Testinar Grade 7 math

Area of a Circle

Students will build Grade 7 mastery of area of a circle using models, equations, and test-style reasoning.

CH07-03 Measurement Geometry

Directions Solve each problem. Show your reasoning on the page or on separate paper. Check the answer key only after finishing.

A circle has a radius of \(4\) centimeters. What is its area? Use \(\pi\approx3.14\).
1. \(12.56\) cm\(^2\)
2. \(25.12\) cm\(^2\)
3. \(37.68\) cm\(^2\)
4. \(50.24\) cm\(^2\)
A circle has a diameter of \(10\) feet. What is its radius?
1. \(5\) feet
2. \(10\) feet
3. \(20\) feet
4. \(31.4\) feet
Which formula should you use to find the area of the circle above? Use \(\pi\approx3.14\).
1. \(A = 2\pi r\)
2. \(A = \pi r^2\)
3. \(A = \pi d\)
4. \(A = 4\pi r\)
A circle has a radius of \(3\) inches. What is its area? Use \(\pi\approx3.14\).
1. \(9.42\) in\(^2\)
2. \(18.84\) in\(^2\)
3. \(28.26\) in\(^2\)
4. \(37.68\) in\(^2\)
A circle has a diameter of \(8\) meters. What is its area? Use \(\pi\approx3.14\).
1. \(25.12\) m\(^2\)
2. \(50.24\) m\(^2\)
3. \(100.48\) m\(^2\)
4. \(200.96\) m\(^2\)
A circle has a radius of \(7\) centimeters. What is its area? Use \(\pi\approx\frac{22}{7}\).
1. \(44\) cm\(^2\)
2. \(88\) cm\(^2\)
3. \(154\) cm\(^2\)
4. \(308\) cm\(^2\)
A circular rug has a radius of \(2.5\) meters. What is the approximate area of the rug? Use \(\pi\approx3.14\).
1. \(7.85\) m\(^2\)
2. \(15.7\) m\(^2\)
3. \(19.625\) m\(^2\)
4. \(39.25\) m\(^2\)
A circular pool has a radius of \(9\) feet. What is the area of the pool's surface? Use \(\pi\approx3.14\).
1. \(56.52\) ft\(^2\)
2. \(113.04\) ft\(^2\)
3. \(254.34\) ft\(^2\)
4. \(1017.36\) ft\(^2\)

Teacher and parent reference

Answer key and explanations

D. \(50.24\) cm\(^2\)
\(A = \pi r^2 = 3.14 \times 4^2 = 3.14 \times 16 = 50.24\) cm\(^2\).
A. \(5\) feet
Radius is half the diameter: \(r = \frac{d}{2} = \frac{10}{2} = 5\) feet.
B. \(A = \pi r^2\)
The area of a circle uses the formula \(A=\pi r^2\). The formula \(2\pi r\) is for circumference.
C. \(28.26\) in\(^2\)
\(A = \pi r^2 = 3.14 \times 3^2 = 3.14 \times 9 = 28.26\) in\(^2\).
B. \(50.24\) m\(^2\)
First find \(r = \frac{d}{2} = 4\) m. Then \(A = 3.14 \times 4^2 = 3.14 \times 16 = 50.24\) m\(^2\).
C. \(154\) cm\(^2\)
\(A = \pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154\) cm\(^2\).
C. \(19.625\) m\(^2\)
\(A = 3.14 \times (2.5)^2 = 3.14 \times 6.25 = 19.625\) m\(^2\).
C. \(254.34\) ft\(^2\)
\(A = \pi r^2 = 3.14 \times 9^2 = 3.14 \times 81 = 254.34\) ft\(^2\).