Square RSTV is inscribed in square WXYZ as shown. If Examine several rectangles, each with an area of 36 in Find the area of the

In Exercise 25 of Section 8.2, we showed that the area of an equilateral triangle whose sides are of length s is given by Following is a picture proof of this area relationship. PICTURE PROOF A = s 2 4 13 Exs. 1–4 GIVEN: The equilateral triangle with sides of length s PROVE: PROOF: Based upon the 30°-60°-90° triangle in Figure 8.34, becomes so A = s 2 4 13. A = 1 2 s s 2 13 A = 1 2 bh A = s 2 4 13 EXAMPLE 1 Find the area of the square whose length of apothem is Solution The apothem is the distance from the center to a side. For the square, s ⫽ 2a; that is, Then becomes and . A = 16 in 2 A = 4 2 A = s 2 s = 4 in. a = 2 in. a ⫽ 2 in. Figure 8.33 쮿 s 30° 60° s 3 s 2 s 2 Figure 8.34 EXAMPLE 2 Find the area of an equilateral triangle not shown in which each side measures 4 inches. Solution becomes or . A = 4 13 in 2 A = 4 2 4 13 A = s 2 4 13 쮿 EXAMPLE 3 Find the area of equilateral triangle ABC in which apothem has a length of 6 cm. Solution See Figure 8.35. If OD ⫽ 6 cm, then in the indicated AD = 6 13 cm OD C A D O B 30° 60° Figure 8.35 30°-60°-90° triangle AOD. In turn, . Now becomes A = 12 13 2 4 13 =

432 4

13 = 10813 cm 2 A = s 2 4 13 AB = 12 13 cm 쮿 Exs. 5–8 We now seek a general formula for the area of any regular polygon. AREA OF A REGULAR POLYGON In Chapter 7 and Chapter 8, we have laid the groundwork for determining the area of a regular polygon. In the proof of Theorem 8.3.1, the figure chosen is a regular pentagon; however, the proof applies to a regular polygon of any number of sides. It is also worth noting that the perimeter P of a regular polygon is the sum of its equal sides. If there are n sides and each has length s, the perimeter of the regular poly- gon is P ⫽ ns. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by A = 1 2 aP THEOREM 8.3.1 GIVEN: Regular polygon ABCDE in Figure 8.36a such that OF ⫽ a and the perimeter of ABCDE is P PROVE: PROOF: From center O, draw radii , , , , and . [See Figure 8.36b.] Now , , , , and are all by SSS. Where s represents the length of each of the congruent sides of the regular polygon and a is the length of an apothem, the area of each triangle is from . Therefore, the area of the pentagon is Because the sum or ns represents the perimeter P of the polygon, we have A ABCDE = 1 2 aP s + s + s + s + s = 1 2 a s + s + s + s + s A ABCDE = a 1 2 sa b + a 1 2 sa b + a 1 2 sa b + a 1 2 sa b + a 1 2 sa b A = 1 2 bh 1 2 sa ⬵ 䉭EOA 䉭DOE 䉭COD 䉭BOC 䉭AOB OE OD OC OB OA A ABCDE = 1 2 aP Exs. 9–11 EXAMPLE 4 Use to find the area of the square whose length of apothem is Solution For this repeat of Example 1, see Figure 8.33 as needed. When the length of apothem of a square is a ⫽ 2, the length of side is s ⫽ 4. In turn, the perimeter is Now becomes , so . NOTE: As expected, the answer from Example 1 is repeated in Example 4. A = 16 in 2 A = 1 2 2 16 A = 1 2 aP P = 16 in. a = 2 in. A = 1 2 aP D E B C F A O a D E B C F A O b Figure 8.36 쮿 쮿