Circle Area - Sample Math Practice Problems

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Complexity=4, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

1.   Area:

2.   Area:

Complexity=6, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

1.   Area:

2.   Area:

Complexity=8, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

1.   Area:

2.   Area:

Complexity=4, Mode=fraction

Find the area. Use 22/7 as an approximation for π, but give the final answer in decimal.
A typed answer would look something like this: 14.5 sq cm.

1.   Area:

2.   Area:

Complexity=4

Find the area. Use 3.14 as an approximation for π.
A typed answer would look something like this: 14.5 sq cm.

1.   Area:

2.   Area:

Answers

Complexity=4, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

#	Problem	Correct Answer	Your Answer
1		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 2 yd, we divide it by 2 to get the radius which then is 1 yd. Plugging values into the equation, we have: A = π(1 yd))2 A = π(1 yd2) A = 1π yd2

#	Problem	Correct Answer	Your Answer
2		Area:
Solution Area = π × (radius)2 A = πr2 Plugging values into the equation, we have: A = π(1 m))2 A = π(1 m2) A = 1π m2

Complexity=6, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

#	Problem	Correct Answer	Your Answer
1		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 2 ft, we divide it by 2 to get the radius which then is 1 ft. Plugging values into the equation, we have: A = π(1 ft))2 A = π(1 ft2) A = 1π ft2

#	Problem	Correct Answer	Your Answer
2		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 1 km, we divide it by 2 to get the radius which then is 0.5 km. Plugging values into the equation, we have: A = π(1 / 2 km))2 A = π(0.25 km2) A = 0.25π km2

Complexity=8, Mode=exact

Find the area in terms of π. Figures are not drawn to scale.
Type "pi" in for π. Example: "7π m²" as "7pi sq m".

#	Problem	Correct Answer	Your Answer
1		Area:
Solution Area = π × (radius)2 A = πr2 Plugging values into the equation, we have: A = π(5 km))2 A = π(25 km2) A = 25π km2

#	Problem	Correct Answer	Your Answer
2		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 3 m, we divide it by 2 to get the radius which then is 1.5 m. Plugging values into the equation, we have: A = π(3 / 2 m))2 A = π(2.25 m2) A = 2.25π m2

Complexity=4, Mode=fraction

Find the area. Use 22/7 as an approximation for π, but give the final answer in decimal.
A typed answer would look something like this: 14.5 sq cm.

#	Problem	Correct Answer	Your Answer
1		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 7 m, we divide it by 2 to get the radius which then is 3.5 m. Plugging values into the equation, we have: A = π(7 / 2 m))2 A ≈ (22/7) × (49 / 4 m2) A ≈ 38.5 m2

#	Problem	Correct Answer	Your Answer
2		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 7 m, we divide it by 2 to get the radius which then is 3.5 m. Plugging values into the equation, we have: A = π(7 / 2 m))2 A ≈ (22/7) × (49 / 4 m2) A ≈ 38.5 m2

Complexity=4

Find the area. Use 3.14 as an approximation for π.
A typed answer would look something like this: 14.5 sq cm.

#	Problem	Correct Answer	Your Answer
1		Area:
Solution Area = π × (radius)2 A = πr2 Since we have that the diameter is 1 in, we divide it by 2 to get the radius which then is 0.5 in. Plugging values into the equation, we have: A = π(1 / 2 in))2 A ≈ 3.14 × (0.25 in2) A ≈ 0.785 in2

#	Problem	Correct Answer	Your Answer
2		Area:
Solution Area = π × (radius)2 A = πr2 Plugging values into the equation, we have: A = π(4 yd))2 A ≈ 3.14 × (16 yd2) A ≈ 50.24 yd2