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SAT Math : Spheres

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Example Questions

Example Question #31 : Spheres

What is the difference between the volume and surface area of a sphere with a radius of 6?

Possible Answers:

144π

720π

216π

133π

288π

Correct answer:

144π

Explanation:

Surface Area = 4πr² = 4 * π * 6² = 144π

Volume = 4πr³/3 = 4 * π * 6³ / 3 = 288π

Volume – Surface Area = 288π – 144π = 144π

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Example Question #32 : Spheres

A sphere is perfectly contained within a cube that has a volume of 216 units. What is the volume of the sphere?

Possible Answers:

$36\pi$

$288\pi$

$9\pi$

$12\pi$

$48\pi$

Correct answer:

$36\pi$

Explanation:

To begin, we must determine the dimensions of the cube. This is done by solving the simple equation: V=s^3

We know the volume is 216, allowing us to solve for the length of a side of the cube.

216=s^3

Taking the cube root of both sides, we get s = 6.

The diameter of the sphere will be equal to side of the cube, since the question states that the sphere is perfectly contained. The diameter of the sphere will be 6, and the radius will be 3.

We can plug this into the equation for volume of a sphere:

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

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

We can cancel out the 3 in the denominator.

$V=4\pi (3)^2$

Simplify.

$V=4\pi (9)$

$V=36\pi$

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Example Question #33 : Spheres

The surface area of a sphere is $36\pi\ mm^2$ . Find the volume of the sphere in cubic millimeters.

Possible Answers:

$4\pi\ mm^3$

$6\pi\ mm^3$

$36\pi\ mm^3$

$3\pi\ mm^3$

$9\pi\ mm^3$

Correct answer:

$36\pi\ mm^3$

Explanation:

$SA=4\pi r^2$

$36\pi=4\pi r^2$

36=4r^2

9=r^2

3=r

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

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

$V=\frac{4}{3}(27)\pi$

$V=4(9)\pi$

$V=36\pi$

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Example Question #34 : Spheres

A solid hemisphere has a radius of length r. Let S be the number of square units, in terms of r, of the hemisphere's surface area. Let V be the number of cubic units, in terms of r, of the hemisphere's volume. What is the ratio of S to V?

Possible Answers:

$\frac{3}{r}$

$\frac{4r}{3}$

$\frac{9r}{2}$

$3r$

$\frac{9}{2r}$

Correct answer:

$\frac{9}{2r}$

Explanation:

First, let's find the surface area of the hemisphere. Because the hemisphere is basically a full sphere cut in half, we need to find half of the surface area of a full sphere. However, because the hemisphere also has a circular base, we must then add the area of the base.

$S = \frac{1}{2}\cdot$ (surface area of sphere) + (surface area of base)

The surface area of a sphere with radius r is equal to $4\pi r^2$ . The surface area of the base is just equal to the surface area of a circle, which is $\pi r^2$ .

$S=\frac{1}{2}\cdot 4\pi r^2+\pi r^2=2\pi r^2+\pi r^2=3\pi r^2$

The volume of the hemisphere is going to be half of the volume of an entire sphere. The volume for a full sphere is $\frac{4}{3}\pi r^3$ .

$V = \frac{1}{2}\cdot$ (volume of sphere)

$V = \frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$

Ultimately, the question asks us to find the ratio of S to V. To do this, we can write S to V as a fraction.

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

In order to simplify this, let's multiply the numerator and denominator both by 3.

$\frac{S}{V}=\frac{3\pi r^2}{\frac{2}{3}\pi r^3}$ = $\frac{9\pi r^2}{2\pi r^3}=\frac{9}{2r}$

The answer is $\frac{9}{2r}$ .

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Example Question #34 : Spheres

Find the volume of a sphere with a radius of $\frac{3}{5}$ .

Possible Answers:

$\frac{36}{5}\pi$

$\frac{36}{125}\pi$

$\frac{36}{25}\pi$

$9\pi$

$\frac{12}{25}\pi$

Correct answer:

$\frac{36}{125}\pi$

Explanation:

Write the formula to find the volume of the sphere.

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

Substitute the radius and solve for the volume.

$V=\frac{4}{3}\pi (\frac{3}{5})^3=\frac{4}{3}\pi (\frac{27}{125})=\frac{36}{125}\pi$

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Example Question #32 : Spheres

Six spheres have volumes that form an arithmetic sequence. The two smallest spheres have radii 4 and 6. Give the volume of the largest sphere.

Possible Answers:

$\frac{3,296}{3} \pi$

$\frac{10, 976 }{3}\pi$

$784 \pi$

$196 \pi$

$464 \pi$

Correct answer:

$\frac{3,296}{3} \pi$

Explanation:

The volume of a sphere with radius can be determined using the formula

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

The smallest sphere, with radius r = 4 , has volume

$V_{1} = \frac{4}{3} \pi \cdot 4 ^{3} = \frac{4}{3} \pi \cdot 64 = \frac{256}{3} \pi$ .

The second-smallest sphere, with radius r = 6 , has volume

$V_{2} = \frac{4}{3} \pi \cdot 6 ^{3} = \frac{4}{3} \pi \cdot 216= \frac{864}{3} \pi$ .

The volumes are in an arithmetic sequence; their common difference is the difference of these two volumes, or

$d = V_{2} - V_{1}= \frac{864}{3} \pi - \frac{256}{3} \pi = \frac{608}{3} \pi$

Since the six volumes are in an arithmetic sequence, the volume of the largest of the six spheres - that is, the sixth-smallest sphere - is

$V_{6}= V_{1}+ 5d$

$= \frac{256}{3} \pi + 5 \cdot \frac{608}{3} \pi$

$= \frac{256}{3} \pi + \frac{3,040}{3} \pi$

$= \frac{3,296}{3} \pi$

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Example Question #12 : How To Find The Volume Of A Sphere

The radii of six spheres form an arithmetic sequence. The smallest and largest spheres have radii 10 and 30, respectively. Give the volume of the second-largest sphere.

Possible Answers:

$\frac{2,048,000}{81} \pi$

None of the other responses gives a correct answer.

$\frac{25,600}{9} \pi$

$\frac{70, 304}{3} \pi$

$2, 704 \pi$

Correct answer:

$\frac{70, 304}{3} \pi$

Explanation:

The radii of the spheres form an arithmetic sequence, with

$r_{1} =1 0$ and $r_{6} =30$

The common difference can be computed as follows:

$r_{6} = (6-1) d + r_{1}$

30=5d + 10

5d= 20

d = 4

The second-largest - or fifth-smallest - sphere has radius:

$r_{5} = (5-1) d + r_{1} = 4 \cdot 4 + 10 = 16 + 10 = 26$

The volume of a sphere with radius can be determined using the formula

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

Set r = 26 :

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

$= \frac{4}{3} \pi \cdot 26^{3}$

$= \frac{4}{3} \pi \cdot 17,576$

$= \frac{70, 304}{3} \pi$

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Example Question #15 : How To Find The Volume Of A Sphere

A sphere with radius fits perfectly inside of a cube so that the sides of the cube are barely touching the sphere. What is the volume of the cube that is not occupied by the sphere?

Possible Answers:

$2744-\frac{4}{3}\pi$

$1000-\frac{4}{3}\pi$

$\frac{1372}{3}\pi - 2744$

$\frac{1372}{3}\pi$

$2744-\frac{1372}{3}\pi$

Correct answer:

$2744-\frac{1372}{3}\pi$

Explanation:

Because the sides of the interior of the cube are tangent to the sphere, we know that the length of each side is equal to the diameter of the sphere. Since the radius of this sphere is , then its diameter is .

To find the volume that is not occupied by the sphere, we will subtract the sphere's volume from the volume of the cube.

The volume of the cube is:

$V=lwh=14\cdot14\cdot14=2744$

The volume of the sphere is:

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\cdot\pi\cdot7^3 = \frac{1372}{3}\pi$

Therefore, with these values, the volume of the cube not occupied by the sphere is:

$2744-\frac{1372}{3}\pi$

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Example Question #811 : Geometry

A sphere fits inside of a cube so that its surface barely touches each side of the cube at any given time. If the volume of the box is 27 cubic centimeters, then what is the volume of the sphere?

Possible Answers:

$\frac{27}{12}\pi$

$\frac{27}{6}\pi$

None of the given answers.

$3\pi$

$6\pi$

Correct answer:

$\frac{27}{6}\pi$

Explanation:

If the volume of the cube is 27 cubic centimeters, then its height, width, and depth are all 3cm. Since the sphere fits perfectly in the cube, then the sphere's diameter is also 3. This means that its radius is $\frac{3}{2}$ .

Substitute this radius value into the equation for the volume of a sphere:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\cdot\pi \cdot(\frac{3}{2})^3=\frac{4}{3}\cdot\pi\cdot\frac{27}{8}$

$=\frac{108}{24}\pi = \frac{27}{6}\pi$

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Example Question #17 : How To Find The Volume Of A Sphere

The surface area of a sphere is 100. Give its volume.

Possible Answers:

$\frac{500 \pi}{3 }$

$\frac{500 \sqrt{\pi}}{3 }$

$\frac{500 \sqrt{\pi}}{3 \pi}$

$\frac{500 }{3 }$

$\frac{500 }{3 \pi}$

Correct answer:

$\frac{500 \sqrt{\pi}}{3 \pi}$

Explanation:

The surface area of a sphere is equal to

$A= 4 \pi r ^{2}$ ,

where is the radius.

Setting A = 100 and solving for :

$4 \pi r ^{2} = 100$

$\frac{4 \pi r ^{2}}{4 \pi } = \frac{100}{4 \pi }$

$r ^{2} = \frac{25 }{\pi }$

$r = \sqrt{\frac{25 }{\pi } }$

Applying the Quotient of Radicals Rule:

$r = \frac{\sqrt{25 } }{\sqrt{\pi}}$

$r = \frac{5 }{\sqrt{\pi}}$

The volume of the sphere can be determined from the radius using the formula

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

so, setting $r = \frac{5 }{\sqrt{\pi}}$ :

$V = \frac{4}{3} \pi \cdot \left ( \frac{5 }{\sqrt{\pi}} \right ) ^{3}$

Applying the Quotient of Radicals Rule:

$V = \frac{4}{3} \pi \cdot \frac{5^{3} }{\left ( \sqrt{\pi}\right ) ^{3}}$

$= \frac{4}{3} \pi \cdot \frac{125 }{\pi \sqrt{\pi} }$

$= \frac{500}{3\sqrt{\pi}}$

Rationalizing the denominator:

$V = \frac{500 \cdot \sqrt{\pi}}{3\sqrt{\pi} \cdot \sqrt{\pi}} = \frac{500 \sqrt{\pi}}{3 \pi}$

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SAT Math : Spheres

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #31 : Spheres

Example Question #32 : Spheres

Example Question #33 : Spheres

Example Question #34 : Spheres

Example Question #34 : Spheres

Example Question #32 : Spheres

Example Question #12 : How To Find The Volume Of A Sphere

Example Question #15 : How To Find The Volume Of A Sphere

Example Question #811 : Geometry

Example Question #17 : How To Find The Volume Of A Sphere

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