## Cylinders - Sample Math Practice Problems

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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 =

### 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 =
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
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
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
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