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These sample problems below for Circle Area were generated by the MathScore.com engine.

Sample Problems For Circle Area


Complexity=4, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
1.   Area:
2.   Area:

Complexity=6, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
1.   Area:
2.   Area:

Complexity=8, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
1.   Area:
2.   Area:

Complexity=4, Mode=fraction

Find the area. Use 22/7 as an approximation for π. A correct answer would look like 14.5 sq cm.
1.   Area:
2.   Area:

Complexity=4

Find the area. Use 3.14 as an approximation for π. A correct answer would look like 4.785 sq in
1.   Area:
2.   Area:

Answers


Complexity=4, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
#ProblemCorrect AnswerYour Answer
1 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 4 m, we divide it by 2 to get the radius which then is 2 m.
Plugging values into the equation, we have:
A = π(2 m)2
A = &pi(4 m2)
A = 4π m2
#ProblemCorrect AnswerYour Answer
2 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 3 mm, we divide it by 2 to get the radius which then is 1.5 mm.
Plugging values into the equation, we have:
A = π(1.5 mm)2
A = &pi(2.25 mm2)
A = 2.25π mm2

Complexity=6, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
#ProblemCorrect AnswerYour Answer
1 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 4 km, we divide it by 2 to get the radius which then is 2 km.
Plugging values into the equation, we have:
A = π(2 km)2
A = &pi(4 km2)
A = 4π km2
#ProblemCorrect AnswerYour 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(1.5 m)2
A = &pi(2.25 m2)
A = 2.25π m2

Complexity=8, Mode=exact

Find the area in terms of π. Type "pi" in for π so that a correct answer would look like 7pi sq m
#ProblemCorrect AnswerYour Answer
1 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 1 cm, we divide it by 2 to get the radius which then is 0.5 cm.
Plugging values into the equation, we have:
A = π(0.5 cm)2
A = &pi(0.25 cm2)
A = 0.25π cm2
#ProblemCorrect AnswerYour Answer
2 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 2 cm, we divide it by 2 to get the radius which then is 1 cm.
Plugging values into the equation, we have:
A = π(1 cm)2
A = &pi(1 cm2)
A = 1π cm2

Complexity=4, Mode=fraction

Find the area. Use 22/7 as an approximation for π. A correct answer would look like 14.5 sq cm.
#ProblemCorrect AnswerYour Answer
1 Area:
Solution
Area = π × (radius)2
A = πr2
Since we have that the diameter is 21 mi, we divide it by 2 to get the radius which then is 10.5 mi.
Plugging values into the equation, we have:
A = π(10.5 mi)2
A = (22/7) × (110.25 mi2)
346.5 mi2
#ProblemCorrect AnswerYour Answer
2 Area:
Solution
Area = π × (radius)2
A = πr2
Plugging values into the equation, we have:
A = π(14 in)2
A = (22/7) × (196 in2)
616 in2

Complexity=4

Find the area. Use 3.14 as an approximation for π. A correct answer would look like 4.785 sq in
#ProblemCorrect AnswerYour Answer
1 Area:
Solution
Area = π × (radius)2
A = πr2
Plugging values into the equation, we have:
A = π(4 m)2
A = 3.14 × (16 m2)
50.24 m2
#ProblemCorrect AnswerYour Answer
2 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 = 3.14 × (1 yd2)
3.14 yd2

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