Spheres

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SAT Math › Spheres

Questions 1 - 10

Find the diameter of a sphere with a surface area of $20\pi$ .

$2\sqrt5$

$\sqrt5$

$\sqrt{10}$

$2\sqrt{10}$

Explanation

Write the formula to find the surface area of a sphere.

$A=4\pi r^2$

Substitute the area and solve for the radius.

$20\pi=4\pi r^2$

$r=\sqrt5$

The diameter is double the radius.

$d=2\sqrt5$

Find the diameter of a sphere with a surface area of $20\pi$ .

$2\sqrt5$

$\sqrt5$

$\sqrt{10}$

$2\sqrt{10}$

Explanation

Write the formula to find the surface area of a sphere.

$A=4\pi r^2$

Substitute the area and solve for the radius.

$20\pi=4\pi r^2$

$r=\sqrt5$

The diameter is double the radius.

$d=2\sqrt5$

A foam ball has a volume of 2 units and has a diameter of x. If a second foam ball has a radius of 2x, what is its volume?

128 units

16 units

8 units

4 units

2 units

Explanation

Careful not to mix up radius and diameter. First, we need to identify that the second ball has a radius that is 4 times as large as the first ball. The radius of the first ball is (1/2)x and the radius of the second ball is 2x. The volume of the second ball will be 43, or 64 times bigger than the first ball. So the second ball has a volume of 2 * 64 = 128.

Find the surface area of a sphere whose radius is $\frac{\pi}{2}$ .

$\pi^3$

$4\pi^2$

$\pi^2$

$\frac{\pi}{4}$

Explanation

The equation for the surface area of a sphere is $A = 4 \pi r^2$ where represents the sphere's radius.

With our radius-value, we find:

$A = 4 \pi r^2 = 4\pi (\frac{\pi}{2})^2 = 4\pi \cdot \frac{\pi^2}{4} = \pi^3$

A foam ball has a volume of 2 units and has a diameter of x. If a second foam ball has a radius of 2x, what is its volume?

128 units

16 units

8 units

4 units

2 units

Explanation

Find the surface area of a sphere whose radius is $\frac{\pi}{2}$ .

$\pi^3$

$4\pi^2$

$\pi^2$

$\frac{\pi}{4}$

Explanation

The equation for the surface area of a sphere is $A = 4 \pi r^2$ where represents the sphere's radius.

With our radius-value, we find:

$A = 4 \pi r^2 = 4\pi (\frac{\pi}{2})^2 = 4\pi \cdot \frac{\pi^2}{4} = \pi^3$

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?

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

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

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

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

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

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:

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

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?

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

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

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

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

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

Explanation

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:

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

Let be a point on a sphere, and be the point on the sphere farthest from . The shortest distance from to along the surface is $32 \pi$ . Give the surface area of the sphere.

$4,096 \pi$

$2,048 \pi$

$1,024 \pi$

$8,192 \pi$

$16,384 \pi$

Explanation

The diagram below shows the sphere with the points in question as well as the curve that connects them.

Sphere

The curve connecting them is a semicircle whose radius coincides with that of the sphere. Given radius , a semicircle has length

$l = \frac{1}{2} C= \frac{1}{2} \cdot 2 \pi r = \pi r$

Setting $l = 32\pi$ and solving for :

$\pi r = 32 \pi$

$\frac{\pi r }{\pi}= \frac{32\pi}{\pi}$

The surface area of a sphere, given its radius , is equal to

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

Setting :

$A= 4 \pi \cdot 32 ^{2}= 4 \cdot 32 ^{2} \cdot \pi = 4 \cdot 1,024\cdot \pi = 4,096 \pi$

Let be a point on a sphere, and be the point on the sphere farthest from . The shortest distance from to along the surface is $32 \pi$ . Give the surface area of the sphere.

$4,096 \pi$

$2,048 \pi$

$1,024 \pi$

$8,192 \pi$

$16,384 \pi$