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Cylinders - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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Complexity=1

Find the surface area and volume. Answer with proper units. To represent m2, use sq m. To answer with π, use 'pi'. Sample area: 5pi sq m. Sample volume: 5pi cu m
1.   Surface Area =
Volume =
2.   Surface Area =
Volume =

Complexity=2

Find the surface area and volume. Answer with proper units. To represent m2, use sq m. To answer with π, use 'pi'. Sample area: 5pi sq m. Sample volume: 5pi cu m
1.   Surface Area =
Volume =
2.   Surface Area =
Volume =

Answers


Complexity=1

Find the surface area and volume. Answer with proper units. To represent m2, use sq m. To answer with π, use 'pi'. Sample area: 5pi sq m. Sample volume: 5pi cu m
#ProblemCorrect AnswerYour Answer
1 Surface Area =
Volume =
Solution
The surface area of a cylinder is the area of the bottom and top circles and the area of the side of the cylinder. This can be expressed as follows:
SA = 2πrh + 2πr2
SA = 2π × (4in) × (2 in) + 2π × (4 in)2
SA = 2π × (8 in2) + 2π × (16 in2)
SA = 16π in2 + 32π in2
SA = 48π in2

The volume of a cylinder is the area of the base circle times the height of the cylinder.
V = bh = πr2h
V = π × (4 in)2 × (2 in)
V = π × (16 in2) × (2 in)
V = π × (32 in3)
V = 32π in3
#ProblemCorrect AnswerYour Answer
2 Surface Area =
Volume =
Solution
The surface area of a cylinder is the area of the bottom and top circles and the area of the side of the cylinder. This can be expressed as follows:
SA = 2πrh + 2πr2
SA = 2π × (2yd) × (1 yd) + 2π × (2 yd)2
SA = 2π × (2 yd2) + 2π × (4 yd2)
SA = 4π yd2 + 8π yd2
SA = 12π yd2

The volume of a cylinder is the area of the base circle times the height of the cylinder.
V = bh = πr2h
V = π × (2 yd)2 × (1 yd)
V = π × (4 yd2) × (1 yd)
V = π × (4 yd3)
V = 4π yd3

Complexity=2

Find the surface area and volume. Answer with proper units. To represent m2, use sq m. To answer with π, use 'pi'. Sample area: 5pi sq m. Sample volume: 5pi cu m
#ProblemCorrect AnswerYour Answer
1 Surface Area =
Volume =
Solution
The surface area of a cylinder is the area of the bottom and top circles and the area of the side of the cylinder. This can be expressed as follows:
SA = 2πrh + 2πr2
SA = 2π × (8in) × (6 in) + 2π × (8 in)2
SA = 2π × (48 in2) + 2π × (64 in2)
SA = 96π in2 + 128π in2
SA = 224π in2

The volume of a cylinder is the area of the base circle times the height of the cylinder.
V = bh = πr2h
V = π × (8 in)2 × (6 in)
V = π × (64 in2) × (6 in)
V = π × (384 in3)
V = 384π in3
#ProblemCorrect AnswerYour Answer
2 Surface Area =
Volume =
Solution
The surface area of a cylinder is the area of the bottom and top circles and the area of the side of the cylinder. This can be expressed as follows:
SA = 2πrh + 2πr2
SA = 2π × (5cm) × (1 cm) + 2π × (5 cm)2
SA = 2π × (5 cm2) + 2π × (25 cm2)
SA = 10π cm2 + 50π cm2
SA = 60π cm2

The volume of a cylinder is the area of the base circle times the height of the cylinder.
V = bh = πr2h
V = π × (5 cm)2 × (1 cm)
V = π × (25 cm2) × (1 cm)
V = π × (25 cm3)
V = 25π cm3
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