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SSAT Upper Level Math : How to find the volume of a cone

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Example Questions

Example Question #1 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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:

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

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

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

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

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

Correct answer:

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

Explanation:

First, convert the dimensions to cubic centimeters by multiplying by 100 : the cone has height 300cm 半径,它的基础 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 #2 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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

2133

6333

4333

5333

Correct answer:

5333

Explanation:

The volume of a cone is:

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

whereis the radius of the circular base, andis 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:

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

whereis 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 andis the volume. So we get:

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

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Example Question #3 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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:

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

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

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

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

$\pi t^3$

Correct answer:

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

Explanation:

The volume of a cone is:

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

whereis the radius of the circular base, andis 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 #4 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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

Possible Answers:

$\sqrt{2}$

$\sqrt[3]{2}$

$\sqrt[3]{3}$

$2\sqrt[3]{2}$

Correct answer:

$\sqrt[3]{2}$

Explanation:

The volume of a cone is given by:

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

whereis the radius of the circular base, andis 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 #5 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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:

$36\pi\:m^3$

$144\pi\:m^3$

$54\pi\:m^3$

$48\pi\:m^3$

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 #1 :How To Find The Volume Of A Cone

A cone has a radius ofinches and a height ofinches. Find the volume of the cone.

Possible Answers:

$108\pi\:in^3$

$36\pi\:in^3$

$12\pi\:in^3$

$24\pi\:in^3$

Correct answer:

$36\pi\:in^3$

Explanation:

The volume of a cone is given by the formula:

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

Now, plug in the values of the radius and height to find the volume of the given cone.

$\text{Volume}=\frac{1}{3}\pi\times3^2\times12=\frac{1}{3}\pi\times9\times12=\frac{108}{3}\pi=36\pi\:in^3$

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SSAT Upper Level Math : How to find the volume of a cone

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

Example Question #2 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

Example Question #3 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

Example Question #4 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

Example Question #5 :Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

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

All SSAT Upper Level Math Resources

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