← Back to the Circumference And Area Of Circles worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: 7.G.B.4 · Level: Grade 7
Problem 2 Diameter from radius
Diameter = 2 × radius = 2 × 2.5.
d = 5 yd
Problem 3 Circumference from diameter
C = πd = 3.14 × 25 = 78.5.
C = 78.5 in
Problem 4 Circumference in terms of π
C = 2πr = 2 × π × 6 = 12π.
C = 12π m
Problem 5 Area from radius
A = πr² = 3.14 × 9² = 3.14 × 81 = 254.34.
A = 254.34 cm²
Problem 6 Area in terms of π
A = πr² = π × 11² = 121π.
A = 121π in²
Problem 7 Area from the diameter
Radius = 30 ÷ 2 = 15 in. A = π × 15² = 225π.
A = 225π in²
Problem 8 Work backward from circumference
d = C ÷ π = 75.36 ÷ 3.14 = 24 ft.
r = 24 ÷ 2 = 12 ft.
d = 24 ft and r = 12 ft
Problem 9 Bicycle wheel
One turn covers the circumference: C = 3.14 × 26 = 81.64.
81.64 in
Problem 10 Garden from the fence
d = 37.68 ÷ 3.14 = 12 ft, so r = 6 ft.
A = 3.14 × 6² = 3.14 × 36 = 113.04.
A = 113.04 ft²
Problem 11 Area of a semicircle
r = 16 ÷ 2 = 8 cm. Full circle: 3.14 × 8² = 3.14 × 64 = 200.96.
Half: 200.96 ÷ 2 = 100.48.
A = 100.48 cm²
Problem 12 Area of a quarter circle
Full circle: π × 12² = 144π. One fourth: 144π ÷ 4 = 36π.
36 × 3.14 = 113.04.
A = 36π ft² ≈ 113.04 ft²
Problem 13 Rearranging a circle
Base = half the circumference = (2π × 5) ÷ 2 = 5π in. Height = radius = 5 in.
Area = 5π × 5 = 25π.
base 5π in, height 5 in, area 25π in²
Problem 14 Effect of doubling
Circumferences: 2π × 4 = 8π and 2π × 8 = 16π, so the larger one is 2 times as long.
Areas: π × 4² = 16π and π × 8² = 64π, so the larger one is 4 times as big.
Circumference doubles (8π to 16π); area quadruples (16π to 64π).
Problem 15 Pizza value
12-in: r = 6, A = 3.14 × 36 = 113.04 in². $9 ÷ 113.04 ≈ $0.080.
16-in: r = 8, A = 3.14 × 64 = 200.96 in². $14 ÷ 200.96 ≈ $0.070.
The 16-inch pizza costs less per square inch (about $0.070 compared with $0.080).
Problem 16 Field with rounded ends
The two semicircles make one circle with d = 24 ft, r = 12 ft.
Perimeter: 2 × 40 + 3.14 × 24 = 80 + 75.36 = 155.36 ft.
Area: 40 × 24 + 3.14 × 12² = 960 + 452.16 = 1,412.16 ft².
Perimeter 155.36 ft; area 1,412.16 ft²
Problem 17 Sprinkler coverage
Lawn: 30 × 30 = 900 ft². Watered: 3.14 × 15² = 3.14 × 225 = 706.5 ft².
900 − 706.5 = 193.5.
193.5 ft² are not watered
Problem 18 Arched window
Semicircle: r = 18 in. Area = 3.14 × 18² ÷ 2 = 3.14 × 324 ÷ 2 = 508.68 in².
Rectangle: 36 × 48 = 1,728 in². Glass: 1,728 + 508.68 = 2,236.68 in².
Trim: 48 + 48 + 36 + (3.14 × 36 ÷ 2) = 132 + 56.52 = 188.52 in.
Area 2,236.68 in²; trim 188.52 in
Problem 19 Path around a pond
Outer radius: 9 + 3 = 12 ft. Outer area: π × 144 = 144π.
Pond: π × 81 = 81π. Path: 144π − 81π = 63π.
63 × 3.14 = 197.82.
63π ft² ≈ 197.82 ft²
Problem 20 Odometer wheel
One turn: 3.14 × 28 = 87.92 in. 500 turns: 87.92 × 500 = 43,960 in.
43,960 ÷ 12 ≈ 3,663.3 ft.
About 3,663 ft