Composite Areas Quiz: Master Composite Areas Quiz

1) A rectangle 10 m by 6 m has a semicircle (diameter 10 m) on top. What is the approximate total area?

80.4

85.7

90.2

99.3

Area = rectangle + semicircle = (10×6) + ½π(5²) = 60 + 39.3 ≈ 99.3 m².

Explanation

Area = rectangle + semicircle = (10×6) + ½π(5²) = 60 + 39.3 ≈ 99.3 m².

2) A playground is shaped like a half circle (diameter 12 m) attached to a rectangle 12 m by 20 m. What is the approximate total area?

296.5

353.7

413.7

500

Area = rectangle + semicircle = (12×20) + ½π(6²) = 240 + 56.5 ≈ 296.5

Explanation

Area = rectangle + semicircle = (12×20) + ½π(6²) = 240 + 56.5 ≈ 296.5

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4) A rectangle 14 m 5 has a triangle (base 14, height 3) attached on top. What is the total area?

90

91

96

98

Area = rectangle + triangle = (14×5) + ½(14×3) = 70 + 21 = 91 m².

Explanation

Area = rectangle + triangle = (14×5) + ½(14×3) = 70 + 21 = 91 m².

5) A right triangle with base 9 m and height 12 m is attached to a rectangle 9 m × 4 m. What is the total area?

80

90

120

132

Area = ½(9×12) + (9×4) = 54 + 36 = 90 m².

Explanation

Area = ½(9×12) + (9×4) = 54 + 36 = 90 m².

6) A rectangle 6 × 4 and a triangle (base 4, height 3) are attached. What is the total area?

28

30

34

36

Area = (6×4) + ½(4×3) = 24 + 6 = 30 m².

Explanation

Area = (6×4) + ½(4×3) = 24 + 6 = 30 m².

7) A semicircle’s area is always half that of a full circle with the same radius.

True

False

Semicircle = ½πr², exactly half of the full circle’s area.

Explanation

Semicircle = ½πr², exactly half of the full circle’s area.

8) A square 12 × 12 has a circle of radius 6 cut out. Approximate remaining area?

44

72.5

113.1

180

Area = 144 − π(6²) ≈ 144 − 113.1 = 30.9, correction: 113.1 is circle area.

Explanation

Area = 144 − π(6²) ≈ 144 − 113.1 = 30.9, correction: 113.1 is circle area.

9) A rectangle 20 × 10 with a semicircle (diameter 20) attached on one side has area __ m².

Area = (20×10) + ½π(10²) = 200 + 157.1 = 357.1 m².

Explanation

Area = (20×10) + ½π(10²) = 200 + 157.1 = 357.1 m².

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10) A running track is shaped like a 60 × 20 rectangle with two semicircles of radius 10 m attached. Approximate total area?

1205.4

1514.2

1828.5

1920

Area = rectangle + 2 semicircles = (60×20) + π(10²) = 1200 + 314.2 = 1514.2 m².

Explanation

Area = rectangle + 2 semicircles = (60×20) + π(10²) = 1200 + 314.2 = 1514.2 m².

11) A 15 × 6 rectangle has a 5 × 6 rectangle shaded inside. What is the unshaded area?

60

70

$80

90

Area = (15×6) − (5×6) = 90 − 30 = 60 m².

Explanation

Area = (15×6) − (5×6) = 90 − 30 = 60 m².

12) A square 8 × 8 has a 4 × 4 square shaded inside. What is the unshaded area?

44

48

52

60

Area = 64 − 16 = 48 cm².

Explanation

Area = 64 − 16 = 48 cm².

13) A rectangular window 15 × 10 cm has a circular pane of radius 5 cm cut out. Remaining area?

70

71.5

75

80

Area = (15×10) − π(5²) ≈ 150 − 78.5 = 71.5 cm².

Explanation

Area = (15×10) − π(5²) ≈ 150 − 78.5 = 71.5 cm².

14) A garden path is 12 × 3 with semicircles on both ends (diameter 3). Approximate total area?

43.1

46

56

74

Area = rectangle + 2 semicircles = 36 + π(1.5²) = 36 + 7.1 = 43.1, correction: full both sides 43.1 m².

Explanation

Area = rectangle + 2 semicircles = 36 + π(1.5²) = 36 + 7.1 = 43.1, correction: full both sides 43.1 m².

15) The total area of a composite figure equals the sum of its non-overlapping parts.

True

False

To find total area, add up all non-overlapping component areas.

Explanation

To find total area, add up all non-overlapping component areas.

16) A fountain base shaped like rectangle 12 × 8 with semicircle (diameter 12) attached. Total area?

120

140

152.5

160

Area = (12×8) + ½π(6²) = 96 + 56.5 = 152.5 m².

Explanation

Area = (12×8) + ½π(6²) = 96 + 56.5 = 152.5 m².

17) A trapezoid base 12 m, top 8 m, height 5 m, with semicircle (diameter 8 m) on top. Approx total area?

75.1

100.6

120.6

140

Area = trapezoid + semicircle = ½(12+8)×5 + ½π(4²) = 50 + 25.1 = 75.1 m².

Explanation

Area = trapezoid + semicircle = ½(12+8)×5 + ½π(4²) = 50 + 25.1 = 75.1 m².

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19) A regular garden path 30 m 2 with semicircle at each end (diameter 2). Total area?

60

63

70

73

Area = (30×2) + π(1²) = 60 + 3.14 = 63.1 m².

Explanation

Area = (30×2) + π(1²) = 60 + 3.14 = 63.1 m².

20) A square 14 × 14 has a quarter circle radius 7 removed. Remaining area?

130.1

145.5

160.8

176

Area = (14²) − ¼π(7²) = 196 − 38.5 = 157.5 ≈ 145.5 cm².

Explanation

Area = (14²) − ¼π(7²) = 196 − 38.5 = 157.5 ≈ 145.5 cm².