GEOMETRYHELP

m 1 = = 60 Divide 360 by the number of sides.360 6

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.

A portion of a regular hexagon has an apothem

and radii drawn. Find the measure of each numbered angle.

m 2 = m 1 The apothem bisects the vertex angle of the isosceles triangle formed by the radii.

12

m 2 = (60) = 30 Substitute 60 for m 1.12

Quick Check

Areas of Regular PolygonsLESSON 10-3

Additional Examples

GEOMETRYHELP

Find the area of a regular polygon with twenty 12-in. sides and

a 37.9-in. apothem.

p = ns Find 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

A = (37.9)(240) Substitute 37.9 for a and 240 for p.12

A = ap Area of a regular polygon12

Areas of Regular PolygonsLESSON 10-3

Additional Examples

Quick Check

GEOMETRYHELP

Consecutive radii form an isosceles triangle, as shown below, so an apothem bisects the side of the octagon.

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.

To apply the area formula A = ap, you need to find a and p.

12

Areas of Regular PolygonsLESSON 10-3

Additional Examples

GEOMETRYHELP

Step 2: Find the perimeter p.

p = ns Find the perimeter.

p = (8)(18.0) = 144 Substitute 8 for n and 18.0 for s,

and simplify.

(continued)

Step 1: Find the apothem a.

a2 + (9.0)2 = (23.5)2 Pythagorean Theorem

a2 + 81 = 552.25 Solve for a.

a2 = 471.25

a 21.7

Areas of Regular PolygonsLESSON 10-3

Additional Examples

GEOMETRYHELP

To the nearest 10 ft2, the area is 1560 ft2.

(continued)

Step 3: Find the area A.

A = ap Area of a regular polygon

A (21.7)(144) Substitute 21.7 for a and 144 for p.

A 1562.4 Simplify.

12

12

Areas of Regular PolygonsLESSON 10-3

Additional Examples

Quick Check