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SAT II Math II : Geometry

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #11 : Volume

Billy has a ice cream cone that consists of a cone and hemisphere. Suppose the cone has a height of 4 inches, and the radius of the hemisphere is 2 inches. Assuming that the combined shape is not irregular, what is the total volume?

Possible Answers:

$\frac{64}{9}\pi$

$\frac{16}{3}\pi$

$\frac{32}{3}\pi$

$\frac{8}{3}\pi$

$\frac{64}{3}\pi$

Correct answer:

$\frac{32}{3}\pi$

Explanation:

Write the volume for a cone.

$V_{cone } = \frac{1}{3}\pi r^2 h$

Substitute the radius and height. The radius is 2.

$V_{cone } = \frac{1}{3}\pi (2)^2 (4) = \frac{16}{3}\pi$

Write the volume for a hemisphere. This should be half the volume of the full sphere.

$V_{hemisphere} = \frac{1}{2}[ \frac{4}{3}\pi r^3] = \frac{2}{3}\pi r^3$

Substitute the radius.

$V_{hemisphere}= \frac{2}{3}\pi (2)^3 = \frac{16}{3}\pi$

Add the volumes of the cone and hemisphere to determine the total volume.

$\frac{16}{3}\pi+\frac{16}{3}\pi$

The answer is: $\frac{32}{3}\pi$

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Example Question #391 : Sat Subject Test In Math Ii

Find the volume of a sphere with a diameter of $\frac{\sqrt[3]{2}}{2}$ .

Possible Answers:

$\frac{1}{24}\pi$

$\frac{1}{9}\pi$

$\frac{1}{3}\pi$

$\frac{1}{6}\pi$

$\frac{1}{12}\pi$

Correct answer:

$\frac{1}{24}\pi$

Explanation:

Divide the diameter by two to get the radius. This is also the same as multiplying the diameter by one-half.

$r =\frac{\sqrt[3]{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt[3]{2}}{4}$

Write the formula for the volume of the sphere.

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

Substitute the radius.

$V=\frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})^3$

Simplify the terms.

$\frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})^3 = \frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})( \frac{\sqrt[3]{2}}{4})( \frac{\sqrt[3]{2}}{4}) = \frac{2}{3(4)(4)}\pi$

The answer is: $\frac{1}{24}\pi$

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Example Question #13 : 3 Dimensional Geometry

If the side of a cube is 3x^3 , what must be the volume?

Possible Answers:

27x^9

$27x^{27}$

9x^9

27x^3

27x^6

Correct answer:

27x^9

Explanation:

Write the formula for the volume of a cube.

V =s^3

Substitute the side length. When we are multiplying common bases with exponents, we are adding the exponents instead.

V =(3x^3)^3 = (3x^3)(3x^3)(3x^3) = 27x^9

The answer is: 27x^9

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Example Question #14 : 3 Dimensional Geometry

Determine the volume of a cube if the side length is $2\sqrt6$ .

Possible Answers:

$8\sqrt6$

$24\sqrt3$

$48\sqrt6$

$72\sqrt3$

Correct answer:

$48\sqrt6$

Explanation:

Write the formula for the volume of a cube.

V= s^3

Substitute the side length into the equation.

$V= (2\sqrt6)^3 = (2\sqrt6)(2\sqrt6)(2\sqrt6) = 8\cdot 6\cdot \sqrt6$

The answer is: $48\sqrt6$

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Example Question #15 : 3 Dimensional Geometry

The radius and the height of a cylinder are equal. If the volume of the cylinder is , what is the diameter of the cylinder?

Possible Answers:

3.21

1.77

2.95

1.61

Correct answer:

3.21

Explanation:

Recall how to find the volume of a cylinder:

$\text{Volume}=\pi r^2 h$

Since we know that the radius and the height are equal, we can rewrite the equation:

$\text{Volume}=\pi r^2 (r)=\pi r^3$

Using the given volume, find the length of the radius.

$13=\pi r^3$

r^3=4.138

r=1.605

Since the question asks you to find the diameter, multiply the radius by two.

d=3.21

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Example Question #392 : Sat Subject Test In Math Ii

Determine the volume of the cube if the side lengths are $\frac{2}{5}$ .

Possible Answers:

$\frac{6}{5}$

$\frac{8}{125}$

$\frac{8}{15}$

$\frac{6}{125}$

Correct answer:

$\frac{8}{125}$

Explanation:

The volume of a cube is: V =LWH

Substitute the dimensions.

$V =(\frac{2}{5})(\frac{2}{5})(\frac{2}{5}) = \frac{8}{125}$

The answer is: $\frac{8}{125}$

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Example Question #91 : Geometry

Determine the volume of a sphere with a diameter of 6.

Possible Answers:

$24\pi$

$296\pi$

$396\pi$

$48\pi$

$36\pi$

Correct answer:

$36\pi$

Explanation:

The radius is half the diameter, which is three.

Write the formula for the volume of a sphere.

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

Substitute the radius.

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

The answer is: $36\pi$

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Example Question #402 : Sat Subject Test In Math Ii

Find the volume of a sphere with a radius of $2\pi^4$ .

Possible Answers:

$\frac{32}{3}\pi ^{13}$

$\frac{16}{3}\pi ^{13}$

$8\pi ^{13}$

$\frac{16}{3}\pi ^{12}$

$\frac{4}{3}\pi ^{13}$

Correct answer:

$\frac{32}{3}\pi ^{13}$

Explanation:

Write the formula for the sphere.

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

Substitute the radius.

$V=\frac{4}{3} \pi (2\pi^4)^3 = \frac{4}{3} \pi (2\pi^4)(2\pi^4)(2\pi^4)$

The answer is: $\frac{32}{3}\pi ^{13}$

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Example Question #21 : 3 Dimensional Geometry

Find the volume of a sphere with a surface area of $36\pi$ .

Possible Answers:

$18\pi$

$27\pi$

$36\pi$

$96\pi$

$72\pi$

Correct answer:

$36\pi$

Explanation:

Write the surface area formula for a sphere.

$A= 4\pi r^2$

Substitute the area.

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

Divide both sides by $4\pi$ .

$\frac{36\pi}{4\pi}=\frac{ 4\pi r^2}{4\pi}$

r^2 =9

r=3

Write the formula for the volume of a sphere.

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

Substitute the radius.

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

The answer is: $36\pi$

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Example Question #1 : Surface Area

A rectangular swimming pool is meters deep throughout and meters wide. Its length is ten meters greater than twice its width. Which of the following expressions gives the total surface area, in square meters, of the inside of the pool?

Possible Answers:

$4W^{2} + 3DW + 10W+ 20D$

$4W^{2} + 6DW + 20W+ 20D$

$2W^{2} + 6DW + 10W+ 20D$

$2DW^{2}+ 10DW$

$2W^{2} + 3DW + 10W+ 10D$

Correct answer:

$2W^{2} + 6DW + 10W+ 20D$

Explanation:

Since the length of the pool is ten meters longer than twice its width , its length is 2W + 10 .

The inside of the pool can be seen as a rectangular prism. The bottom, or the base, has dimensions and 2W + 10 , so its area is the product of these:

$W (2W + 10) = 2W^{2}+ 10W$

The sides of the pool have depth . Two sides have width and therefore have area .

Two sides have width 2W + 10 and therefore have area

$D \left (2W + 10 \right ) = 2DW + 10D$

The total area of the inside of the pool is

$2W^{2}+ 10W + 2 (DW ) + 2 (2DW + 10D)$

$=2W^{2}+ 10W + 2 DW + 4DW + 20D$

$=2W^{2} + 6DW + 10W+ 20D$

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SAT II Math II : Geometry

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #11 : Volume

Example Question #391 : Sat Subject Test In Math Ii

Example Question #13 : 3 Dimensional Geometry

Example Question #14 : 3 Dimensional Geometry

Example Question #15 : 3 Dimensional Geometry

Example Question #392 : Sat Subject Test In Math Ii

Example Question #91 : Geometry

Example Question #402 : Sat Subject Test In Math Ii

Example Question #21 : 3 Dimensional Geometry

Example Question #1 : Surface Area

All SAT II Math II Resources

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