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ISEE Upper Level Math : Spheres

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Example Question #1 : How To Find The Diameter Of A Sphere

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its diameter?

Possible Answers:

15ft

25 ft

30ft

22.5ft

Correct answer:

30ft

Explanation:

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its diameter?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$900 \pi ft^2=4 \pi r^2$

First divide both sides by $4\pi$ .

225ft^2=r^2

Then square root both sides to get our radius:

r=15ft

Now, because the question is asking for our diameter and not our radius, we need to double our radius to get our answer:

d=2r=2*15ft=30 ft

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Example Question #1 : How To Find The Radius Of A Sphere

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Possible Answers:

Cannot be determined from the information provided

12.5m

25 m

15m

Correct answer:

15m

Explanation:

A wooden ball has a surface area of $900\pi m^2$ .

What is its radius?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4\pi r^2$

Now, plug in our surface area and solve with algebra:

$900 \pi m^2=4\pi r^2$

Get rid of the pi

900 m^2=4 r^2

Divide by 4

225 m^2= r^2

Square root both sides to get our answer:

r=15m

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Example Question #2 : How To Find The Radius Of A Sphere

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its radius?

Possible Answers:

15 ft

25 ft

225 ft

45 ft

Correct answer:

15 ft

Explanation:

There is a perfectly spherical weather balloon with a surface area of $900 \pi ft^2$ , what is its radius?

Begin with the formula for surface area of a sphere:

$SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$900 \pi ft^2=4 \pi r^2$

First divide both sides by $4\pi$ .

225ft^2=r^2

Then square root both sides to get our answer:

r=15ft

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Example Question #2 : Volume Of A Sphere

In terms of $\pi$ , give the volume, in cubic inches, of a spherical water tank with a diameter of 20 feet.

Possible Answers:

$307,200\pi\textrm{ in}^{3}$

$230,400 \pi\textrm{ in}^{3}$

$2,304,000 \pi \textrm{ in}^{3}$

$57,600 \pi\textrm{ in}^{3}$

$18,432,000\pi\textrm{ in}^{3}$

Correct answer:

$2,304,000 \pi \textrm{ in}^{3}$

Explanation:

20 feet = $20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set r = 120 , substitute in the volume formula, and solve for :

$V = \frac{4}{3} \pi r^{3}$

$V = \frac{4}{3} \pi \cdot 120^{3}$

$V = \frac{4}{3} \pi \cdot 1,728,000$

$V = 2,304,000 \pi$ cubic inches

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Example Question #21 : Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

A sphere has diameter 3 meters. Give its volume in cubic centimeters (leave in terms of $\pi$ ).

Possible Answers:

$3,375,000\pi \textrm{ cm}^{3}$

$27,000,000\pi \textrm{ cm}^{3}$

$4,500,000\pi \textrm{ cm}^{3}$

$36,000,000\pi \textrm{ cm}^{3}$

$1,125,000\pi \textrm{ cm}^{3}$

Correct answer:

$4,500,000\pi \textrm{ cm}^{3}$

Explanation:

The diameter of 3 meters is equal to $3 \times 100 = 300$ centimeters; the radius is half this, or 150 centimeters. Substitute r = 150 in the volume formula:

$V= \frac{4}{3} \pi r^{3}$

$V= \frac{4}{3} \cdot \pi \cdot 150^{3}$

$V= \frac{4}{3} \cdot 150 \cdot 150 \cdot 150\cdot \pi$

$V= 4,500,000\pi$ cubic centimeters

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Example Question #1 : Spheres

A spherical buoy has a radius of 5 meters. What is the volume of the buoy?

Possible Answers:

$166\frac{2}{3}\pi m^3$

$166\frac{2}{3} m^3$

$165 \pi m^3$

$16\frac{2}{3}\pi m^3$

Correct answer:

$166\frac{2}{3}\pi m^3$

Explanation:

A spherical buoy has a radius of 5 meters. What is the volume of the buoy?

To find the volume of a sphere, use the following formula:

$V_{sphere}=\frac{4}{3}\pi r^3$

All we have to do is plug in 5 meters and simplify:

$V_{sphere}=\frac{4}{3}\pi (5m)^3=\frac{4}{3}\pi*125m^3=166\frac{2}{3}\pi m^3$

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

You have a ball with a radius of 12 cm, what is its volume?

Possible Answers:

$144 \pi cm^3$

$404 \pi cm^3$

$192 \pi cm^3$

$2304 \pi cm^3$

Correct answer:

$2304 \pi cm^3$

Explanation:

You have a ball with a radius of 12 cm, what is its volume?

The volume of a sphere can be found via the following formula:

$V_{sphere}=\frac{4}{3}\pi r^3$

We know our radius, so all we need to do is plug in and simplify:

$V_{sphere}=\frac{4}{3}\pi (12cm)^3=2304 \pi cm^3$

So we have our answer:

$2304 \pi cm^3$

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

You have a wooden ball which you are going to paint. If the radius is 12 inches, what is the volume of the ball?

Possible Answers:

$576 \pi in^3$

$1728\pi in^3$

2304 in^3

$2304\pi in^3$

Correct answer:

$2304\pi in^3$

Explanation:

You have a wooden ball which you are going to paint. If the radius is 12 inches, what is the volume of the ball?

Alright, let's begin with the volume of a sphere formula:

$V_{sphere}=\frac{4}{3}\pi r^3$

Now, plug in 12 and simplify:

$V_{sphere}=\frac{4}{3}\pi(12in)^3=\frac{4}{3}\pi*1728in^3=2304\pi in^3$

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Example Question #4 : How To Find The Volume Of A Sphere

Find the volume of a sphere with a diameter of 6cm.

Possible Answers:

$216\pi \text{ cm}^3$

$108\pi \text{ cm}^3$

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

$42\pi \text{ cm}^3$

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

Correct answer:

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

Explanation:

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

$V = \frac{4}{3} \cdot \pi \cdot r^3$

where r is the radius of the sphere.

Now, we know the diameter of the sphere is 6cm. We also know that the diameter is two times the radius. Therefore, the radius is 3cm.

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

$V = \frac{4}{3} \cdot \pi \cdot (3\text{cm})^3$

$V = \frac{4}{3} \cdot \pi \cdot 3\text{cm} \cdot 3\text{cm} \cdot 3\text{cm}$

$V = \frac{4}{3} \cdot \pi \cdot 27\text{cm}^3$

Now, before we continue, we can simplify. The 3 and the 27 can both be divided by 3. We get

$V = \frac{4}{1} \cdot \pi \cdot 9\text{cm}^3$

$V = 4 \cdot \pi \cdot 9\text{cm}^3$

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

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Example Question #5 : How To Find The Volume Of A Sphere

Find the volume of a sphere with a diameter of 12in.

Possible Answers:

$256\pi \text{ in}^3$

$144\pi \text{ in}^3$

$200\pi \text{ in}^3$

$288\pi \text{ in}^3$

$96\pi \text{ in}^3$

Correct answer:

$288\pi \text{ in}^3$

Explanation:

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

$V = \frac{4}{3} \cdot \pi \cdot r^3$

where r is the radius of the sphere.

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

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

$V = \frac{4}{3} \cdot \pi \cdot (6\text{in})^3$

$V = \frac{4}{3} \cdot \pi \cdot 6\text{in} \cdot 6\text{in} \cdot 6\text{in}$

Now, we can simplify before we multiply to make things easier. The 3 and a 6 can both be divided by 3. So, we get

$V = \frac{4}{1} \cdot \pi \cdot 2\text{in} \cdot 6\text{in} \cdot 6\text{in}$

$V = 4 \cdot \pi \cdot 72\text{in}^3$

$V = 288\pi \text{ in}^3$

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ISEE Upper Level Math : Spheres

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 Diameter Of A Sphere

Example Question #1 : How To Find The Radius Of A Sphere

Example Question #2 : How To Find The Radius Of A Sphere

Example Question #2 : Volume Of A Sphere

Example Question #21 : Know And Use The Formulas For The Volumes Of Cones, Cylinders, And Spheres: Ccss.Math.Content.8.G.C.9

Example Question #1 : Spheres

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

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

Example Question #4 : How To Find The Volume Of A Sphere

Example Question #5 : How To Find The Volume Of A Sphere

All ISEE Upper Level Math Resources

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