5-Minute Check APK. 11.3 Perimeters and Areas of Similar Polygons.
11.3 Areas of Regular Polygons and Circles
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Transcript of 11.3 Areas of Regular Polygons and Circles
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11.3 Areas of Regular Polygons and Circles
By Alysa Smith and Erin McCoy!!!!!!
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Find areas of regular polygons Find areas of circles
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Areas of Regular PolygonsRecall 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 apothemapothem.
GH is an apothem of hexagonABCDEF
So, GH is perpendicular to ED
A B
C
DE
FG
H
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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
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Example 1:
Find the area of a regular pentagon with a perimeter of 40 cm.
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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.
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Areas of Circles
If a circle has an area of A sq. units and a radius of r units then
A = r sq.
r
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Example 2:Example 2:
Find the area of circle P with a circumference of 52 in.
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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.
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Find the area of the shaded region. Assume the triangle is equilateral.
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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.
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Assignment
Pg. 613 # 8-22, 23-27