Develop and use formulas for finding the areas of regular polygons and composite shapes.

1. 5.13.2 Polygons & Composites
The student is able to (I can):
• Develop and use formulas to find the areas of regular
polygons and composite shapes

2. apothem
central angle
The distance from the center of a polygon
to one of the sides.
An angle whose vertex is at the center and
whose sides go through consecutive
vertices.
•
central angle

3. Like circles, for regular polygons, we can cut out a regular
polygon and arrange the wedges as shown:
apothem
1
2
perimeter
A bh=
1
A perimeter apothem
2
= i
1
A aP
2
=

4. Since all of the central angles have to add
up to 360˚, the measure of any central
angle is
apothem
central angle
360
n

5. If we look at the isosceles triangle created
by a central angle (the apothem is the
height), one of the triangle’s base angle is
½ the interior angle.
apothem
interior angle central angle
½ interior angle

6. What this means is that on any regular
polygon, we can set up a right triangle, with
one leg as the apothem and one leg as half
the length of a side.
apothem
½ side
½ central angle =
When you are calculating areas, you will
almost always be using the tangent ratio
(opposite/adjacent).
180
n
180 ½s
tan
n a
1
s
2a
180
tan
n
 
= 
 
=
 
 
 

7. Let’s review our table of interior angles and calculate
what the values will be for the angles of their right
triangles:
SidesSidesSidesSides NameNameNameName Each Int.Each Int.Each Int.Each Int. ½ Int.½ Int.½ Int.½ Int. ∠∠∠∠ CentralCentralCentralCentral ∠∠∠∠ ½ Central½ Central½ Central½ Central ∠∠∠∠
3333 Triangle 60º 30º 120º 60º
4444 Quadrilateral 90º 45º 90º 45º
5555 Pentagon 108º 54º 72º 36º
6666 Hexagon 120º 60º 60º 30º
7777 Heptagon ≈128.6º ≈64.3º ≈51.4º ≈25.7º
8888 Octagon 135º 67.5º 45º 22.5º
9999 Nonagon 140º 70º 40º 20º
10101010 Decagon 144º 72º 36º 18º
12121212 Dodecagon 150º 75º 30º 15º
nnnn n-gon ( )n 2 180
n
− ( )n 2 90
n
− 360
n
180
n

8. Examples 1. Find the area of the regular pentagon:
•
10 Let’s sketch in our
right triangle:
5
a
36º
To find the apothem (a), we can use the
tangent ratio:
5
tan 36
a
° =
Thus, a ≈ 6.88. The perimeter = 50, so
1
A (6.88)(50) 172
2
= =
° =
a
tan54
5
54°
or
( )P 5 10 50= =

9. (For those who like algebra)
You can put all of the previous information
together into one equation by using a little
algebra (n = # of sides; s = side length):
1
A aP
2
=
( )
1
s
1 2 n s
3602
tan
2n
 
 
=  
  
  
  
i
2
ns
180
4 tan
n
=
 
 
 

10. Example 1b. Find the area of a regular octagon with
side length 15 in.
n = 8; s = 15
2
ns
A
180
4 tan
n
=
 
 
 
( )2
8 15
180
4 tan
8
=
 
 
 
2
1086.4 in≈
(when you enter
this, be sure to
use () on the
denominator)

11. Examples 2. Find the exact area of the equilateral
triangle:
12
6
60º
This is a 30-60-90 triangle, so the
apothem is going to be 6
3 2 3
3
=
2 3
The perimeter will be 3(12) = 36. So the
area is
( )( )
1 1
A aP 2 3 36 36 3
2 2
= = =
•

12. Because the central angle of an equilateral
triangle forms a special right triangle, we
also have a special formula for the exact
area of an equilateral triangle:
Looking at the previous example, if we plug
in 12 for s, we get:
The advantage of this is that we don’t have
to try to draw in that tiny 30-60-90
triangle. Remember this as “s-2-3-4”.
2
s 3
A
4
=
2
12 3 144 3
A 36 3
4 4
= = =
(Where s is the
side length.)

13. Likewise, because the hexagon is composed
of 6 equilateral triangles, we have a special
formula for the exact area of a hexagon:
Example: What is the exact area of a
hexagon with a side length of 14?
2
s 3
A 6
4
 
=  
 
2
14 3 196 3
A 6 6
4 4
   
= =   
   
( )6 49 3 294 3= =

14. Summary • Regular polygon formulas
— or
— Equilateral triangle:
— Regular hexagon:
=
1
A aP
2
=
2
s 3
A
4
 
=  
 
2
s 3
A 6
4
2
ns
A
180
4 tan
n
=
 
 
 

15. composite
figure
A shape made up of simple shapes such as
triangles, rectangles, trapezoids, and
circles.
To find the area of the composite figure,
divide the shape into the simple shapes,
find the areas of each, and add (or
subtract) them.

16. Examples Find the area of each shape. Round to the
nearest tenth if necessary.
1.
38
20
70
40
3238
(20)(38) = 760

17. Examples Find the area of each shape. Round to the
nearest tenth if necessary.
1.
38
20
70
40
760
3238
− =2 2
40 32 24
24
( )( ) =
1
32 24 384
2
384
Area = 760 + 384 = 1144
(20)(38) = 760

18. Examples 2.
17" 42"
8"
Area = Triangle — ½ Circle
= =
1
Triangle (17)(42) 357 sq. in.
2
= π = π211 Circle (4 ) 8 sq. in.2 2
= − π ≈A 357 8 331.9 sq. in.

19. Examples 3. Find the area of the shaded portion to
the nearest tenth.
6 ft.
( )= π = π2 2
Circle 3 9 ft
( )( )= = 21
Square 6 6 18 ft
2
= π − ≈ 2
Area 9 18 10.3 ft