Volume of Cylinders, Cones and Spheres: math test solutions, Grade 8 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

a. \(\pi \times 36 \times 5 = 180\pi \text{ cm}^3\). (1 pt)

b. \(\dfrac{1}{3} \times 180\pi = 60\pi \text{ cm}^3\). (1 pt)

c. \(\dfrac{4}{3}\pi \times 12^3 = \dfrac{4}{3} \times 1{,}728\pi = 2{,}304\pi \text{ mm}^3\). (1 pt)

d. \(180\pi \div 60\pi = 3\). Three cones fill the cylinder. (1 pt)

a. The radius is 7 in. \(V = \pi \times 49 \times 9 = 441\pi \approx 1{,}384.74\). The volume is about \(1{,}384.7 \text{ in}^3\). (2 pts)

b. \(1{,}384.74 \div 231 \approx 5.99\). The barrel holds about 6.0 gallons. (1 pt)

\(\dfrac{1}{3}\pi r^2 \times 18 = 150\pi\). (1 pt)

\(6r^2 = 150\), so \(r^2 = 25\). (1 pt)

\(r = 5\). The radius is 5 cm. (1 pt)

Cylinder: \(\pi \times 25 \times 12 = 300\pi\). (1 pt)

Hemisphere: \(\dfrac{2}{3}\pi \times 125 = \dfrac{250\pi}{3}\). (1 pt)

Total: \(300\pi + \dfrac{250\pi}{3} = \dfrac{1{,}150\pi}{3}\). (1 pt)

With \(\pi \approx 3.14\): \(\dfrac{1{,}150 \times 3.14}{3} \approx 1{,}203.7\). The volume is about \(1{,}203.7 \text{ cm}^3\). (1 pt)

Ball: \(\dfrac{4}{3}\pi \times 27 = 36\pi \text{ cm}^3\). (1 pt)

Layer of water: \(\pi \times 16 \times h = 16\pi h\), and it equals the ball’s volume, so \(16\pi h = 36\pi\). (1 pt)

\(h = \dfrac{36}{16} = 2.25\). The water rises 2.25 cm. (1 pt)

a. True. \(V = \dfrac{1}{3}\pi r^2 h\) is proportional to \(h\). (1 pt)

b. False. \(V = \pi r^2 h\) depends on \(r^2\), so the volume is multiplied by 4. (1 pt)

c. True. A hemisphere is half a sphere, so \(\dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3\). (1 pt)

Test yourself: quick challenge for Grade 8