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11.3 Areas of Regular Polygons and Circles. By Alysa Smith and Erin McCoy!!!!!!. Objectives. Find areas of regular polygons Find areas of circles. Areas of Regular Polygons. Recall that in a regular polygon, all angles are congruent and all sides are congruent.
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11.3 Areas of Regular Polygons and Circles By Alysa Smith and Erin McCoy!!!!!!
Objectives • Find areas of regular polygons • Find areas of circles
Areas of Regular Polygons Recall that in a regular polygon, all angles are congruent and all sides are congruent. 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 B A G F C E D H
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
Example 1: Find the area of a regular pentagon with a perimeter of 40 cm.
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.
Areas of Circles If a circle has an area of A sq. units and a radius of r units then A = r sq. r
Example 2: Find the area of circle P with a circumference of 52 in.
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.
Find the area of the shaded region. Assume the triangle is equilateral. Example 3:
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.
Assignment Pg. 613 # 8-22, 23-27