NAME
DATE
11-4
PERIOD
Study Guide and Intervention
Areas of Regular Polygons and Composite Figures
Areas of Regular Polygons In a regular polygon, the segment drawn from the center
of the polygon perpendicular to the opposite side is called the apothem. In the figure at the
−−
−−−
right, AP is the apothem and AR is the radius of the circumscribed circle.
U
Area of a Regular Polygon
If a regular polygon has an area of A square units,
a perimeter of P units, and an apothem of a units,
1
then A = −
aP.
V
2
R
Example 1
Verify the formula
1
aP for the regular pentagon above.
A= −
2
For ∆RAS, the area is
2
( 2)
1
pentagon is A = 5 −
(RS)(AP). Substituting
P for 5RS and substituting a for AP, then
1
A=−
aP.
S
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Example 2
Find
the area of regular
pentagon RSTUV above if its perimeter
is 60 centimeters.
RP
tan m∠RAP = −
2
AP
6
tan 36° = −
AP
6
AP = −
tan 36°
≈ 8.26
1
1( )
So, A = −
aP = −
60 (8.26) or 247.8.
2
2
Exercises
The area is about 248 square centimeters.
Find the area of each regular polygon. Round to the nearest tenth.
1.
2.
14 m
4.
15 in.
10 in.
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84.9 m
3.
Geo-SG11-03-03-860188
172.0 in
2
5 √3 cm
Chapter 11
225.0 in2
10.9 m
5.
6.
10 in.
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Geo-SG11-03-04-860188
2
Geo-SG11-03-06-860188
25
7.5 m
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Glencoe Geometry
Lesson 11-4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
P
First find the apothem.
360°
The measure of central angle RAS is −
5
or
72°. Therefore, m∠RAP = 36. The perimeter
is 60, so RS = 12 and RP = 6.
1
1
A=−
bh = −
(RS)(AP). So the area of the
2
T
A
NAME
DATE
11-4
PERIOD
Study Guide and Intervention (continued)
Areas of Regular Polygons and Composite Figures
Areas of Composite Figures A composite figure is a figure that can be seprated into
regions that are basic figures. To find the area of a composite figure, separate the figure
into basic figures of which we can find the area. The sum of the areas of the basic figures is
the area of the figure.
Example
a.
Find the area of the shaded region.
b.
50 ft
5 cm
30 ft
The figure is a rectangle minus one half of
a circle. The radius of the circle is one half
of 30 or 15.
1
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πr2
A = lw - −
2
= 50(30) - 0.5π(15)2
≈ 1146.6 or about 1147 ft2
The dimensions of the rectangle are
10 centimeters and 30 centimeters. The area
of the shaded region is
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(10)(30) - 3π(52) = 300 - 75π
≈ 64.4 cm2
Exercises
Find the area of each figure. Round to the nearest tenth if necessary.
34 ft
1.
2.
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3.
14 cm
24 in.
24 in.
598.4 ft2
960 in2
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10 cm
38 cm
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
15 ft
40 in.
4.
22 cm
40 cm
42 cm
466 cm2
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5.
40 m
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6.
64 m
20 m
704 cm2
15 yd
20 m
1920 m2
262.5 yd2
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Geo-SG11-04-07-860188
Chapter 11
35 yd
26
Glencoe Geometry