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Common Core: 8th Grade Math : Know and Use the Formulas for the Volumes of Cones, Cylinders, and Spheres: CCSS.Math.Content.8.G.C.9

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Example Questions

Example Question #1 : How To Find The Volume Of A Cone

Chestnut wood has a density of about $0.45\frac{g}{cm^3}$ . A right circular cone made out of chestnut wood has a height of three meters, and a base with a radius of two meters. What is its mass in kilograms (nearest whole kilogram)?

Possible Answers:

$5,755 \; \textrm{kg}$

$5,855 \; \textrm{kg}$

$6,055 \; \textrm{kg}$

$5,955 \; \textrm{kg}$

$5,655 \; \textrm{kg}$

Correct answer:

$5,655 \; \textrm{kg}$

Explanation:

First, convert the dimensions to cubic centimeters by multiplying by 100 : the cone has height 300cm , and its base has radius 200cm .

$3m*\frac{100cm}{1m}=300cm\ \ \2m*\frac{100cm}{1m}=200cm$

Its volume is found by using the formula and the converted height and radius.

$V=\frac{1}{3} \pi r^{2}h$

$V=\frac{1}{3} \pi (200cm)^{2} ( 300cm )\approx 12,566,371 \; \textrm{cm}^{3}$

Now multiply this by $0.45\frac{g}{cm^3}$ to get the mass.

$m=V*\rho$

$12,566,371cm^3*0.45\frac{g}{cm^3} \approx 5,654,867g$

Finally, convert the answer to kilograms.

$5654867g *\frac{1kg}{1,000g} \approx 5,655kg$

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Example Question #1 : How To Find The Volume Of A Cone

A cone has the height of 4 meters and the circular base area of 4 square meters. If we want to fill out the cone with water (density = $1000\frac{Kg}{m^3}$ ), what is the mass of required water (nearest whole kilogram)?

Possible Answers:

10,000

5333

6333

4333

2133

Correct answer:

5333

Explanation:

The volume of a cone is:

$Volume=\frac{1}{3}\pi r^2h$

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

As the circular base area is $\pi r^2$ , so we can rewrite the volume formula as follows:

$Volume =\frac{A\times h}{3}$

where is the circular base area and known in this problem. So we can write:

$Volume =\frac{A\times h}{3}=\frac{4\times 4}{3}=\frac{16}{3}\approx 5.333 m^3$

We know that density is defined as mass per unit volume or:

$\rho =\frac{m}{v}$

Where $\rho$ is the density; is the mass and is the volume. So we get:

$m=\rho v=1000\times 5.333=5333 Kg$

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Example Question #2 : How To Find The Volume Of A Cone

The vertical height (or altitude) of a right cone is . The radius of the circular base of the cone is . Find the volume of the cone in terms of .

Possible Answers:

$\frac{1}{3}\pi t^3$

$\pi t^3$

$\frac{2}{3}\pi t^3$

$\frac{2}{3}\pi t^2$

$\frac{1}{3}\pi t^2$

Correct answer:

$\frac{2}{3}\pi t^3$

Explanation:

The volume of a cone is:

$Volume=\frac{1}3{}\pi r^2h$

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

$Volume=\frac{1}3{}\pi r^2h =\frac{1}3{}\pi t^2\times 2t=\frac{2}{3}\pi t^3$

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Example Question #3 : How To Find The Volume Of A Cone

A right cone has a volume of $8\pi$ , a height of and a radius of the circular base of . Find .

Possible Answers:

$\sqrt[3]{2}$

$\sqrt{2}$

$2\sqrt[3]{2}$

$\sqrt[3]{3}$

Correct answer:

$\sqrt[3]{2}$

Explanation:

The volume of a cone is given by:

$Volume=\frac{1}{3}\pi r^2h$

where is the radius of the circular base, and is the height; the perpendicular distance from the base to the vertex. Substitute the known values in the formula:

$Volume=\frac{1}{3}\pi r^2h\Rightarrow 8\pi=\frac{1}{3}\pi\times (2t)^2\times 3t$

$\Rightarrow 8\pi=4\pi t^3\Rightarrow t^3=2\Rightarrow t=\sqrt[3]{2}$

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Example Question #2 : How To Find The Volume Of A Cone

A cone has a diameter of $12\:m$ and a height of $4\:m$ . In cubic meters, what is the volume of this cone?

Possible Answers:

$54\pi\:m^3$

$36\pi\:m^3$

$48\pi\:m^3$

$144\pi\:m^3$

Correct answer:

$48\pi\:m^3$

Explanation:

First, divide the diameter in half to find the radius.

$\frac{d}{2}=r$

$\frac{12}{2}=6\:m$

Now, use the formula to find the volume of the cone.

$\text{Volume}=\frac{1}{3}\pi r^2 h$

$\text{Volume}=\frac{1}{3}\pi \times 6^2\times4=\frac{1}{3}\pi \times 36\times4=\frac{144}{3}\pi=48\pi\:m^3$

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Example Question #143 : Geometry

The height of a cylinder is 3 inches and the radius of the circular end of the cylinder is 3 inches. Give the volume and surface area of the cylinder.

Possible Answers:

V=84.78 in^3, A=123.04in^2

V=74.78 in^3, A=103.04in^2

V=74.78 in^3, A=113.10in^2

V=84.78 in^3, A=103.04in^2

V=84.82 in^3, A=113.10in^2

Correct answer:

V=84.82 in^3, A=113.10in^2

Explanation:

The volume of a cylinder is found by multiplying the area of one end of the cylinder (base) by its height or:

$Volume=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder. So we can write:

$Volume=\pi r^2h=\pi \times 3^2\times 3=84.82in^3$

The surface area of the cylinder is given by:

$A=2\pi r^2+2\pi rh$

where is the surface area of the cylinder, is the radius of the cylinder and is the height of the cylinder. So we can write:

$A=2\pi r^2+2\pi rh$

$A=2\pi (3)^2+2\pi \times 3\times 3$

$A=18\pi+18\pi$

$A=36\pi$

A=113.10

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Example Question #1 : How To Find The Volume Of A Cylinder

The height of a cylinder is two times the length of the radius of the circular end of a cylinder. If the volume of the cylinder is $16\pi$ , what is the height of the cylinder?

Possible Answers:

$2\sqrt{2}$

Correct answer:

Explanation:

The volume of a cylinder is:

$Volume=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder.

Since h=2r , we can substitute that into the volume formula. So we can write:

$Volume=\pi r^2h$

$16\pi=\pi r^2\cdot (2r)$

$16\pi =2\pi r^{3}$

$16=2r^{3}$

$8=r^{3}$

r=2

So we get:

$h=2r=2\times 2=4$

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Example Question #144 : Geometry

The end (base) of a cylinder has an area of $16\pi$ square inches. If the height of the cylinder is half of the radius of the base of the cylinder, give the volume of the cylinder.

Possible Answers:

$16\pi$

$32\pi$

$8\pi$

Correct answer:

$32\pi$

Explanation:

The area of the end (base) of a cylinder is $\pi r^2$ , so we can write:

$\pi r^2=16\pi\Rightarrow r^2=16\Rightarrow r=4\ inches$

The height of the cylinder is half of the radius of the base of the cylinder, that means:

$h=\frac{r}{2}=\frac{4}{2}=2\ inches$

The volume of a cylinder is found by multiplying the area of one end of the cylinder (base) by its height:

$Volume=Area\times h=16\pi\times 2=32\pi$

$Volume=\pi r^2h=\pi\times 4^2\times 2=32\pi$

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Example Question #145 : Geometry

We have two right cylinders. The radius of the base Cylinder 1 is $\sqrt{3}$ times more than that of Cylinder 2, and the height of Cylinder 2 is 4 times more than the height of Cylinder 1. The volume of Cylinder 1 is what fraction of the volume of Cylinder 2?

Possible Answers:

$V_{1}=\frac{3}{2}V_{2}$

$V_{1}=\frac{2}{3}V_{2}$

$V_{1}=\frac{4}{3}V_{2}$

$V_{1}=\frac{3}{4}V_{2}$

$V_{1}=V_{2}$

Correct answer:

$V_{1}=\frac{3}{4}V_{2}$

Explanation:

The volume of a cylinder is:

$V=\pi r^2h$

where is the volume of the cylinder, is the radius of the circular end of the cylinder, and is the height of the cylinder.

So we can write:

$V_{1}=\pi (r_{1})^2h_{1}$

and

$V_{2}=\pi (r_{2})^2h_{2}$

Now we can summarize the given information:

$r_{1}=\sqrt{3}\cdot r_{2}$

$h_{2}=4\cdot h_{1}\Rightarrow h_{1}=\frac{h_{2}}{4}$

Now substitute them in the $V_{1}$ formula:

$V_{1}=\pi (\sqrt{3}r_{2})^2\times \frac{h_{2}}{4}\Rightarrow V_{1}=\frac{3}{4}\pi (r_{2})^2\times h_{2}\Rightarrow V_{1}=\frac{3}{4}V_{2}$

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Example Question #146 : Geometry

Two right cylinders have the same height. The radius of the base of the first cylinder is two times more than that of the second cylinder. Compare the volume of the two cylinders.

Possible Answers:

$V_{1}=2\cdot V_{2}$

$V_{1}=3\cdot V_{2}$

$V_{1}=4\cdot V_{2}$

$V_{1}=\sqrt{2}\cdot V_{2}$

$V_{1}=V_{2}$

Correct answer:

$V_{1}=4\cdot V_{2}$

Explanation:

The volume of a cylinder is:

$V=\pi r^2h$

where is the radius of the circular end of the cylinder and is the height of the cylinder. So we can write:

$V_{1}=\pi (r_{1})^2h_{1}$

$V_{2}=\pi (r_{2})^2h_{2}$

We know that

$h_{1}=h_{2}$

and

$r_{1}=2r_{2}$ .

So we can write:

$V_{1}=\pi (r_{1})^2h_{1}=\pi (2r_{2})^2\cdot h_{2}=4\pi (r_{2})^2h_{2}\Rightarrow V_{1}=4V_{2}$

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Common Core: 8th Grade Math : Know and Use the Formulas for the Volumes of Cones, Cylinders, and Spheres: CCSS.Math.Content.8.G.C.9

Study concepts, example questions & explanations for Common Core: 8th Grade Math

All Common Core: 8th Grade Math Resources

Example Questions

Example Question #1 : How To Find The Volume Of A Cone

Example Question #1 : How To Find The Volume Of A Cone

Example Question #2 : How To Find The Volume Of A Cone

Example Question #3 : How To Find The Volume Of A Cone

Example Question #2 : How To Find The Volume Of A Cone

Example Question #143 : Geometry

Example Question #1 : How To Find The Volume Of A Cylinder

Example Question #144 : Geometry

Example Question #145 : Geometry

Example Question #146 : Geometry

All Common Core: 8th Grade Math Resources

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