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

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

Like circles, for regular polygons, we can cut out a regular polygon and arrange the wedges as shown:

apothem

1

2perimeter

A bh=1

A perimeter apothem2

= i

1A aP

2=

Since all of the central angles have to add up to 360, the measure of any central angle is

apothem

central angle

360

n

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

apothem

interior angle central angle

interior angle

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

side

central angle =

When you are calculating areas, you will almost always be using the tangent ratio (opposite/adjacent).

180

n

180 stan

n a

1s

2a180

tann

=

=

Lets 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. Int. Int. Int. Int. Central Central Central Central Central Central Central Central

3333 Triangle 60 30 120 60

4444 Quadrilateral 90 45 90 45

5555 Pentagon 108 54 72 36

6666 Hexagon 120 60 60 30

7777 Heptagon 128.6 64.3 51.4 25.7

8888 Octagon 135 67.5 45 22.5

9999 Nonagon 140 70 40 20

10101010 Decagon 144 72 36 18

12121212 Dodecagon 150 75 30 15

nnnn n-gon ( )n 2 180n

( )n 2 90n

360n

180

n

Examples 1. Find the area of the regular pentagon:

10 Lets sketch in our right triangle:

5

a

36

To find the apothem (a), we can use the tangent ratio:

5tan 36

a =

Thus, a 6.88. The perimeter = 50, so1

A (6.88)(50) 1722

= =

=a

tan545

54

or

( )P 5 10 50= =

(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):

1A aP

2=

( )

1s

1 2 n s3602

tan2n

=

i

2ns

1804 tan

n

=

Example 1b. Find the area of a regular octagon with side length 15 in.

n = 8; s = 15

2nsA

1804 tan

n

=

( )28 15180

4 tan8

=

21086.4 in

(when you enter this, be sure to use () on the denominator)

Examples 2. Find the exact area of the equilateral triangle:

12

6

60

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 1A aP 2 3 36 36 32 2

= = =

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 dont have to try to draw in that tiny 30-60-90 triangle. Remember this as s-2-3-4.

2s 3A

4=

212 3 144 3A 36 3

4 4= = =

(Where s is the side length.)

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?

2s 3A 6

4

=

214 3 196 3A 6 6

4 4

= =

( )6 49 3 294 3= =

Summary Regular polygon formulas

or

Equilateral triangle:

Regular hexagon:

=1

A aP2

=2s 3

A4

=

2s 3A 6

4

2nsA

1804 tan

n

=

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.

Examples Find the area of each shape. Round to the nearest tenth if necessary.

1.38

20

70

40

3238

(20)(38) = 760

Examples Find the area of each shape. Round to the nearest tenth if necessary.

1.38

20

70

40

760

3238 =2 240 32 24

24

( )( ) =1

32 24 3842

384

Area = 760 + 384 = 1144

(20)(38) = 760

Examples 2.

17" 42"

8"

Area = Triangle Circle

= =1

Triangle (17)(42) 357 sq. in.2

= = 211 Circle (4 ) 8 sq. in.2 2

= A 357 8 331.9 sq. in.

Examples 3. Find the area of the shaded portion to the nearest tenth.

6 ft.

( )= = 2 2Circle 3 9 ft

( )( )= = 21

Square 6 6 18 ft2

= 2Area 9 18 10.3 ft