Python Notes for mca i year students osmania university.docx
5.13.2 Area of Regular Polygons and Composite Shapes
1. Polygons & Composites
The student is able to (I can):
• Develop and use formulas to find the areas of regular
polygons and composite shapes
2. apothemapothemapothemapothem – the distance from the center of a polygon to one
of the sides.
centralcentralcentralcentral angleangleangleangle – and 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
perimeter apothem
2
A = i
1
2
A aP=
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
tan
1
2
180
tan
s
n a
s
a
n
=
=
½
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 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 ( )2 180n
n
− ( )2 90n
n
− 360
n
180
n
9. Examples
1. Find the area of the regular pentagon:
•
12 Let’s sketch in our
right triangle:
6
a
36°
To find the apothem (a), we can use the
tangent ratio:
6
tan 36
a
° =
Thus, a ≈ 8.258... The perimeter = 60, so
1
(8.26)(60) 247.8
2
A = =
tan54
6
a
° =
54°
or
( )5 12 60P = =
10. (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
2
A aP=
( )
1
1 2
3602
tan
2
s
n s
n
=
i
2
180
4tan
ns
n
=
11. Example
1b. Find the area of a regular octagon with side length 15 in.
n = 8; s = 15
2
180
4tan
ns
A
n
=
( )2
8 15
180
4tan
8
=
2
1086.4 in≈
When you enter
this, be sure to
use () on the
denominator!
12. 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 1
2 3 36 36 3
2 2
= = =A aP
•
13. 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
3
4
s
A =
2
12 3 144 3
36 3
4 4
A = = =
(Where s is the
side length.)
14. 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
3
6
4
s
A
=
2
14 3 196 3
6 6
4 4
A
= =
( )6 49 3 294 3= =
15. Summary • Regular polygon formulas
– or
– Equilateral triangle:
– Regular hexagon:
1
2
A aP=
2
3
4
s
A =
2
3
6
4
s
A
=
2
180
4tan
ns
A
n
=
16. compositecompositecompositecomposite figurefigurefigurefigure – 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.
17. Examples
Find the area of each shape. Round to the nearest tenth if
necessary.
1.
38
20
70
40
18. Examples
Find the area of each shape. Round to the nearest tenth if
necessary.
1.
38
20
70
40
760
3238
(20)(38) = 760
19. 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
20. Examples
2.
17" 42"
8"
Area = Triangle – ½ Circle
1
Triangle (17)(42) 357 sq. in.
2
= =
211 Circle (4 ) 8 sq. in.
2 2
= π = π
357 8 331.9 sq. in.= − π ≈A
21. Examples
3. Find the area of the shaded portion to the nearest tenth.
6 ft.
( )2 2
Circle 3 9 ft= π = π
( )( ) 21
Square 6 6 18 ft
2
= =
2
Area 9 18 10.3 ft= π− ≈
