Solved Examples and Worksheet for Solving Problems on Area of a Circles

Q1What is the area of the circle with a circumference of 45 ft? [Use π = 3.14.]

A. 160.97 sq. ft.
B. 168.97 sq. ft.
C. 175.97 sq. ft.
D. 165.97 sq. ft.

Solutions
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Step: 1
Circumference of the circle, c = 45 ft.
  [Given.]
Step: 2
Circumference of a circle = 2 × π × r.
  [Formula.]
Step: 3
= 2 × 3.14 × r = 45
  [Substitute.]
Step: 4
⇒ r = 45(2×3.14)
  [Divide each side by (2 × 3.14).]
Step: 5
= 7.16 ft
  [Simplify.]
Step: 6
Area of the circle = πr2 = 3.14 × 7.16 × 7.16
  [Substitute r = 7.16.]
Step: 7
= 160.97 sq.ft
  [Simplify.]
Step: 8
The area of the circle is 160.97 sq. ft.
Correct Answer is :   160.97 sq. ft.
Q2The diameter of a small pizza is 8 inches. What is its area? [Use π = 3.14.]
A. 54.2 in.2
B. 52.2 in.2
C. 48.2 in.2
D. 50.2 in.2

Solutions
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Step: 1
The radius of the pizza = Diameter2 = 82
  [Substitute d = 8.]
Step: 2
= 4 inches
Step: 3
Area of a circle = π r2.
  [Formula.]
Step: 4
= π × 42
  [Substitute the values.]
Step: 5
= 3.14 × 16
  [Substitute π = 3.14.]
Step: 6
= 50.2 in.2
Correct Answer is :   50.2 in.2
Q3What is the area of a circular field of radius 1.6 miles? [Use π = 3.14.]
A. 10.03 miles2
B. 11.43 miles2
C. 6.03 miles2
D. 8.03 miles2

Solutions
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Step: 1
The area of a circle = π r2
Step: 2
π × (1.6)2
  
Step: 3
= 3.14 × 2.56
  [Substitute π = 3.14.]
Step: 4
= 8.03 miles2.
  
Correct Answer is :   8.03 miles2
Q4The circumference of the bottom of a circular swimming pool is 83 ft. What is the area of the bottom of the pool? [Use π = 3.14.]
A. 21631.46 ft
B. 547.94 ft3
C. 547.94 ft2
D. 5407.86 ft3

Solutions
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Step: 1
Circumference of a circle = 2πr
Step: 2
2 x 3.14 x r = 83
  [Substitute π = 3.14.]
Step: 3
r = 832x3.14
  [Divide each side by (2 x 3.14).]
Step: 4
= 13.21 ft.
  
Step: 5
Area of the bottom = πr2
Step: 6
= 3.14 x 13.21 x 13.21
  [Substitute r = 13.21.]
Step: 7
= 547.94 ft.2
  
Step: 8
The area of the bottom is about 547.94 ft2.
Correct Answer is :   547.94 ft2
Q5A goat is tied to a pole with a 3 m rope. The goat can graze to the full length of the rope and 360o around the pole. How much area does the goat have to graze? [Use π = 3.14.]

A. 9 m2
B. 1080 m2
C. 18.84 m2
D. 28.26 m2

Solutions
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Step: 1
Area of the circular region where the goat can graze = πr2
Step: 2
= 3.14 × 3 × 3
  [Substitute r = 3 and π = 3.14.]
Step: 3
= 28.26 m.2
  
Step: 4
The area the goat has to graze is 28.26 m2.
Correct Answer is :   28.26 m2
Q6A wire 40 cm long is bent to form a circle. Find the area of the circle formed.
A. 400π cm2
B. π400 cm2
C. 400π2
D. 400π cm2

Solutions
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Step: 1
Circumference of the circle = 2πr = 40 cm.
Step: 2
2πr = 40
  
Step: 3
πr = 402
  [Divide each side by 2.]
Step: 4
r = 20π
  [Divide each side by π.]
Step: 5
Area of the circle = πr2
Step: 6
= π × (20π)2
  [Substitute r = 50π.]
Step: 7
= 400π cm2.
Correct Answer is :   400π cm2
Q7A circular park of radius 39 m has a path of width 2 m along the inner circumference. What is the area of the path?

A. 1521π m2
B. 152π m2
C. 157π m2
D. 1369π m2

Solutions
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Step: 1
Radius of the outer circle = 39 m
Step: 2
Area of the outer circle = πr2
Step: 3
= π x 392
  [Substitute r = 39.]
Step: 4
= 1521π m2
  [Multiply.]
Step: 5
Radius of the inner circle = 39 - 2 = 37 m
Step: 6
Area of the inner circle = πr2
Step: 7
= π x 372
  [Substitute r = 37.]
Step: 8
= 1369π m2
  [Multiply.]
Step: 9
Area of the path = Area of the outer circle - Area of the inner circle
Step: 10
= 1521π - 1369π
  [Substitute the values.]
Step: 11
= 152π
  [Subtract.]
Step: 12
Area of the path is 152π m2.
Correct Answer is :   152π m2
Q8The diameter of a small pizza is 10 inches. What is its area? [Use π = 3.14.]
A. 78.5 in.2
B. 76.5 in.2
C. 82.5 in.2
D. 80.5 in.2

Solutions
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Step: 1
The radius of the pizza = Diameter2 = 102
  [Substitute d = 10.]
Step: 2
= 5 inches
  
Step: 3
Area of a circle = πr2.
Step: 4
= π × 52
  [Substitute the values.]
Step: 5
= 3.14 × 25
  [Substitute π = 3.14.]
Step: 6
= 78.5 in.2
  
Correct Answer is :   78.5 in.2
Q9What is the area of a circular field with a radius of 1.6 yd?
[Use π = 3.14.]

A. 10.04 yd2
B. 8.04 yd2
C. 11.44 yd2
D. 6.04 yd2

Solutions
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Step: 1
The area of a circle = π r2
Step: 2
= π × (1.6)2
  
Step: 3
= 3.14 × 2.56
  [Substitute π = 3.14.]
Step: 4
= 8.04 yd2
  
Correct Answer is :   8.04 yd2
Q10The circumference of a circular swimming pool at its bottom is 73 ft. What is the area of the bottom of the pool?[Use π = 3.14.]

A. About 423.98 ft3
B. About 423.98 ft2
C. About 423.98 ft
D. About 403.98 ft3

Solutions
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Step: 1
Circumference of a circle = 2πr
Step: 2
2 × 3.14 × r = 73
  [Substitute π = 3.14.]
Step: 3
r = 732 × 3.14
  [Divide each side by (2 × 3.14).]
Step: 4
= 11.62 ft
  
Step: 5
Area of the bottom = πr2
Step: 6
= 3.14 × 11.62 × 11.62
  [Substitute r = 11.62.]
Step: 7
≈ 423.98 ft2
  [Simplify.]
Step: 8
The area of the bottom is about 423.98 ft2.
Correct Answer is :   About 423.98 ft2
Q11A goat is tethered to a stake with a 3 m rope. The goat can graze to the full length of the rope and 360° around the stake. What is the area of the circle in which the goat can graze?
[Use π = 3.14.]

A. 30.26 m2
B. 23.26 m2
C. 28.26 m2
D. 33.26 m2

Solutions
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Step: 1
Area of the circular region where the goat can graze = πr2
Step: 2
= 3.14 × 3 × 3
  [Substitute r = 3 and π = 3.14.]
Step: 3
= 28.26 m2
  
Step: 4
The area of the circle the goat can graze is 28.26 m2.
Correct Answer is :   28.26 m2
Q12What is the area of a disk, if its diameter is 8 in.?
A. 50.24 in.2
B. 71.44 in.2
C. 52.44 in.
D. None of the above

Solutions
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Step: 1
Diameter of the disk = 8 in.
Step: 2
Radius of the disk = diameter2 = 82 = 4 in.
Step: 3
Area of the disk = πr2
  [Formula.]
Step: 4
= 3.14 × 42
  [Substitute the value of π and r.]
Step: 5
= 3.14 × 16
  
Step: 6
= 50.24
  [Simplify.]
Step: 7
So, area of the disk of diameter 8 in. is 50.24 in.2
Correct Answer is :   50.24 in.2
Q13The diameter of a nickel is 22 mm. Estimate the area of the nickel to the nearest millimeter. [Use π = 3.]
A. 368 mm2
B. 351 mm2
C. 375 mm2
D. 363 mm2

Solutions
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Step: 1
Diameter of the nickel = 22 mm
Step: 2
Radius of the nickel = diameter2 = 222 = 11 mm
Step: 3
Area of the nickel = πr2
  [Formula.]
Step: 4
= 3 × ( 11 )2
  [Substitute the values of π and r.]
Step: 5
= 3 × 121
  [Simplify.]
Step: 6
= 363
  
Step: 7
= 363 mm2
  [Round to the nearest millimeter.]
Correct Answer is :   363 mm2

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