How to find the radius of a sphere - PSAT Math

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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?

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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.

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

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S = 4π(r2)

100π = 4π(r2)

100 = 4r2

25 = r2

5 = r

Find the radius of a sphere whose volume is $\small \small 36\pi$ .

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

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

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

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

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

$\small r = 3$

So, the radius of the sphere is 3

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.

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Given radius , the surface area of a sphere is given by the formula

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

Set and solve for :

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

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

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

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

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

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?

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Given the radius , the volume of a sphere is given by the formula

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

We find the inner radius by setting :

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

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

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

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

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.

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Given radius , the volume of a sphere is given by the formula

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

Since 1,000 liters = 1 cubic meter, 50,000 liters = 50 cubic meters, and we find by setting :

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

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

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

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

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.

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

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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 $273 + 10 = 283 \textup{ K}$

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

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

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

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

$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}$$

so set and solve for :

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

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

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

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

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 , a spherical balloon is filled with helium until its radius is 100 meters. At 2:00 PM, the temperature is . To the nearest tenth of a meter, what is the radius of the balloon at this time?

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The volume of the balloon is given by the formula

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

Also, by the variation relationship,

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

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}$

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

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

The initial radius is

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

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

Substitute to find the final radius:

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

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

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

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The volume of a sphere is equal to

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

Therefore,

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

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?

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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.

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

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S = 4π(r2)

100π = 4π(r2)

100 = 4r2

25 = r2

5 = r

Find the radius of a sphere whose volume is $\small \small 36\pi$ .

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

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

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

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

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

$\small r = 3$

So, the radius of the sphere is 3

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.

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Given radius , the surface area of a sphere is given by the formula

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

Set and solve for :

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

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

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

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

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

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?

Tap to see back →

Given the radius , the volume of a sphere is given by the formula

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

We find the inner radius by setting :

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

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

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

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

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.

Tap to see back →

Given radius , the volume of a sphere is given by the formula

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

Since 1,000 liters = 1 cubic meter, 50,000 liters = 50 cubic meters, and we find by setting :

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

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

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

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

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.

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

Tap to see back →

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 $273 + 10 = 283 \textup{ K}$

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

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

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

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

$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}$$

so set and solve for :

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

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

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

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

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 , a spherical balloon is filled with helium until its radius is 100 meters. At 2:00 PM, the temperature is . To the nearest tenth of a meter, what is the radius of the balloon at this time?

Tap to see back →

The volume of the balloon is given by the formula

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

Also, by the variation relationship,

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

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}$

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

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

The initial radius is

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

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

Substitute to find the final radius:

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

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

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

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The volume of a sphere is equal to

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

Therefore,

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

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?

Tap to see back →

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.

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

Tap to see back →

S = 4π(r2)

100π = 4π(r2)

100 = 4r2

25 = r2

5 = r