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360 6. m 1 = = 60 Divide 360 by the number of sides. 1 2. m 2 = ( 60 ) = 30 Substitute 60 for m 1. m 2 = m 1 The apothem bisects the vertex angle of the isosceles triangle formed by the radii. 1 2.
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360 6 m 1 = = 60 Divide 360 by the number of sides. 1 2 m 2 = (60) = 30 Substitute 60 for m 1. m 2 = m 1The apothem bisects the vertex angle of the isosceles triangle formed by the radii. 1 2 m 3 = 180 – (90 + 30) = 60 The sum of the measures of the angles of a triangle is 180. m 1 = 60, m 2 = 30, and m 3 = 60. Areas of Regular Polygons LESSON 10-3 Additional Examples A portion of a regular hexagon has an apothem and radii drawn. Find the measure of each numbered angle. Quick Check
1 2 A = apArea of a regular polygon 1 2 A = (37.9)(240) Substitute 37.9 for a and 240 for p. Areas of Regular Polygons LESSON 10-3 Additional Examples Find the area of a regular polygon with twenty 12-in. sides and a 37.9-in. apothem. p = nsFind the perimeter. p = (20)(12) = 240 Substitute 20 for n and 12 for s. A = 4548 Simplify. The area of the polygon is 4548 in.2 Quick Check
Consecutive radii form an isosceles triangle, as shown below, so an apothem bisects the side of the octagon. 1 2 To apply the area formula A = ap, you need to find a and p. Areas of Regular Polygons LESSON 10-3 Additional Examples A library is in the shape of a regular octagon. Each side is 18.0 ft. The radius of the octagon is 23.5 ft. Find the area of the library to the nearest 10 ft2.
Step 1: Find the apothem a. a2 + (9.0)2 = (23.5)2Pythagorean Theorem a2 + 81 = 552.25 Solve for a. a2 = 471.25 a 21.7 Areas of Regular Polygons LESSON 10-3 Additional Examples (continued) Step 2: Find the perimeter p. p = nsFind the perimeter. p = (8)(18.0) = 144 Substitute 8 for n and 18.0 for s, and simplify.
Step 3: Find the area A. A = apArea of a regular polygon A (21.7)(144) Substitute 21.7 for a and 144 for p. A 1562.4 Simplify. 1 2 1 2 Areas of Regular Polygons LESSON 10-3 Additional Examples (continued) To the nearest 10 ft2, the area is 1560 ft2. Quick Check