 . As shown in Figure 8.22b, Also, has an altitude of length h and a base of length b 1 . [See Figure 8.22c.] Then Thus, = 1 2 h b 1 + b 2 = 1 2 hb 1 + 1 2 hb 2 A ABCD = A ABC + A ADC A ABC = 1 2 hb 1 䉭ABC A ADC = 1 2 hb 2 䉭ADC AC A ABCD = 1 2 h b 1 + b 2 AB 7 DC AREA OF A TRAPEZOID Recall that the two parallel sides of a trapezoid are its bases. The altitude is any line segment that is drawn perpendicular from one base to the other. In Figure 8.21, and are bases and is an altitude for the trapezoid. We use the more common formula for the area of a triangle namely, to develop our remaining theorems. In Theorem 8.2.3, b 1 and b 2 represent the lengths of the bases of the trapezoid. In some textbooks, b represents the length of the shorter base and B represents the length of the longer base. A = 1 2 bh AE DC AB Theorem Subscripted Symbol Meaning Theorem 8.2.3 b 1 Length of first base of trapezoid Corollary 8.2.5 d 2 Length of second diagonal of rhombus Theorem 8.2.7 A 1 Area of first triangle The area A of a trapezoid whose bases have lengths b 1 and b 2 and whose altitude has length h is given by A = 1 2 h b 1 + b 2 THEOREM 8.2.3 General Rule: Many area relationships depend upon the use of the Area-Addition Postulate. Illustration: In the proof of Theorem 8.2.3, the area of the trapezoid is developed as the sum of areas of two triangles. STRATEGY FOR PROOF 왘 Proving Area Relationships A B C E D Figure 8.21 D A 1 b h C B 2 b a Figure 8.22 D A 1 b h C B 2 b b D A 1 b h C B 2 b h c 쮿 The following activity reinforces the formula for the area of a trapezoid. EXAMPLE 4 Given that , find the area of the trapezoid in Figure 8.23. Note that RS ⫽ 5, Solution Let RS ⫽ 5 ⫽ b 1 and TV ⫽ 13 ⫽ b 2 . Also, RW ⫽ h ⫽ 6. Now, becomes = 3 18 = 54 units 2 = 1 2 6 18 A = 1 2 65 + 13 A = 1 2 h b 1 + b 2 RS 7 VT 쮿 Discover Cut out two trapezoids that are copies of each other and place one next to the other to form a parallelogram. a How long is the base of the parallelogram? b What is the area of the parallelogram? c What is the area of the trapezoid? a b c 1 2 hb 1 +b 2 hb 1 +b 2 b 1 +b 2 R S T W V RS || VT — — — — Figure 8.23 TV ⫽ 13, and RW ⫽ 6. h b 1 b 2 h b 1 b 2 b 2 b 1 ANSWERS QUADRILATERALS WITH PERPENDICULAR DIAGONALS The following theorem leads to Corollaries 8.2.5 and 8.2.6, where the formula found in Theorem 8.2.4 is also used to find the area of a rhombus and kite. Exs. 9–12 The area of any quadrilateral with perpendicular diagonals of lengths d 1 and d 2 is given by A = 1 2 d 1 d 2 THEOREM 8.2.4 GIVEN: Quadrilateral ABCD with [see Figure 8.24a on page 368.] PROVE: A ABCD = 1 2 d 1 d 2 AC ⬜ BD PROOF: Through points A and C, draw lines parallel to . Likewise, draw lines parallel to through points B and D. Let the points of intersection of these lines be R, S, T, and V, as shown in Figure 8.24b. Because each of the quadrilaterals ARDE, ASBE, BECT, and CEDV is a parallelogram containing a right angle, each is a rectangle. Furthermore, , A 䉭ADE = 1 2 A ARDE AC DB In problems involving the rhombus, we often utilize the fact that its diagonals are perpendicular. If the length of a side and the length of either diagonal are known, the length of the other diagonal can be found by applying the Pythagorean Theorem. AREA OF A KITE For a kite, we proved in Exercise 27 of Section 4.2 that one diagonal is the perpendicu- lar bisector of the other. See Figure 8.27. Corollary 8.2.5 and Corollary 8.2.6 are immediate consequences of Theorem 8.2.4. Example 5 illustrates Corollary 8.2.5. C B D d 2 d 1 E a A Figure 8.24 C A B T V S R D d 2 d 1 E b 1 d 2 d Figure 8.25 P N M Q R Figure 8.26 The area A of a rhombus whose diagonals have lengths d 1 and d 2 is given by A = 1 2 d 1 d 2 COROLLARY 8.2.5 The area A of a kite whose diagonals have lengths d 1 and d 2 is given by A = 1 2 d 1 d 2 COROLLARY 8.2.6 EXAMPLE 5 Find the area of the rhombus MNPQ in Figure 8.26 if MP ⫽ 12 and NQ ⫽ 16. Solution By Corollary 8.2.5, A MNPQ = 1 2 d 1 d 2 = 1 2 12 16 = 96 units 2 쮿 V S R T Figure 8.27 , , and . Then . But RSTV is a rectangle, because it is a parallelogram containing a right angle. Because RSTV has dimensions d 1 and d 2 [see Figure 8.24b], its area is . By substitution, AREA OF A RHOMBUS Recall that a rhombus is a parallelogram with two congruent adjacent sides; in turn, we proved that all four sides were congruent. Because the diagonals of a rhombus are per- pendicular, we have the following corollary of Theorem 8.2.4. See Figure 8.25. A ABCD = 1 2 d 1 d 2 . d 1 d 2 A ABCD = 1 2 A RSTV A 䉭DEC = 1 2 A CEDV A 䉭BEC = 1 2 A BECT A 䉭ABE = 1 2 A ASBE 쮿 We apply Corollary 8.2.6 in Example 6. AREAS OF SIMILAR POLYGONS The following theorem compares the areas of similar triangles. In Figure 8.29, we refer to the areas of the similar triangles as A 1 and A 2 . The triangle with area A 1 has sides of lengths a 1 , b 1 , and c 1 , and the triangle with area A 2 has sides of lengths a 2 , b 2 , and c 2 . Where a 1 corresponds to a 2 , b 1 to b 2 , and c 1 to c 2 , Theorem 8.2.7 implies that We prove only the first relationship; the other proofs are analogous. A 1 A 2 = a a 1 a 2 b 2 or A 1 A 2 = a b 1 b 2 b 2 or A 1 A 2 = a c 1 c 2 b 2 Exs. 13–17 The ratio of the areas of two similar triangles equals the square of the ratio of the lengths of any two corresponding sides; that is, A 1 A 2 = a a 1 a 2 b 2 THEOREM 8.2.7 EXAMPLE 6 Find the length of in Figure 8.28 if the area of the kite RSTV is 360 in. 2 and Solution becomes , in which 360 = 1 2 30d A = 1 2 d 1 d 2 SV = 30 in. RT S V R T W Figure 8.28 d is the length of the remaining diagonal . Then 360 ⫽ 15d, which means that d ⫽ 24. Then RT = 24 in. RT 쮿 Reminder Corresponding altitudes of similar triangles have the same ratio as any pair of corresponding sides. GIVEN: Similar triangles as shown in Figure 8.29 PROVE: PROOF: For the similar triangles, h 1 and h 2 are the respective lengths of altitudes to the corresponding sides of lengths b 1 and b 2 . Now and , so Simplifying, we have Because the triangles are similar, we know that . Because corresponding alti- tudes of similar triangles have the same ratio as a pair of corresponding sides Theorem b 1 b 2 = a 1 a 2 A 1 A 2 = b 1 b 2 h 1 h 2 A 1 A 2 = 1 2 b 1 h 1 1 2 b 2 h 2 or A 1 A 2 = 1 2 1 2 b 1 b 2 h 1 h 2 A 2 = 1 2 b 2 h 2 A 1 = 1 2 b 1 h 1 A 1 A 2 = a a 1 a 2 b 2 b 1 a 1 c 1 a 2 c 2 b 2 b 1 a 1 h 1 c 1 a 2 h 2 c 2 b 2 Figure 8.29 5.3.2, we also know that . Through substitution, becomes . Then . Because Theorem 8.2.7 can be extended to any pair of similar polygons, we could also prove that the ratio of the areas of two squares equals the square of the ratio of the lengths of any two sides. We apply this relationship in Example 7. A 1 A 2 = A a 1 a 2 B 2 A 1 A 2 = a 1 a 2 a 1 a 2 A 1 A 2 = b 1 b 2 h 1 h 2 h 1 h 2 = a 1 a 2 s 1 s 2 Figure 8.30 Exs. 18–21

1. 2.

7. 8.

3. 4.

is the area of the second triangle. 1 4 EXAMPLE 7 Use the ratio to compare the areas of: a Two similar triangles in which the sides of the first triangle are as long as the sides of the second triangle b Two squares in which each side of the first square is 3 times as long as each side of the second square Solution a , so . See Figure 8.30. Now , so that or . That is, the area of the first triangle A 1 A 2 = 1 4 A 1 A 2 = A 1 2 B 2 A 1 A 2 = A s 1 s 2 B 2 s 1 s 2 = 1 2 s 1 = 1 2 s 2 1 2 A 1 A 2 쮿 , so that or . That is, the area of the first square is 9 times the area of the second square. NOTE: For Example 7, Figures 8.30 and 8.31 provide visual evidence of the relationship described in Theorem 8.2.7. A 1 A 2 = 9 A 1 A 2 = A 3 B 2 A 1 A 2 = A s 1 s 2 B 2 s 2 Exercises 8.2 In Exercises 1 to 8, find the perimeter of each figure. 5. 6. 5 in. 12 in. B C D A 8 in. 13 in. 7 in. ABCD B 1 d C D A 2 d ABCD with AB ≅ BC 1 d = 4 m 2 d = 10 m B 5 C A D 4 O ABCD in O A 7 ft 4 ft D B C 13 ft Trapezoid ABCD with AB ≅ DC x √5 3 2 x D 13 A C 13 B 10 12 √11 √11 AB ≅ BC in concave quadrilateral ABCD 20 cm 16 cm 5 cm b , so . See Figure 8.31. s 1 s 2 = 3 s 1 = 3s 2 s 1 Figure 8.31 쮿 In Exercises 9 and 10, use Heron’s Formula.

9. Find the area of a triangle whose sides measure 13 in.,

14 in., and 15 in.

10. Find the area of a triangle whose sides measure 10 cm,

17 cm, and 21 cm. For Exercises 11 and 12, use Brahmagupta’s Formula.

11. For cyclic quadrilateral ABCD,

find the area if AB ⫽ 39 mm, BC ⫽ 52 mm, CD ⫽ 25 mm, and DA ⫽ 60 mm.

12. For cyclic quadrilateral ABCD, find

the area if AB ⫽ 6 cm, BC ⫽ 7 cm, CD ⫽ 2 cm, and DA ⫽ 9 cm. In Exercises 13 to 18, find the area of the given polygon. C B A D A 7 ft 4 ft D B C 13 ft Trapezoid ABCD with AB ≅ DC a The ratio of corresponding sides is . b The lengths of the sides of the first triangle are 6, 8, and 10 in., and those of the second triangle are 3, 4, and 5 in.

24. Find the ratio of the areas of two similar rectangles if:

A 1 A 2 s 1 s 2 = 3 2 a The ratio of corresponding sides is . b The length of the first rectangle is 6 m, and the length of the second rectangle is 4 m. In Exercises 25 and 26, give a paragraph form of proof. Provide drawings as needed.

25. Given: Equilateral

with each side of length s Prove: HINT: Use Heron’s Formula.

26. Given: Isosceles with

and Prove: NOTE: . In Exercises 27 to 30, find the area of the figure shown.

27. Given: In ,

OA ⫽ 5, BC ⫽ 6, and CD ⫽ 4 Find: A ABCD }O s 7 a A MNQ = a2s 2 - a 2 MN = 2a QM = QN = s 䉭MNQ A ABC = s 2 4 13 䉭ABC s 1 s 2 = 2 5

22. The numerical difference between the area of a square and

the perimeter of that square is 32. Find the length of a side of the square.

23. Find the ratio of the areas of two similar triangles if:

A 1 A 2

19. In a triangle of perimeter 76 in., the length of the first side

is twice the length of the second side, and the length of the third side is 12 in. more than the length of the second side. Find the lengths of the three sides.

20. In a triangle whose area is 72 in

2 , the base has a length of 8 in. Find the length of the corresponding altitude.

21. A trapezoid has an area of 96 cm

2 . If the altitude has a length of 8 cm and one base has a length of 9 cm, find the length of the other base.

13. 14.

15. 16.

17. 18.

12 m 20 m 15 m B 5 ABCD 8 D A C B 5 ABCD with BC ≅ CD 6 C D A A 45° Kite ABCD with BD = 12 m ∠ BAC = 45° , m ∠ BCA = 30° 12 B D C 30° B 12 Kite ABCD 20 D A C D C B A O

28. Given: Hexagon RSTVWX with

RS ⫽ 10 ST ⫽ 8 TV ⫽ 5 WV ⫽ 16 Find: A RSTVWX

29. Given: Pentagon ABCDE

with AE ⫽ AB ⫽ 5 BC ⫽ 12 Find: A ABCDE DC ⬵ DE WX ⬵ VT WV 7 XT 7 RS W V T X R S D E A C B

30. Given: Pentagon RSTVW

with , , and Find: A RSTVW

31. Mary Frances has a rectangular garden plot that encloses

an area of 48 yd 2 . If 28 yd of fencing are purchased to enclose the garden, what are the dimensions of the rectangular plot?

32. The perimeter of a right triangle is 12 m. If the