1. What does the word “polygon” mean?
Let's Discuss
What does the word “polygon” mean?
What is the smallest number of sides a polygon can have?
What is the largest number of sides a polygon can have?

2. Polygons Poly means many and gon means angles.
The word ‘polygon’ is a Greek word.
Poly means many and gon means angles.

3. Polygons
The word polygon means “many angles”
A two dimensional object
A closed figure

4. More about Polygons
Made up of three or more straight line segments
There are exactly two sides that meet at a vertex
The sides do not cross each other

5. EXAMPLES

6. Triangle Octagon Quadrilateral Nonagon Pentagon Decagon Dodecagon
Hexagon
n-gon
Heptagon
Hendecagon

7. More Examples of Polygons

8. These are not Polygons

9. Terminology
Side: One of the line segments that make up a polygon.
Vertex: Point where two sides meet.

10. Vertex
Side

11. Interior angle: An angle formed by two adjacent sides inside the polygon.
Exterior angle: An angle formed by two adjacent sides outside the polygon.

12. Exterior angle
Interior angle

13. Let us recapitulate
Interior angle
Diagonal
Vertex
Side
Exterior angle

14. Types of Polygons
Equiangular Polygon: a polygon in which all of the angles are equal
Equilateral Polygon: a polygon in which all of the sides are the same length

15. Regular Polygon: a polygon where all the angles are equal and all of the sides are the same length. They are both equilateral and equiangular

16. Examples of Regular Polygons

17. A convex polygon: A polygon whose each of the interior angle measures less than 180°.
If one or more than one angle in a polygon measures more than 180° then it is known as concave polygon. (Think: concave has a "cave" in it)

18. IN TERIOR ANGLES OF A POLYGON

19. Let us find the connection between the number of sides, number of diagonals and the number of triangles of a polygon.

20. 3 x 180o = 540o 4 sides Quadrilateral 5 sides Pentagon 2 1 diagonal
2 diagonals
180o
180o
180o
180o
180o
180o
6 sides
Hexagon
7 sides
Heptagon/Septagon 4
4 x 180o = 720o 5
5 x 180o = 900o
3 diagonals
4 diagonals

21. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600

22. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900

23. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900
Pentagon 5
3 x1800
= 5400
5400/5
= 1080

24. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900
Pentagon 5
3 x1800
= 5400
5400/5
= 1080
Hexagon 6
4 x1800
= 7200
7200/6
= 1200

25. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900
Pentagon 5
3 x1800
= 5400
5400/5
= 1080
Hexagon 6
4 x1800
= 7200
7200/6
= 1200
Heptagon 7
5 x1800
= 9000
9000/7 =

26. Sum of the interior angles Each interior angle
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900
Pentagon 5
3 x1800
= 5400
5400/5
= 1080
Hexagon 6
4 x1800
= 7200
7200/6
= 1200
Heptagon 7
5 x1800
= 9000
9000/7 =
“n” sided polygon n
Association with no. of sides
Association with no. of triangles
Association with sum of interior angles

27. Sum of the interior angles
Regular
Polygon
No. of sides
No. of diagonals
No. of
Sum of the interior angles
Each interior angle
Triangle 3 1
1800
1800/3
= 600
Quadrilateral 4 2
2 x1800
= 3600
3600/4
= 900
Pentagon 5
3 x1800
= 5400
5400/5
= 1080
Hexagon 6
4 x1800
= 7200
7200/6
= 1200
Heptagon 7
5 x1800
= 9000
9000/7 =
“n” sided polygon n
n - 3
n - 2
(n - 2) x1800
(n - 2) x1800 / n

28. Polygons with 3 sides… Triangles Polygons with 4 sides… Quadrilaterals
Polygons with 5 sides.. Pentagons
But wait we have more polygons
Polygons with 6 sides… Hexagons
Polygons with 7 sides… Heptagons
Polygons with 8 sides… Octagons
But still we have more polygons
Polygons with 9 sides… Nonagons
Polygons with 10 sides… Decagons
Polygons with 12 sides… Dodecagons
And now we have our polygons

29. 1
Calculate the Sum of Interior Angles and each interior angle of each of these regular polygons.
7 sides
Septagon/Heptagon
Sum of Int. Angles 900o
Interior Angle 128.6o 2 3 4
9 sides
10 sides
11 sides
Nonagon
Decagon
Hendecagon
Sum 1260o
I.A. 140o
Sum 1440o I.A. 144o
Sum 1620o
I.A o

30. Find the unknown angles below.
Diagrams not drawn accurately.
Find the unknown angles below. x
115o
110o
75o
95o
75o
100o
70o w
2 x 180o = 360o
3 x 180o = 540o
360 – 245 = 115o
540 – 395 = 145o y
117o
121o
100o
125o
140o z
133o
137o
138o
125o
105o
4 x 180o = 720o
5 x 180o = 900o
720 – 603 = 117o
900 – 776 = 124o

31. EXTERIOR ANGLES OF A POLYGON

32. An exterior angle of a regular polygon is
formed by extending one side of the polygon.
Angle CDY is an exterior angle to angle CDE D E Y B C A F 1 2
Exterior Angle + Interior Angle of a regular polygon =1800

33. Sum of exterior angles = 360º
No matter what type of polygon we have, the sum of the exterior angles is ALWAYS equal to 360º.
  Sum of exterior angles = 360º
Polygons

34. In a regular polygon with ‘n’ sides
Sum of interior angles = (n -2) x 1800
i.e. 2(n – 2) x right angles
Exterior Angle + Interior Angle =1800
Each exterior angle = 3600/n
No. of sides = 3600/exterior angle               
Polygons

35. Let us explore few more problems
Find the measure of each interior angle of a polygon with 9 sides.
Ans : 1400
Find the measure of each exterior angle of a regular decagon.
Ans : 360
How many sides are there in a regular polygon if each interior angle measures 1650?
Ans : 24 sides
Is it possible to have a regular polygon with an exterior angle equal to 400 ?
Ans : Yes
Polygons

36. Review

37. Important Terms CONSECUTIVE VERTICES are two endpoints of any side.
A VERTEX is the point of intersection of two sides
CONSECUTIVE VERTICES are two endpoints of any side. F A B C D E
A segment whose endpoints are two
nonconsecutive vertices is called a DIAGONAL.
Sides that share a vertex are called CONSECUTIVE SIDES.

38. Polygons are named by listing its vertices consecutively.F E D

39. More Important Terms EQUILATERAL - All sides are congruent
EQUIANGULAR - All angles are congruent
REGULAR - All sides and angles are congruent

40. Three more important terms
Interior Angles
Exterior Angles
SUPPLEMENTARY
the SUM of an interior angle and it’s corresponding exterior angle = 180o

41. Polygons can be CONCAVE or CONVEX

42. Classify each polygon as convex or concave.

43. It’s what’s inside that counts 
Finding and using the interior angle measures of polygons

44. What is the sum of the measures of the interior angles of a triangle?
180°
180°
180°
What is the sum of the measures of the interior angles of any quadrilateral?
360°

45. Sum of measures of interior angles
# of triangles
Sum of measures of interior angles
# of sides
1(180) = 180 3 1
2(180) = 360 4 2 3
3(180) = 540 5 6 4
4(180) = 720
n-2
(n-2) 180 n

46. If a convex polygon has n sides, then the sum of the measure of the interior angles is (n – 2)(180°)

47. If you know the sum of the angles of a regular polygon, how can you find the measure of one of the congruent angles?
Regular Polygon
Interior angle sum
Measure of one angle
Triangle
180o
Quadrilateral
360o
Pentagon
Hexagon
Heptagon
Octagon
Decagon
Dodecagon
n-gon

48. If a regular convex polygon has n sides, then the measure of one of the interior angles is

49. Perimeter and Area

50. Area
Definition:
The number of square units needed to cover a surface. (INSIDE)
Length x width

51. Perimeter Definition:
The sum of the length of all the sides of a polygon.

52. Base
Definition:
A side or face of a figure on which the figure stands.

53. Height
Definition:
The distance from the bottom to the top of a figure.

54. Width
Definition:
How wide a figure is from side to side.

55. Length
Definition:
The measure of the distance across an figure.

56. Warm Up
Graph the line segment for each set of ordered pairs. Then find the length of the line segment.
1. (–7, 0), (0, 0)
2. (0, 3), (0, 6)
3. (–4, –2), (1, –2)
4. (–5, 4), (–5, –2)
7 units
3 units
5 units
6 units

57. Vocabulary
perimeter
area
base
height
composite figure

58. Perimeter is the distance around a polygon
Perimeter is the distance around a polygon. To find the perimeter of any polygon, you add the lengths of all its sides.
Since opposite sides of a parallelogram are equal in length, you can find a formula for the perimeter of a parallelogram.
P = w + l + w + l
= w + w + l + l
= 2w + 2l w l

60. Additional Example 1: Finding the Perimeter of Parallelograms
A. Find the perimeter of the figure. 5 14
P = 2w + 2l
Perimeter of a parallelogram.
= 2(5) + 2(14)
Substitute 5 for w and 14 for l.
= = 38 units

61. Additional Example 1: Finding the Perimeter of Parallelograms
B. Find the perimeter of the figure. 20 16
P = 2w + 2l
Perimeter of a parallelogram.
= 2(16) + 2(20)
Substitute 16 for w and 20 for l.
= = 72 units

62. Check It Out! Example 1
A. Find the perimeter of the figure. 6 11
P = 2w + 2l
Perimeter of a parallelogram.
= 2(6) + 2(11)
Substitute 6 for w and 11 for l.
= = 34 units

63. Check It Out! Example 1
B. Find the perimeter of the figure. 5 13
P = 2w + 2l
Perimeter of a parallelogram.
= 2(5) + 2(13)
Substitute 5 for w and 13 for l.
= = 36 units

64. The area of a plane figure is the number of unit squares needed to cover the figure. The base of a parallelogram is the length of one side. The height is the perpendicular distance from the base to the opposite side.
Height
Side
Base

65. While perimeter is expressed in linear units, such as inches (in
While perimeter is expressed in linear units, such as inches (in.) or meters (m), area is expressed in square units, such as square feet (ft2).
You can cut a parallelogram and shift the cut piece to form a rectangle whose base and height are the same as those of the original parallelogram. The same number of unit squares are needed to cover the two figures. So a parallelogram and a rectangle that have the same base and height have the same area.

67. Since the base and height of a rectangle are the same as its length and width, the formula for the area of a rectangle can also be written as A = lw.
Helpful Hint

68. Additional Example 2: Using a Graph to Find Area
Graph and find the area of the figure with the given vertices.
A. (–1, –2), (2, –2), (2, 3), (–1, 3)
Area of a rectangle.
A = bh
Substitute 3 for b and 5 for h.
A = 3 • 5
A = 15 units2

69. The height of a parallelogram is not the length of its slanted side
The height of a parallelogram is not the length of its slanted side. The height of a figure is always perpendicular to the base.
Caution!

70. Additional Example 2: Using a Graph to Find Area
Graph and find the area of the figure with the given vertices.
B. (0, 0), (5, 0), (6, 4), (1, 4)
Area of a parallelogram.
A = bh
Substitute 5 for b and 4 for h.
A = 5 • 4
A = 20 units2

71. Graph and find the area of the figure with the given vertices.
Check It Out! Example 2
Graph and find the area of the figure with the given vertices.
A. (–3, –2), (1, –2), (1, 3), (–3, 3) x y
(–3, –2)
(1, –2)
(1, 3)
(–3, 3) 4 5
Area of a rectangle.
A = bh
Substitute 4 for b and 5 for h.
A = 4 • 5
A = 20 units2

72. Area of a parallelogram.
Check It Out! Example 2
Graph the figure with the given vertices. Then find the area of the figure.
B. (–1, –1), (3, –1), (5, 3), (1, 3)
(5, 3) x y
(–1, –1)
(3, –1)
(1, 3) 4
Area of a parallelogram.
A = bh
Substitute 4 for b and 4 for h.
A = 4 • 4
A = 16 units2

73. A composite figure is made up of basic geometric shapes such as rectangles, triangles, trapezoids, and circles. To find the area of a composite figure, find the areas of the geometric shapes and then add the areas.

74. Additional Example 3: Finding Area and Perimeter of a Composite Figure
Find the perimeter and area of the figure. 6 6 3 3 6 5 5
The length of the side that is not labeled is the same as the sum of the lengths of the sides opposite, 18 units.
P =
= 52 units

75. Additional Example 3 Continued6 6 3 3 6 5 5
A = 6 • • • 5
Add the areas together. =
= 72 units2

76. Find the perimeter of the figure.
Check It Out! Example 3
Find the perimeter of the figure.
The length of the side that is not labeled is 2. 2 4 6 7 7 2 6 2
P = ?
= 42 units 4

77. Check It Out! Example 3 Continued2
Find the area of the figure. 4 6 7
Add the areas together.
A = 2 • • • • 2 7 2 2 6 = 2 2
= 38 units2 2 6 4 2 4 7 2 2 + + +

78. Lesson Quiz: Part I
1. Find the perimeter of the figure.
44 ft
2. Find the area of the figure.
108 ft2

79. Lesson Quiz: Part II
Graph and find the area of each figure with the given vertices.
3. (–4, 2), (6, 2), (6, –3), (–4, –3)
50 units2

80. Lesson Quiz: Part III
Graph and find the area of each figure with the given vertices.
4. (4, –2), (–2, –2), (–3, 5), (3, 5)
42 units2

81. Warm Up
A rectangle has sides lengths of 12 ft and 20 ft.
1. Find the perimeter.
64 ft
2. Find the area.
240 ft2

82. Additional Example 1: Using Perimeter
Find the missing measurement when the perimeter is 71 in.
18 in.
15 in. d
22 in.
P = d
71 = 55 + d
Substitute 71 for P.
Subtract 55 from both sides.
16 = d
d = 16 in.

83. Check It Out! Example 1
Find the missing measurement when the perimeter is 58 in.
14 in.
7 in. d
28 in.
P = d
58 = 49 + d
Substitute 58 for P.
Subtract 49 from both sides.
9 = d
d = 9 in.

84. Additional Example 2: Multi-Step Application
A homeowner wants to plant a border of shrubs around her yard that is in the shape of a right triangle. She knows that the length of the shortest side of the yard is 12 feet and the length of the longest side is 20 feet. How long will the border be?
Step 1: Find the length of the third side.
a2 + b2 = c2
Use the Pythagorean Theorem.
122 + b2 = 202
Substitute 12 for a and 20 for c.
144 + b2 = 400
b2 = 256
b = 16
√256 = 16.

85. Additional Example 3 Continued
Step 2: Find the perimeter of the yard.
P = a + b + c =
Add all sides.
= 48
The border will be 48 feet long.

86. Check It Out! Example 2
A gardener wants to plant a border of flowers around the building that is in the shape of a right triangle. He knows that the length of the shortest sides of the building are 38 feet and 32 feet. How long will the border be?
Step 1: Find the length of the third side.
a2 + b2 = c2
Use the Pythagorean Theorem.
= c2
Substitute 38 for a and 32 for b.
= c2
2468 = c2
c ≈ 49.68
√2468 

87. Check It Out! Example 2 Continued
Step 2: Find the perimeter of the yard.
P = a + b + c 
Add all sides. 
The border will be about feet long.

88. area of a triangle or trapezoid have as a factor.
A triangle or trapezoid can be thought of as half of a parallelogram. Therefore, the formulas for the
area of a triangle or trapezoid have as a factor.
1 2

90. Additional Example 3: Finding the Area of Triangles and Trapezoids
Graph and find the area of the figure with the given vertices.
A. (–2, 2), (4, 2), (0, 5)
Area of a triangle y
A = bh 1 2
(0, 5)
Substitute for b and h.
= • 6 • 3 1 2 3
(–2, 2)
(4, 2) 6
= 9 units2 x

91. Additional Example 3: Finding the Area of Triangles and Trapezoids
Graph and find the area of the figure with the given vertices.
B. (–1, 1), (4, 1), (4, 4), (0, 4) y
Area of a trapezoid
A = h(b1 + b2) 1 2
(0, 4) 4
(4, 4) 3
Substitute for h, b1, and b2.
(–1, 1)
(4, 1) x
= • 3(5 + 4) 1 2 5
= 13.5 units2

92. 1 A = h(b1 + b2) 2 1 = • 6(8 + 4) 2 Check It Out! Example 3
Graph and find the area of the figure with the given vertices.
A. (–1, –2), (5, –2), (5, 2), (–1, 6) y
(–1, 6)
Area of a trapezoid
A = h(b1 + b2) 1 2
(5, 2)
Substitute for h, b1, and b2. 8 6 4 x
= • 6(8 + 4) 1 2
(–1, –2)
(5, –2)
= 36 units2

93. 1 A = bh 2 1 = • 6 • 4 2 Check It Out! Example 3
Graph and find the area of the figure with the given vertices.
B. (–1, 1), (5, 1), (1, 5)
Area of a triangle y
A = bh 1 2
(1, 5)
Substitute for b and h.
= • 6 • 4 1 2 4
(–1, 1)
(5, 1)
= 12 units2 x 6

94. 1. the perimeter of the triangle 36 cm
Lesson Quiz
Use the figure to find the following measurements.
1. the perimeter of the triangle
36 cm
2. the perimeter of the trapezoid
44 cm
3. the perimeter of the combined figure
64 cm
4. the area of the triangle
54 cm2
5. the area of the trapezoid
104 cm2

95. Circles

96. Warm Up
1. Two angles are supplementary. One angle measures 61°. Find the measure of the other angle.
2. Two angles are complementary. One angle measures 19°. Find the measure of the other angle.
3. Two angles are supplementary and congruent. Find the measure of each angle.
119°
71°
90°

97. Vocabulary circle center of a circle arc radius diameter chord
central angle
sector

98. A circle is the set of all points in a plane that are the same distance from a given point, called the center of a circle. This distance is called the radius of the circle.
A circle is named by its center. For example, if point A is the center of a circle, then the name of the circle is circle A. There are special names for the different parts of a circle.

99. Diameter
Arc
Part of a circle named by its endpoints
Line segment that passes
through the center of a circle, and whose endpoints lie on the circle
Radius
Line segment whose endpoints are the center
of a circle and any point on the circle
Chord
Line segment whose
endpoints are any two
points on a circle

100. Additional Example 1: Identifying Parts of Circles
Name the parts of circle M. N
A. radii:
MN, MR, MQ, MO O P
B. diameters:
NR, QO M Q
C. chords:
NR, QO, QN, NP R
Radii is the plural form of radius.
Reading Math

101. Name the parts of circle M.
Check It Out! Example 1
Name the parts of circle M. B A C
A. radii:
GB, GA, GF, GD G
B. diameters:
BF, AD D H F
C. chords:
AH, AB, CE,
BF, AD E

102. A central angle of a circle is an angle formed by two
radii. A sector of a circle
is the part of the circle
enclosed by two radii and an
arc connecting them.
Sector )
The sum of the measures of all
of the central angles in a circle
is 360°. We say that there are
360° in a circle.
Central angle

103. Additional Example 2: Problem Solving Application
The circle graph shows the results of a survey about favorite types of muffins. Find the central angle measure of the sector that shows the percent of people whose favorite type of muffin is blueberry.

104. Additional Example 2 Continued1
Understand the Problem
List the important information:
• The percent of people whose favorite
muffin is blueberry is 40%.

105. Additional Example 2 Continued
Make a Plan
The central angle measure of the sector that represents this group is 40% of the angle measure of the whole circle. The angle measure of a circle is 360°. Since the sector is 40% of the circle graph, the central angle is 40% of the 360° in the circle.
40% of 360° = 0.40 · 360°
Solve 3
0.40 · 360° = 144°
Multiply.
The central angle of the sector measures 144°.

106. Additional Example 2 Continued4
Look Back
The 40% sector is less than half the graph, and 144° is less than half of 360°. Since 144° is close to 180°, the answer is reasonable.

107. Check It Out! Example 2
The circle graph shows the results of a survey about favorite types of muffins. Find the central angle measure of the sector that shows the percent of people whose favorite type of muffin is banana nut.

108. Check It Out! Example 2 Continued1
Understand the Problem
List the important information:
• The percent of people whose favorite
muffin is banana nut is 35%.

109. Check It Out! Example 2 Continued
Make a Plan
The central angle measure of the sector that represents this group is 35% of the angle measure of the whole circle. The angle measure of a circle is 360°. Since the sector is 35% of the circle graph, the central angle measure is 35% of the 360° in the circle.
35% of 360° = 0.35 · 360°
Solve 3
0.35 · 360° = 126°
Multiply.
The central angle of the sector measures 126°.

110. Check It Out! Example 2 Continued4
Look Back
The 35% sector is about one-third the graph, and 126° is about one-third of 360°. Since 126° is close to 120°, the answer is reasonable.

111. Lesson Quiz
Name the parts of circle B.
1. radii
2. diameter(s)
3. chord(s) 4.
BA, BC AC
DE, FE, AC
A pie is cut into 8 equal sectors. Find the
measure of the central angle of one slice of
pie.
45°

112. Vocabulary
circumference

113. d = 2r Radius Center The diameter d is twice the radius r. Diameter
Circumference
The circumference of a circle is the distance around the circle.

115. What is pi? Pi= …
Pi is the mathematical constant whose value is the ratio of a circle’s circumference to its diameter.
Circumference = The distance around a circle
Diameter = The width of a circle.

116. Remember!
Pi () is an irrational number that is often approximated by the rational numbers 3.14 and 22 7

117. Additional Example 1: Finding the Circumference of a Circle
Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
A. circle with a radius of 4 m
C = 2r
= 2(4)
= 8 m  25.1 m
B. circle with a diameter of 3.3 ft
C = d
=  (3.3)
= 3.3 ft  10.4 ft

118. Check It Out! Example 1
Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
A. circle with a radius of 8 cm
C = 2r
= 2(8)
= 16 cm  50.2 cm
B. circle with a diameter of 4.25 in.
C = d
= (4.25)
= 4.25 in.  13.3 in.

120. Additional Example 2: Finding the Area of a Circle
Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
A. circle with a radius of 4 in.
A = r2 = (42)
= 16 in2  50.2 in2
B. circle with a diameter of 3.3 m d 2
= 1.65
A = r2 = (1.652)
=  m2  8.5 m2

121. Check It Out! Example 2
Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
A. circle with a radius of 8 cm
A = r2 =  (82)
= 64 cm2  cm2
B. circle with a diameter of 2.2 ft d 2
= 1.1
A = r2 =  (1.12)
= 1.21 ft2  3.8 ft2

122. Additional Example 3: Finding the Area and Circumference on a Coordinate Plane
Graph the circle with center (–2, 1) that passes through (1, 1). Find the area and circumference, both in terms of  and to the nearest tenth. Use 3.14 for .
A = r2
C = d
= (32)
= (6)
= 9 units2
= 6 units
 28.3 units2
 18.8 units

123. Check It Out! Example 3
Graph the circle with center (–2, 1) that passes through (–2, 5). Find the area and circumference, both in terms of  and to the nearest tenth. Use 3.14 for . y
A = r2
C = d
(–2, 5)
= (42)
= (8) 4
= 16 units2
= 8 units
 50.2 units2
 25.1 units x
(–2, 1)

124. Additional Example 4: Measurement Application
A Ferris wheel has a diameter of 56 feet and makes 15 revolutions per ride. How far
would someone travel during a ride? Use for . 22 7
C = d = (56)
Find the circumference.
 (56)  22 7 56 1 22 7
 176 ft
The distance is the circumference of the wheel times the number of revolutions, or about 176  15 = 2640 ft.

125. 22 7 22 22 14  (14)   44 in. 7 7 1 Check It Out! Example 4
A second hand on a clock is 7 in. long. What is
the distance it travels in one hour? Use for . 22 7
C = d =  (14)
Find the circumference. 14 1 22 7 12 3 6 9
 (14)  22 7
 44 in.
The distance is the circumference of the clock times the number of revolutions, or about 44  60 = 2640 in.

126. Lesson Quiz
Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
1. radius 5.6 m
11.2 m; 35.2 m
2. diameter 113 m
113 mm; mm
Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for .
3. radius 3 in.
9 in2; 28.3 in2
4. diameter 1 ft
0.25 ft2; 0.8 ft2

127. Warm Up Find each area. Round to the nearest tenth, if necessary.
1. a rectangle with length 10 cm and width 4 cm
2. a parallelogram with base 18 ft and height 12 ft
3. a triangle with base 16 cm and height 8 cm
4. a circle with radius 5 in.
40 cm2
216 ft2
64 cm2
78.5 in2

128. Additional Example 1A: Finding the Area of Composite Figures by Adding
Find the shaded area. Round to the nearest tenth, if necessary.
2 m
Divide the figure into a rectangle and a trapezoid.
8 m
2 m
4 m
8 m
6 m
6 m
area of the rectangle:
12 m
A = bh
A = 12 • 6
A = 72 m2

129. Additional Example 1A Continued
Find the shaded area. Round to the nearest tenth, if necessary.
area of the trapezoid:
2 m
A = h(b1 + b2) 1 2 __
8 m
2 m
4 m
A = 2(4 + 2) 1 2 __
6 m
6 m
A = (12) 1 2 __
A = 6 m2
Add the area of the rectangle and the area of the trapezoid.
total area: A = = 78 m2.

130. Additional Example 1B: Finding the Area of Composite Figures by Adding
Find the shaded area. Round to the nearest tenth, if necessary.
Divide the figure into a rectangle and a semicircle.
12 in.
8 in.
8 in.
8 in.
area of the rectangle:
20 in.
A = bh
A = 20 • 8
A = 160 in2

131. Additional Example 1B Continued
Find the shaded area. Round to the nearest tenth, if necessary.
area of the semicircle:
12 in.
8 in.
A = (pr2) 1 2 __
8 in.
A  (3.14 • 42) 1 2 __
20 in.
A  (50.24) 1 2 __
A  25.1 in2
Add the area of the rectangle and the area of the semicircle.
total area: A = = in2.

132. Find the shaded area. Round to the nearest tenth, if necessary.
Check It Out! Example 1A
Find the shaded area. Round to the nearest tenth, if necessary.
10 yd
9 yd
8 yd
Divide the figure into a rectangle and a triangle.
area of the rectangle:
A = bh
A = 8 • 9
A = 72 yd2

133. Check It Out! Example 1A Continued
Find the shaded area. Round to the nearest tenth, if necessary.
area of the triangle:
A = bh 1 2 __
10 yd
9 yd
8 yd
A = (2 • 9) 1 2 __
A = (18) 1 2 __
A = 9 yd2
Add the area of the rectangle and the area of the triangle.
total area: A = = 81 yd2

134. Find the shaded area. Round to the nearest tenth, if necessary.
Check It Out! Example 1B
Find the shaded area. Round to the nearest tenth, if necessary.
Divide the figure into a rectangle and a semicircle.
12 m
10 m
10 m
area of the rectangle:
22 m
A = bh
A = 22 • 10
A = 220 m2

135. Check It Out! Example 1B Continued
Find the shaded area. Round to the nearest tenth, if necessary.
area of the semicircle:
A = (pr2) 1 2 __
12 m
10 m
A  (3.14 • 52) 1 2 __
10 m
A  (78.5) 1 2 __
22 m
A  39.3 m2
Add the area of the rectangle and the area of the semicircle.
total area: A = = m2

136. Subtract the area of the triangle from the area of the rectangle.
Additional Example 2: Finding the Area of Composite Figures by Subtracting
5 ft
Find the shaded area.
6 ft
9 ft
Subtract the area of the triangle from the area of the rectangle.
12 ft
Area of the rectangle:
Area of the triangle:
A = bh 1 2 __
A = bh
A = 12 • 9 = 108 ft2
A = • (6)(7) 1 2 __
A = • 42 = 21 ft2 1 2 __
Shaded area:
A = 108 – 21 = 87 ft2

137. Subtract the area of the triangle from the area of the rectangle.
Check It Out! Example 2
9 in.
Find the shaded area.
5 in.
8 in.
Subtract the area of the triangle from the area of the rectangle.
16 in.
Area of the rectangle:
Area of the triangle:
A = bh 1 2 __
A = bh
A = 16 • 8 = 128 in2
A = • (5)(7) 1 2 __
A = • 35 = 17.5 in2 1 2 __
Shaded area:
A = 128 – 17.5 = in2

138. Additional Example 3: Landscaping Application
6 ft
What is the area of the room floor shown in the figure? Round to the nearest tenth.
12 ft
12 ft
6 ft
18 ft
To find the area, divide the composite figure into a square, a rectangle, and a semicircle.

139. Additional Example 3 Continued
6 ft
Area of the square:
12 ft
12 ft
A = s2
6 ft
A = 62 = 36 ft2
18 ft
Area of the rectangle:
Area of the semicircle:
A = pr2 1 2 __
A = bh
A = 18 • 6 = 108 ft2
A  • 3.14 • (9)2 1 2 __
A  (254.34)
 ft2 1 2 __

140. Additional Example 3 Continued
6 ft
What is the area of the room floor shown in the figure? Round to the nearest tenth.
12 ft
12 ft
6 ft
18 ft
Area of the room:
A = = ft2
The area of the room is approximately ft2.

141. Check It Out! Example 3
5 ft
What is the area of the stage floor shown in the figure? Round to the nearest tenth.
10 ft
10 ft
5 ft
15 ft
To find the area, divide the composite figure into a square, a rectangle, and a semicircle.

142. Check It Out! Example 3 Continued
5 ft
Area of the square:
10 ft
10 ft
A = s2
5 ft
A = 52 = 25 ft2
15 ft
Area of the rectangle:
Area of the semicircle:
A = pr2 1 2 __
A = bh
A = 15 • 5 = 75 ft2
A  • 3.14 • (7.52) 1 2 __
A  (314)
 88.3 ft2 1 2 __

143. Check It Out! Example 3 Continued
5 ft
What is the area of the room floor shown in the figure? Round to the nearest tenth.
10 ft
10 ft
5 ft
15 ft
Area of the room:
A = = ft2
The area of the room is ft2.

144. Lesson Quiz: Part l
Find the perimeter and area of each figure. Round to the nearest tenth, if necessary. 1. 2. 3.
111 ft2
5 ft
45.9 in2
188.9 mm2

145. Lesson Quiz: Part ll
4. The figure gives the dimensions of a backdrop for a school play. Find the area of the backdrop.
97 ft2

146. Warm Up
Find the area of each figure. Round to the nearest tenth, if necessary.
1. a trapezoid with bases 10 cm and 12 cm and height 5 m
2. a parallelogram with base 9 in. and height 8 in.
3. a semicircle with radius 5 m
4. a semicircle with diameter 8 cm
55 cm2
72 in2
39.3 m2
25.1 cm2

147. One method of estimating the area of an irregular figure is to count the number of squares the figure covers.

148. Additional Example 1: Finding Area by Counting
Find the area of each figure. A.
Count the full squares: 10
Count the half-full squares: 4
Add the number of full squares plus half the number of half-full squares:
10 + ( • 4) = =12
1 2
The area of the figure is 12 square units.

149. For shapes such as the one in Example 1B you can only estimate the area.
Helpful Hint

150. Additional Example 1: Finding Area by Counting
Find the area of each figure. B.
Count the full and almost-full squares: 11
Count the half-full squares: 4
Add the number of full squares plus half the number of half-full squares:
11 + ( • 4) = =13
1 2
The area of the figure is approximately 13 square units.

151. Check It Out! Example 1
Find the area of each figure. A.
Count the full squares: 11
Count the half-full squares: 8
Add the number of full squares plus half the number of half-full squares:
11 + ( • 8) = = 15
1 2
The area of the figure is 15 square units.

152. Check It Out! Example 1
Find the area of each figure. B.
Count the full and almost-full squares: 11
Count the half-full squares: 6
Add the number of full squares plus half the number of half-full squares:
11 + ( • 6) = =14
1 2
The area of the figure is about 14 square units.

153. Additional Example 2A: Estimating Area Using Composite Figures
Use a composite figure to estimate the shaded area.
Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

154. Additional Example 2A Continued
Use a composite figure to estimate the shaded area.
area of the trapezoid:
A = h(b1 + b2)
1 2
= 1(4 + 2) = 3
1 2
area of the triangle:
A = bh 1 2 __
= (4 • 2) = 4 1 2 __
The shaded area is approximately 7 square units.

155. Additional Example 2B: Estimating Area Using Composite Figures
Use a composite figure to estimate the shaded area.
Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

156. Additional Example 2B Continued
Use a composite figure to estimate the shaded area.
area of the rectangle:
A = bh
= 1 • 4 = 4
area of the semicircle:
A = pr2 1 2 __
 • 3.14(22)
 6.28 1 2 __
The shaded area is approximately 10.3 square units.

157. Check It Out! Example 2A
Use a composite figure to estimate the shaded area.
Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

158. Check It Out! Example 2A Continued
Use a composite figure to estimate the shaded area.
area of the rectangle:
A = bh
= 3 • 2 = 6
area of the triangle:
A = bh 1 2 __
= (3 • 2) = 3 1 2 __
The shaded area is approximately 9 square units.

159. Check It Out! Example 2B
Use a composite figure to estimate the shaded area.
Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

160. Check It Out! Example 2B Continued
Use a composite figure to estimate the shaded area.
area of the square:
A = s2
= 2 • 2 = 4
area of the semicircle:
A = pr2 1 2 __
 • 3.14(12)
 1.57 1 2 __
The shaded area is approximately 5.5 square units.

161. Lesson Quiz: Part l
Find the area of each figure. 1. 2.
8 square units
approximately 12 square units

162. Lesson Quiz: Part ll
Use a composite figure to estimate the shaded area. 3. 4.
1 2
approximately 4 square units
approximately 6 square units

163. Competition Problems

164. The length of a rectangle is 4 inches longer than the width
The length of a rectangle is 4 inches longer than the width. If the perimeter is 28 inches, what is the width?

165. Answer: x + x + (x+4) + (x+4) = 28 4x + 8 = 28 4x = 20 x = 5

166. A regular pentagon has a side measure of 1 ft 4 in
A regular pentagon has a side measure of 1 ft 4 in. What is its perimeter?

167. 16in
16in
Answer: in + 16in + 16in + 16in + 16in = 5·16in = 80 inches = 6ft 8in
16in
1ft 4in = 12in + 4in = 16in
16in

168. Andrew and Ted built a snowman 98 inches tall by stacking three balls of snow. The largest ball had a diameter that was twice as long as the middle ball’s diameter. The smallest ball had a diameter that was half of the middle ball’s diameter. What was the circumference of the middle ball of snow? Use 3.14 for π.

169. Answer: 4d + 2d + d = 98 7d = 98 d = 14 = 87.92 inches d
Middle Circle diameter
2d = 2(14) = 28 inches
Circumference = π· d
= π (28)
= 3.14 (28)
= inches 2d
Answer: d + 2d + d = 98 7d = 98 d = 14
98” 4d

170. Which of the following points would be in Quadrant III. A. (1,3) B
Which of the following points would be in Quadrant III?   A.(1,3) B.(-1,4) C.(4,-3) D.(-4,-1) E.NOTA

171. Answer: D. (-4,-1)

172. If the perimeter of a square is 56 inches, what is the area of the square?  

173. Answer: P = 4S = 56 P = 56/4 P = 14 inches A = S·S = 14in·14in = 196 in2

174. Carter and Alex are building a circular garden
Carter and Alex are building a circular garden. A four-foot high fence will encircle the garden, except for a three-foot opening for a gate. If the radius of the garden is 15 feet, how much fencing will they need? Use 3.14 for pi.  

175. Answer: radius = 15 diameter = 30 circumference = πd = 3. 14(30) = 94
Answer: radius = 15 diameter = 30 circumference = πd = 3.14(30) = less 3’ opening for gate =94.2 – 3 =91.2ft

176. A rectangle has an area of 48 sq. cm. and a width of 3 cm
A rectangle has an area of 48 sq. cm. and a width of 3 cm. If its length is reduced by 5 cm., and its width is increased by 5 cm., what is the resulting area?  

177. Answer: A = L·W = 11(8) = 88 cm2 A = 48 W = 3cm A = L·W 48=L (3)
Reduced Length = 16-5 = 11cm
A = 48
W = 3cm
Increase Width = = 8cm
A = L·W
48=L (3)
16cm=L
A = L·W = 11(8) = 88 cm2