This grade 8 volume worksheet answer key solves all 23 problems of the volume of cylinders, cones and spheres worksheet step by step.
Each solution below shows the formula, the substitution, and the final answer with cubic units. Compare your work step by step, and look especially at whether you halved a diameter before squaring or cubing the radius.
Grade 8 Volume Worksheet Answer Key: problems 1–5
Problem 1: Cylinder with radius and height
Use \(V=\pi r^{2}h\).
Substitute: \(V=\pi(3)^{2}(10)\).
Square first: \(V=\pi\cdot 9\cdot 10=90\pi\).
\(90\pi\text{ cm}^{3}\)
Problem 2: Cylinder using 3.14
Use \(V=\pi r^{2}h\).
Substitute: \(V=\pi(5)^{2}(4)\).
Square first: \(V=\pi\cdot 25\cdot 4=100\pi\).
Use 3.14: \(V\approx 3.14\cdot 25\cdot 4=314.00\).
\(V\approx 314.00\text{ in}^{3}\)
Problem 3: Cylinder from the diameter
The radius is half the diameter: \(r=12: 2=6\text{ ft}\).
Use \(V=\pi r^{2}h\).
Substitute: \(V=\pi(6)^{2}(7)\).
Square first: \(V=\pi\cdot 36\cdot 7=252\pi\).
\(252\pi\text{ ft}^{3}\)
Problem 4: Cone with radius and height
Use \(V=\frac{1}{3}\pi r^{2}h\).
Substitute: \(V=\frac{1}{3}\pi(3)^{2}(7)\).
Multiply, then divide by 3: \(V=\frac{1}{3}\pi\cdot 63=21\pi\).
\(21\pi\text{ in}^{3}\)
Problem 5: Cone using 3.14
Use \(V=\frac{1}{3}\pi r^{2}h\).
Substitute: \(V=\frac{1}{3}\pi(6)^{2}(9)\).
Multiply, then divide by 3: \(V=\frac{1}{3}\pi\cdot 324=108\pi\).
Use 3.14: \(V\approx 3.14\cdot 108=339.12\).
\(V\approx 339.12\text{ cm}^{3}\)
Grade 8 Volume Worksheet Answer Key: problems 6–10
Problem 6: Cone from the diameter
The radius is half the diameter: \(r=10: 2=5\text{ m}\).
Use \(V=\frac{1}{3}\pi r^{2}h\).
Substitute: \(V=\frac{1}{3}\pi(5)^{2}(12)\).
Multiply, then divide by 3: \(V=\frac{1}{3}\pi\cdot 300=100\pi\).
\(100\pi\text{ m}^{3}\)
Problem 7: Sphere with a given radius
Use \(V=\frac{4}{3}\pi r^{3}\).
Substitute: \(V=\frac{4}{3}\pi(3)^{3}\).
Cube first: \(V=\frac{4}{3}\pi\cdot 27=36\pi\).
\(36\pi\text{ cm}^{3}\)
Problem 8: Sphere using 3.14
Use \(V=\frac{4}{3}\pi r^{3}\).
Substitute: \(V=\frac{4}{3}\pi(6)^{3}\).
Cube first: \(V=\frac{4}{3}\pi\cdot 216=288\pi\).
Use 3.14: \(V\approx\frac{4}{3}(3.14)(216)=904.32\).
\(V\approx 904.32\text{ in}^{3}\)
Problem 9: Sphere from the diameter
The radius is half the diameter: \(r=10: 2=5\text{ ft}\).
Use \(V=\frac{4}{3}\pi r^{3}\).
Substitute: \(V=\frac{4}{3}\pi(5)^{3}\).
Cube first: \(V=\frac{4}{3}\pi\cdot 125=\frac{500}{3}\pi\).
\(\frac{500}{3}\pi\text{ ft}^{3}\)
Use 3.14: \(V\approx\frac{4}{3}(3.14)(125)=523.33\).
\(V\approx 523.33\text{ ft}^{3}\)
Problem 10: Find the height of a cylinder
Use \(V=\pi r^{2}h\) and substitute: \(150\pi=\pi(5)^{2}h\), so \(150\pi=25\pi h\).
Divide both sides by \(25\pi\): \(h=150: 25=6\).
\(h=6\text{ cm}\)
Check: \(\pi(25)(6)=150\pi\).
Grade 8 Volume Worksheet Answer Key: problems 11–15
Problem 11: Find the radius of a cylinder
Substitute into \(V=\pi r^{2}h\): \(245\pi=\pi r^{2}(5)\).
Divide both sides by \(5\pi\): \(r^{2}=49\).
Take the positive square root: \(r=7\text{ in}\).
Check: \(\pi(49)(5)=245\pi\).
Problem 12: Find the height of a cone
Substitute into \(V=\frac{1}{3}\pi r^{2}h\): \(96\pi=\frac{1}{3}\pi(6)^{2}h=12\pi h\).
Divide both sides by \(12\pi\): \(h=96: 12=8\).
\(h=8\text{ ft}\)
Check: \(\frac{1}{3}\pi(36)(8)=96\pi\).
Problem 13: Find the radius of a sphere
Substitute into \(V=\frac{4}{3}\pi r^{3}\): \(288\pi=\frac{4}{3}\pi r^{3}\).
Multiply both sides by \(\frac{3}{4}\) and divide by \(\pi\): \(r^{3}=288\cdot\frac{3}{4}=216\).
Since \(6\cdot 6\cdot 6=216\), \(r=6\text{ cm}\).
Problem 14: Cylinder versus cone
Cylinder: \(V=\pi(4)^{2}(9)=\pi\cdot 16\cdot 9=144\pi\text{ cm}^{3}\).
Cone: \(V=\frac{1}{3}\pi(4)^{2}(9)=\frac{1}{3}\pi\cdot 144=48\pi\text{ cm}^{3}\).
Cylinder: \(144\pi\text{ cm}^{3}\); cone: \(48\pi\text{ cm}^{3}\). Since \(144: 48=3\), the cylinder holds 3 times as much as the cone.
Problem 15: Volume of a hemisphere
First find the volume of the whole sphere: \(V=\frac{4}{3}\pi(9)^{3}=\frac{4}{3}\pi\cdot 729=972\pi\).
A hemisphere is half of that: \(972\pi: 2=486\pi\).
\(486\pi\text{ m}^{3}\)
Grade 8 Volume Worksheet Answer Key: problems 16–20
Problem 16: Soup can
The radius is half the diameter: \(r=3: 2=1.5\text{ in}\).
Use \(V=\pi r^{2}h\).
Substitute: \(V=\pi(1.5)^{2}(4)\).
Square first: \(V=\pi\cdot 2.25\cdot 4=9\pi\).
Use 3.14: \(V\approx 3.14\cdot 2.25\cdot 4=28.26\).
\(V\approx 28.26\text{ in}^{3}\)
The can holds about 28.26 cubic inches of soup.
Problem 17: Water storage tank
(a) \(V=\pi(4)^{2}(10)=160\pi\approx 3.14\cdot 160=502.4\). The tank holds about \(502.4\text{ ft}^{3}\).
(b) Time \(=502.4: 40=12.56\), which rounds to 12.6 minutes.
Problem 18: Pile of sand
(a) \(V=\frac{1}{3}\pi(6)^{2}(5)=\frac{1}{3}\pi\cdot 180=60\pi\text{ ft}^{3}\).
(b) \(60\cdot 3.14=188.4\), so \(V\approx 188.4\text{ ft}^{3}\).
(c) \(188.4: 20=9.42\). Nine loads are not enough, so the truck needs 10 loads.
Problem 19: Ice cream cone with a scoop
Cone: \(\frac{1}{3}\pi(1)^{2}(4)=\frac{4}{3}\pi\).
Hemisphere: \(\frac{1}{2}\cdot\frac{4}{3}\pi(1)^{3}=\frac{2}{3}\pi\).
Total: \(\frac{4}{3}\pi+\frac{2}{3}\pi=\frac{6}{3}\pi=2\pi\). \(2\pi\text{ in}^{3}\).
With 3.14: \(2\cdot 3.14=6.28\), so about \(6.28\text{ in}^{3}\).
Problem 20: Basketball
The radius is half the diameter: \(r=9: 2=4.5\text{ in}\).
Use \(V=\frac{4}{3}\pi r^{3}\).
Substitute: \(V=\frac{4}{3}\pi(4.5)^{3}\).
Cube first: \(V=\frac{4}{3}\pi\cdot 91.125=\frac{243}{2}\pi\).
Use 3.14: \(V\approx\frac{4}{3}(3.14)(91.125)=381.51\).
\(V\approx 381.51\text{ in}^{3}\)
The basketball holds about 381.51 cubic inches of air.
Grade 8 Volume Worksheet Answer Key: problems 21–23
Problem 21: Tennis ball can
Can: \(\pi(1.5)^{2}(9)=\pi\cdot 2.25\cdot 9=20.25\pi\).
One ball: \(\frac{4}{3}\pi(1.5)^{3}=\frac{4}{3}\pi\cdot 3.375=4.5\pi\). Three balls: \(13.5\pi\).
Empty space: \(20.25\pi-13.5\pi=6.75\pi\text{ in}^{3}\).
With 3.14: \(6.75\cdot 3.14=21.195\), so about \(21.2\text{ in}^{3}\) of empty space.
Problem 22: Filling paper cups
Jug: \(\pi(3)^{2}(8)=72\pi\text{ in}^{3}\).
Cup: \(\frac{1}{3}\pi(2)^{2}(3)=4\pi\text{ in}^{3}\).
Number of cups: \(72\pi: 4\pi=18\). 18 cups.
Problem 23: Reshaping a ball of clay
Ball: \(\frac{4}{3}\pi(3)^{3}=36\pi\text{ cm}^{3}\).
Cylinder with the same volume: \(\pi(3)^{2}h=9\pi h=36\pi\), so \(h=36: 9=4\).
The cylinder is 4 cm tall.
How to use this grade 8 volume worksheet answer key
Read this grade 8 volume worksheet answer key after trying each problem yourself, and redo any problem you missed.
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Grade 8 Volume of Cylinders, Cones and Spheres worksheet answer key
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