Transcript 11.3 Areas of Regular Polygons and Circles

11.3 Areas of Regular Polygons and
Circles
By Alysa Smith and Erin McCoy!!!!!!
 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.
A
GH is an apothem of hexagon
ABCDEF
B
G
F
C
So, GH is perpendicular to ED
E
H
D
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.
EXAMPLE 3:
Find the area of the shaded region. Assume the triangle is
equilateral.
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