1. 11.3 Areas of Regular Polygons and Circles By Alysa Smith and Erin McCoy!!!!!!

2.  Find areas of regular polygons  Find areas of circles

3. Areas of Regular Polygons Recall that in a regular polygon, all angles are congruent and all sides are congruent. apothem A segment drawn from the center of a regular polygon to a side of the polygon and is perpendicular to the side is called an apothem. GH is an apothem of hexagon ABCDEF So, GH is perpendicular to ED A B C DE F G H

4. If a regular polygon has an area of A square units, a perimeter of P units, and an apothem of a units, then… A=1/2Pa

5. Example 1: Find the area of a regular pentagon with a perimeter of 40 cm.

6. Because it is a regular pentagon, all angles add up to equal 360, and all angles are congruent. Therefore, the measure of each angle is 360 divided by 5 or 72. FG is an apothem of pentagon ABCDE. It bisects < EFD and is a perpendicular bisector of ED. The m< DFE = ½ (72) or 36. Because the perimeter is 40 cm, each side is 8 cm and GD is 4 cm. tan < DFG = GD/FG tan 36 = 4/FG (FG) tan 36 = 4 FG = 4/tan 36 FG ≈ 5.5 A = ½ Pa A= ½ (40)(5.5) A ≈110 The area of the pentagon is ≈110 cm sq.

7. Areas of Circles If a circle has an area of A sq. units and a radius of r units then A = r sq. r

8. Example 2: Find the area of circle P with a circumference of 52 in.

9. To find the radius of circle, find the diameter. d = Circumference/pi or d = 52/pi The diameter of circle P is 16.6. Since the radius is half the diameter, the radius is 8.4. A = pi r squared A = pi(8.4)squared The area of circle P is 221.7 in sq.

10. Find the area of the shaded region. Assume the triangle is equilateral.

11. The area of the shaded region is the difference of the area of the circle and the area of the triangle. A = pi r sq. A= pi (4) sq. A= about 50.3 To find the area of the triangle, use properties of 30-60-90 triangles. First, find the length of the base. The hypotenuse of triangle ABC is 4. BC = 2 √3, so EC = 4√3 Next, find the height of the triangle, DB. Since m<DCB is 60, DB = 2 √3 (√3) or 6. A = ½ bh A = ½ (4√3) (6) A = about 20.8; the area of the shaded region is 50.3 – 20.8 or 29.5 m sq.

12. Assignment Pg. 613 # 8-22, 23-27