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PSAT Math : How to find the radius of a sphere

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Example Questions

Example Question #1 : Spheres

A cube with volume 27 cubic inches is inscribed inside a sphere such that each vertex of the cube touches the sphere. What is the radius, in inches, of the sphere?

Possible Answers:

(3√3)/2 (approximately 2.60)

8.5

√3/2 (approximately 1.73)

Correct answer:

(3√3)/2 (approximately 2.60)

Explanation:

We know that the cube has a volume of 27 cubic inches, so each side of the cube must be ∛27=3 inches. Since the cube is inscribed inside the sphere, the diameter of the sphere is the diagonal length of the cube, so the radius of the sphere is half of the diagonal length of the cube. To find the diagonal length of the cube, we use the distance formula d=√(3²+3²+3² )=√(3*3² )=3√3, and then divide the result by 2 to find the radius of the sphere, (3√3)/2.

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Example Question #1 : Spheres

The surface area of a sphere is 100π square feet. What is the radius in feet?

Possible Answers:

100

Correct answer:

Explanation:

S = 4π(r²)

100π = 4π(r²)

100 = 4r²

25 = r²

5 = r

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

A spherical water tank has a surface area of 400 square meters. To the nearest tenth of a meter, give the radius of the tank.

Possible Answers:

$11.3\textup{ m}$ $\displaystyle 11.3\textup{ m}$

$3.2\textup{ m}$ $\displaystyle 3.2\textup{ m}$

$9.1\textup{ m}$ $\displaystyle 9.1\textup{ m}$

$4.6\textup{ m}$ $\displaystyle 4.6\textup{ m}$

$5.6\textup{ m}$ $\displaystyle 5.6\textup{ m}$

Correct answer:

$5.6\textup{ m}$ $\displaystyle 5.6\textup{ m}$

Explanation:

Given radius $\displaystyle r$, the surface area of a sphere $\displaystyle S$ is given by the formula

$S = 4 \pi r^{2}$ $\displaystyle S = 4 \pi r^{2}$

Set S = 400 $\displaystyle S = 400$ and solve for $\displaystyle r$:

$4 \pi r^{2} = 400$ $\displaystyle 4 \pi r^{2} = 400$

$4 \pi r^{2}\div 4 = 400 \div 4$ $\displaystyle 4 \pi r^{2}\div 4 = 400 \div 4$

$\pi r^{2} = 100$ $\displaystyle \pi r^{2} = 100$

$\pi r^{2} \div \pi = 100 \div \pi$ $\displaystyle \pi r^{2} \div \pi = 100 \div \pi$

$r^{2} \approx 100 \div 3.14159 \approx 31.831$ $\displaystyle r^{2} \approx 100 \div 3.14159 \approx 31.831$

$r \approx \sqrt{31.831} \approx 5.6$ $\displaystyle r \approx \sqrt{31.831} \approx 5.6$ meters.

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

A spherical tank is to hold 50,000 liters of water. Given that one cubic meter is equal to 1,000 liters, give the radius of this tank to the nearest tenth of a meter.

Possible Answers:

$4.0 \textup{ m}$ $\displaystyle 4.0 \textup{ m}$

$4.6\textup{ m}$ $\displaystyle 4.6\textup{ m}$

$2.3\textup{ m}$ $\displaystyle 2.3\textup{ m}$

$3.5\textup{ m}$ $\displaystyle 3.5\textup{ m}$

$2.0 \textup{ m}$ $\displaystyle 2.0 \textup{ m}$

Correct answer:

$2.3\textup{ m}$ $\displaystyle 2.3\textup{ m}$

Explanation:

Given radius $\displaystyle r$, the volume $\displaystyle V$ of a sphere is given by the formula

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

Since 1,000 liters = 1 cubic meter, 50,000 liters = 50 cubic meters, and we find $\displaystyle r$ by setting V= 50 $\displaystyle V= 50$:

$\frac{4}{3} \pi r^{3} = 50$ $\displaystyle \frac{4}{3} \pi r^{3} = 50$

$\frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 50$ $\displaystyle \frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 50$

$\pi r^{3} = 37.5$ $\displaystyle \pi r^{3} = 37.5$

$r^{3} =\frac{ 37.5}{\pi} \approx 11.9366$ $\displaystyle r^{3} =\frac{ 37.5}{\pi} \approx 11.9366$

$r \approx\sqrt[3]{ 11.9366} \approx 2.3$ $\displaystyle r \approx\sqrt[3]{ 11.9366} \approx 2.3$ meters.

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

The Kelvin temperature scale is basically the same as the Celsius scale except with a different zero point; to convert degrees Celsius to Kelvins, add 273. Also, by Charles's law, the volume of a given mass of gas varies directly as its temperature, expressed in Kelvins.

In the early morning, when the temperature is $5^{\circ }C$ $\displaystyle 5^{\circ }C$, a spherical balloon is filled with helium until its radius is 100 meters. At 2:00 PM, the temperature is $20^{\circ }C$ $\displaystyle 20^{\circ }C$. To the nearest tenth of a meter, what is the radius of the balloon at this time?

Possible Answers:

$101.8 \textup{ m}$ $\displaystyle 101.8 \textup{ m}$

$107.2\textup{ m}$ $\displaystyle 107.2\textup{ m}$

$98.3 \textup{ m}$ $\displaystyle 98.3 \textup{ m}$

$158.7 \textup{ m}$ $\displaystyle 158.7 \textup{ m}$

$63.0\textup{ m}$ $\displaystyle 63.0\textup{ m}$

Correct answer:

$101.8 \textup{ m}$ $\displaystyle 101.8 \textup{ m}$

Explanation:

The volume of the balloon is given by the formula

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

Also, by the variation relationship,

$\frac{V'}{T'} = \frac{V}{T}$ $\displaystyle \frac{V'}{T'} = \frac{V}{T}$

We do not need to calculate the original volume of the balloon; we can simply replace the volume with the formula:

$\frac{ \frac{4}{3} \pi r'^{3}}{T'} = \frac{ \frac{4}{3} \pi r^{3}}{T}$ $\displaystyle \frac{ \frac{4}{3} \pi r'^{3}}{T'} = \frac{ \frac{4}{3} \pi r^{3}}{T}$

Dividing both sides by $\frac{4}{3} \pi$ $\displaystyle \frac{4}{3} \pi$ yields a new equation:

$\frac{ r'^{3}}{T'} = \frac{ r^{3}}{T}$ $\displaystyle \frac{ r'^{3}}{T'} = \frac{ r^{3}}{T}$

The initial radius is r = 100 $\displaystyle r = 100$

The initial temperature is $T = 273 + 5 = 278 \textup{ K}$ $\displaystyle T = 273 + 5 = 278 \textup{ K}$

The final (noon) temperature is $T ' = 273 + 20= 293 \textup{ K}$ $\displaystyle T ' = 273 + 20= 293 \textup{ K}$

Substitute to find the final radius:

$\frac{ r'^{3}}{293} = \frac{ 100^{3}}{278}$ $\displaystyle \frac{ r'^{3}}{293} = \frac{ 100^{3}}{278}$

$r'^{3} = \frac{ 1,000,000}{278} \cdot 293 \approx 1,053,957$ $\displaystyle r'^{3} = \frac{ 1,000,000}{278} \cdot 293 \approx 1,053,957$

$r' = \approx\sqrt[3]{ 1,053,957} \approx 101.8$ $\displaystyle r' = \approx\sqrt[3]{ 1,053,957} \approx 101.8$ meters.

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

The city of Washington wants to build a spherical water tank for the town hall. The tank is to have capacity 120 cubic meters of water.

To the nearest tenth, what will the radius of the tank be?

Possible Answers:

$6.1\textup{ m}$ $\displaystyle 6.1\textup{ m}$

$5.4\textup{ m}$ $\displaystyle 5.4\textup{ m}$

The correct answer is not given among the other responses.

$10.7\textup{ m}$ $\displaystyle 10.7\textup{ m}$

$3.1\textup{ m}$ $\displaystyle 3.1\textup{ m}$

Correct answer:

$3.1\textup{ m}$ $\displaystyle 3.1\textup{ m}$

Explanation:

Given the radius $\displaystyle r$, the volume $\displaystyle V$ of a sphere is given by the formula

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

We find the inner radius $\displaystyle r$ by setting V= 120 $\displaystyle V= 120$:

$\frac{4}{3} \pi r^{3} = 120$ $\displaystyle \frac{4}{3} \pi r^{3} = 120$

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

$\pi r^{3} = 90$ $\displaystyle \pi r^{3} = 90$

$r^{3} =\frac{ 90}{\pi} \approx 28.65$ $\displaystyle r^{3} =\frac{ 90}{\pi} \approx 28.65$

$r \approx\sqrt[3]{ 28.65} \approx 3.1$ $\displaystyle r \approx\sqrt[3]{ 28.65} \approx 3.1$ meters.

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Example Question #201 : Geometry

A spherical balloon is filled with 10,000 cubic meters of gas in the morning when the temperature is $10^{\circ }C$ $\displaystyle 10^{\circ }C$. The temperature increases to $25^{\circ }C$ $\displaystyle 25^{\circ }C$ by noon, with no other conditions changing. What is the radius of the balloon at noon, to the nearest tenth of a meter?

Possible Answers:

$13.6 \textup{ m}$ $\displaystyle 13.6 \textup{ m}$

$18.1\textup{ m}$ $\displaystyle 18.1\textup{ m}$

$13.1\textup{ m}$ $\displaystyle 13.1\textup{ m}$

$9.8\textup { m}$ $\displaystyle 9.8\textup { m}$

None of the other responses gives the correct answer.

Correct answer:

$13.6 \textup{ m}$ $\displaystyle 13.6 \textup{ m}$

Explanation:

First, we use the variation equation to figure out the volume of the balloon at noon. First, we add 273 to each of the temperatures.

The initial temperature is $273 + 10 = 283 \textup{ K}$ $\displaystyle 273 + 10 = 283 \textup{ K}$

The final (noon) temperature is $273 + 25= 298 \textup{ K}$ $\displaystyle 273 + 25= 298 \textup{ K}$

Since volume varies directly as temperature, we can set up the equation

$\frac{V'}{T'} = \frac{V}{T}$ $\displaystyle \frac{V'}{T'} = \frac{V}{T}$

$\frac{V'}{298} = \frac{10,000}{283}$ $\displaystyle \frac{V'}{298} = \frac{10,000}{283}$

$V' = \frac{10,000}{283} \cdot 298 \approx 10,530$ $\displaystyle V' = \frac{10,000}{283} \cdot 298 \approx 10,530$

The volume of a sphere is given by the formula

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

so set $\displaystyle V = 10,530$ and solve for $\displaystyle r$:

$\frac{4}{3} \pi r^{3} = 10,530$ $\displaystyle \frac{4}{3} \pi r^{3} = 10,530$

$\frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 10,530$ $\displaystyle \frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 10,530$

$\pi r^{3} \approx 7,897.5$ $\displaystyle \pi r^{3} \approx 7,897.5$

$r^{3} \approx 7,897.5 \div \pi \approx 2,513.8$ $\displaystyle r^{3} \approx 7,897.5 \div \pi \approx 2,513.8$

$r \approx\sqrt[3]{ 2,513.8} \approx 13.6$ $\displaystyle r \approx\sqrt[3]{ 2,513.8} \approx 13.6$ meters.

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

Find the radius of a sphere whose volume is $\small \small 36\pi$ $\displaystyle \small \small 36\pi$ $\text{in}^3$ $\displaystyle \text{in}^3$.

Possible Answers:

$\small 3\pi$ $\displaystyle \small 3\pi$ $\text{in}$ $\displaystyle \text{in}$

$\small 9\pi$ $\displaystyle \small 9\pi$ $\text{in}$ $\displaystyle \text{in}$

$\small 9$ $\displaystyle \small 9$ $\text{in}$ $\displaystyle \text{in}$

$\small 3$ $\displaystyle \small 3$ $\text{in}$ $\displaystyle \text{in}$

Correct answer:

$\small 3$ $\displaystyle \small 3$ $\text{in}$ $\displaystyle \text{in}$

Explanation:

Use the equation for the volume of a sphere to find the radius.

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

$\small 36\pi = \frac{4}{3}\pi r^{3}$ $\displaystyle \small 36\pi = \frac{4}{3}\pi r^{3}$

$\small \small 36 = \frac{4}{3} r^{3}$ $\displaystyle \small \small 36 = \frac{4}{3} r^{3}$

$\small 27 = r^{3}$ $\displaystyle \small 27 = r^{3}$

$\small r = 3$ $\displaystyle \small r = 3$

So, the radius of the sphere is 3

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

If a sphere has a volume of $36\pi\textup{ cubic inches}$ $\displaystyle 36\pi\textup{ cubic inches}$, then what is the radius of the sphere?

Possible Answers:

$3\textup{ in}$ $\displaystyle 3\textup{ in}$

$4.5\textup{ in}$ $\displaystyle 4.5\textup{ in}$

$9\textup{ in}$ $\displaystyle 9\textup{ in}$

$18\textup{ in}$ $\displaystyle 18\textup{ in}$

$6\textup{ in}$ $\displaystyle 6\textup{ in}$

Correct answer:

$3\textup{ in}$ $\displaystyle 3\textup{ in}$

Explanation:

The volume of a sphere is equal to

$v=\left ( \frac{4}{3} \right )\pi r^3$ $\displaystyle v=\left ( \frac{4}{3} \right )\pi r^3$

Therefore,

$r=\sqrt[3]{(3/4)v/\pi}=\sqrt[3]{(3/4)36\pi/\pi}=\sqrt[3]{27}=3$ $\displaystyle r=\sqrt[3]{(3/4)v/\pi}=\sqrt[3]{(3/4)36\pi/\pi}=\sqrt[3]{27}=3$

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PSAT Math : How to find the radius of a sphere

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Spheres

Example Question #1 : Spheres

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

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

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

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

Example Question #201 : Geometry

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

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

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