Given: Hexagon RSTVWX with Given: Pentagon ABCDE

52. Trapezoid ABCD not shown is inscribed in 䉺O so that

side is a diameter of . If DC ⫽ 10 and AB ⫽ 6, find the exact area of trapezoid ABCD. }O DC

51. The shaded region is that of a trapezoid. Determine the

height of the trapezoid. 8 A and B are midpoints. 6 A B

53. Each side of square RSTV has length 8. Point W lies on

and point Y lies on in such a way to form parallelogram VWSY, which has an area of 16 units 2 . Find x, the length of . VW TS VR W Y R S V T Regular Polygon Center and Central Angle of a Regular Polygon Radius and Apothem of a Regular Polygon Area of a Regular Polygon Regular Polygons and Area 8.3 KEY CONCEPTS Regular polygons are, of course, both equilateral and equiangular. As we saw in Section 7.3, we can inscribe a circle within any regular polygon and we can circum- scribe a circle about any regular polygon. For regular hexagon ABCDEF shown in Figure 8.32, suppose that and bisect the interior angles of ABCDEF as shown. In terms of hexagon ABCDEF, recall these terms and theorems.

1. Point Q is the center of regular hexagon ABCDEF. This

point Q is the common center of both the inscribed and circumscribed circles for regular hexagon ABCDEF. 2. is a radius of regular hexagon ABCDEF. A radius joins the center of the regular polygon to a vertex. 3. is an apothem of regular hexagon ABCDEF. An apothem is a line segment drawn from the center of a regular polygon so that it is perpendicular to a side of the polygon. 4. is a central angle of regular hexagon ABCDEF. Center point Q is the vertex of a central angle, whose sides are consecutive radii of the polygon. The measure of a central angle of a regular polygon of n sides is . 5. Any radius of a regular polygon bisects the interior angle to which it is drawn. 6. Any apothem of a regular polygon bisects the side to which it is drawn. Among regular polygons are the square and the equilateral triangle. As we saw in Section 8.1, the area of a square whose sides have length s is given by . A = s 2 c = 360° n ∠EQD QG QE QD QE Q E G D A F C B Figure 8.32 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