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ISEE Upper Level Math : Cones

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Example Questions

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

A cone has height 18 inches; its base has radius 4 inches. Give its volume in cubic feet (leave in terms of $\pi$ )

Possible Answers:

$\frac{1}{3} \pi \textrm{ ft}^{3}$

$\frac{1}{4} \pi \textrm{ ft}^{3}$

$\frac{1}{18} \pi \textrm{ ft}^{3}$

$\frac{1}{6} \pi \textrm{ ft}^{3}$

$\frac{1}{12} \pi \textrm{ ft}^{3}$

Correct answer:

$\frac{1}{18} \pi \textrm{ ft}^{3}$

Explanation:

Convert radius and height from inches to feet by dividing by 12:

Height: 18 inches = $18 \div 12 = \frac{18}{12} = \frac{3}{2}$ feet

Radius: 4 inches = $4 \div 12 = \frac{4}{12} = \frac{1}{3}$

The volume of a cone is given by the formula

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

Substitute $r = \frac{1}{3}, h = \frac{3}{2}$ :

$V = \frac{1}{3}\cdot \left ( \frac{1}{3} \right ) ^{2}\cdot \frac{3}{2} \cdot \pi$

$V = \frac{1}{3}\cdot\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{3}{2} \cdot \pi$

$V = \frac{1}{1}\cdot\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \pi$

$V = \frac{1}{18} \pi$

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

Give the volume of a cone whose height is 10 inches and whose base is a circle with circumference $6 \pi$ inches.

Possible Answers:

$30 \pi \textrm{ in}^{3}$

$45 \pi \textrm{ in}^{3}$

$120 \pi \textrm{ in}^{3}$

$360 \pi \textrm{ in}^{3}$

$90 \pi \textrm{ in}^{3}$

Correct answer:

$30 \pi \textrm{ in}^{3}$

Explanation:

A circle with circumference $6 \pi$ inches has as its radius

$r = \frac{C }{2\pi }=\frac{6\pi }{2\pi } = 3$ inches.

The area of the base is therefore

$B = \pi r^{2} = \pi \cdot 3^{2} = 9 \pi$ square inches.

To find the volume of the cone, substitute $B = 9 \pi , h = 10$ in the formula for the volume of a cone:

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 9 \pi \cdot 10 = 30 \pi$ cubic inches

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Example Question #501 : Grade 8

The height of a cone and the radius of its base are equal. The circumference of the base is $10 \pi$ inches. Give its volume.

Possible Answers:

$\frac{5,000}{3} \pi \textrm{ in}^{3}$

$1,000 \textrm{ in}^{3}$

$\frac{125}{3} \pi \textrm{ in}^{3}$

$\frac{1,000}{3} \pi \textrm{ in}^{3}$

$125 \textrm{ in}^{3}$

Correct answer:

$\frac{125}{3} \pi \textrm{ in}^{3}$

Explanation:

A circle with circumference $10 \pi$ inches has as its radius

$r = \frac{C }{2\pi }=\frac{10\pi }{2\pi } = 5$ inches.

The height is also inches, so substitute h = r = 5 in the volume formula for a cone:

$V = \frac{1}{3} \pi r ^{2}h = \frac{1}{3} \pi \cdot 5 ^{2} \cdot 5 = \frac{125}{3} \pi$ cubic inches

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

You are an architect designing a cone shaped structure. If the structure will be 30 ft tall and 10 feet wide at the base, what will the volume of the structure be?

Possible Answers:

$100 \pi ft^3$

$22.5 \pi ft^3$

250 ft^3

$250 \pi ft^3$

Correct answer:

$250 \pi ft^3$

Explanation:

You are an architect designing a cone shaped structure. If the structure will be 30 ft tall and 10 feet wide at the base, what will the volume of the structure be?

Begin by using the formula for volume of a cone:

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

Now, we simply need to plug in our knowns.

We know the height is 30 ft.

We know that the diameter is 10ft, however, we need the radius.

Divide 10 by 2 to get a radius of 5 ft.

Now, let's go....

$V=\pi (5ft)^2\frac{30ft}{3}=\pi25ft^2*10ft=250 \pi ft^3$

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

Find the volume of a cone with the following measurements:

diameter = 12in
height = 6in

Possible Answers:

$72\pi \text{ in}^3$

$54\pi \text{ in}^3$

$36\pi \text{ in}^3$

$84\pi \text{ in}^3$

$48\pi \text{ in}^3$

Correct answer:

$72\pi \text{ in}^3$

Explanation:

To find the volume of a cone, we will use the following formula:

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

where r is the radius and h is the height of the cone.

Now, we know the diameter of the cone is 12in. We also know the diameter is two times the radius. Therefore, the radius is 6in.

We know the height of the cone is 6in.

Knowing all of this, we can substitute into the formula. We get

$V = \pi \cdot (6\text{in})^2 \cdot \frac{6\text{in}}{3}$

$V = \pi \cdot 36\text{in}^2 \cdot 2\text{in}$

$V = \pi \cdot 72\text{in}^3$

$V = 72\pi \text{ in}^3$

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Example Question #413 : Geometry

Find the volume of a cone with the following measurements:

Diameter: 14in
Height: 9in

Possible Answers:

$147\pi \text{ in}^3$

$441\pi \text{ in}^3$

$252\pi \text{ in}^3$

$63\pi \text{ in}^3$

$126\pi \text{ in}^3$

Correct answer:

$147\pi \text{ in}^3$

Explanation:

To find the volume of a cone, we will use the following formula:

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

where r is the radius and h is the height of the cone.

Now, we know the diameter of the cone is 14in. We also know the diameter is two times the radius. Therefore, the radius is 7in.

We know the height of the cone is 9in.

Knowing all of this, we can substitute into the formula. We get

$V = \pi \cdot (7\text{in})^2 \cdot \frac{9\text{in}}{3}$

$V = \pi \cdot 49\text{in}^2 \cdot 3\text{in}$

$V = \pi \cdot 147\text{in}^3$

$V = 147\pi \text{ in}^3$

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Example Question #414 : Geometry

Find the volume of a cone with the following measurements:

height: 12in
diameter: 6in

Possible Answers:

$24\pi \text{ in}^3$

$32\pi \text{ in}^3$

$36\pi \text{ in}^3$

$28\pi \text{ in}^3$

$48\pi \text{ in}^3$

Correct answer:

$36\pi \text{ in}^3$

Explanation:

To find the volume of a cone, we will use the following formula:

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

where r is the radius and h is the height of the cone.

Now, we know the height of the cone is 12in. We also know the diameter of the cone is 6in. We know the diameter is two times the radius. Therefore, the radius is 3in.

So, we get

$V = \pi \cdot (3\text{in})^2 \cdot \frac{12\text{in}}{3}$

$V = \pi \cdot 9\text{in}^2 \cdot 4\text{in}$

$V = \pi \cdot 36\text{in}^3$

$V = 36\pi \text{ in}^3$

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Example Question #101 : Solid Geometry

Find the volume of a cone with a base diameter of 2, and a height of 10.

Possible Answers:

$10\pi$

$\frac{10}{3}\pi$

$\frac{5}{3}\pi$

$\frac{40}{3}\pi$

$\frac{20}{3}\pi$

Correct answer:

$\frac{10}{3}\pi$

Explanation:

Write the formula to find the volume of a cone.

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

The diameter is 2, which means the radius is 1. Substitute the known dimensions.

$V=\frac{1}{3}\pi (1)^2(10)$

The answer is: $\frac{10}{3}\pi$

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Example Question #101 : Solid Geometry

A cone has height 240 centimeters; its base has radius 80 centimeters. Give its volume in cubic meters.

Possible Answers:

$0.768 \pi \textrm{ m}^{3}$

$1.536 \pi \textrm{ m}^{3}$

$0.512 \pi \textrm{ m}^{3}$

$4.608 \pi \textrm{ m}^{3}$

Correct answer:

$0.512 \pi \textrm{ m}^{3}$

Explanation:

Convert both dimensions from centimeters to meters by dividing by 100:

Height: 240 centimeters = $240 \div 100 = 2.4$ meters.

Radius: 80 centimeters = $80 \div 100 = 0.8$ meters.

Substitute r = 0.8, h = 2.4 in the volume formula:

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

$V = \frac{1}{3} \cdot \pi \cdot 0.8 ^{2} \cdot 2.4$

$V = \frac{1}{3}\cdot 0.8 \cdot 0.8 \cdot 2.4 \cdot \pi$

$V =0.512 \pi$

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Example Question #102 : Solid Geometry

What is the surface area of a cone with the following measurements?

r=4;d=8;h=10

Possible Answers:

$5+8\pi$

$5+64\pi$

$5+4\pi$

$5+16\pi$

Correct answer:

$5+16\pi$

Explanation:

The surface area of a cone can be found with the following equation:

$SA=\frac{1}{2}h+r^2\pi$

$SA=\frac{1}{2}(10)+4^2\pi=5+16\pi$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Cones

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

All ISEE Upper Level 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 #501 : Grade 8

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

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

Example Question #413 : Geometry

Example Question #414 : Geometry

Example Question #101 : Solid Geometry

Example Question #101 : Solid Geometry

Example Question #102 : Solid Geometry

All ISEE Upper Level Math Resources

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