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SAT II Math II : Geometry

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

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Surface Area

A water tank takes the shape of a closed rectangular prism whose exterior has height 30 feet, length 20 feet, and width 15 feet. Its walls are one foot thick throughout. What is the total surface area of the interior of the tank?

Possible Answers:

$2,204 \textrm{ ft}^{2}$

$1,223 \textrm{ ft}^{2}$

$1,102\textrm{ ft}^{2}$

$2,446\textup{ ft}^{2}$

$2,694\textup{ ft}^{2}$

Correct answer:

$2,204 \textrm{ ft}^{2}$

Explanation:

The height, length, and width of the interior tank are each two feet less than the corresponding dimension of the exterior of the tank, so the dimensions of the interior are 28, 18, and 13 feet. The surface area of the interior is what we are looking for here. It comprises six rectangles:

Two with area $28 \times 18 = 504$ square feet;

Two with area $28 \times 13 = 364$ square feet;

Two with area $18 \times 13 = 234$ square feet.

Add:

$2 \times 504 + 2 \times 364 + 2 \times 234 = 1,008 + 728 + 468 = 2,204$ square feet.

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Example Question #1 : Surface Area

What is the surface area of a cube with a side length of 5?

Possible Answers:

150

125

Correct answer:

150

Explanation:

If you were to take apart a cube so that you could lay it flat on a surface, you would be able to see that a cube is just made up of 6 identical squares. The area of one square is the length of the side squared, so the surface area of the cube would be denoted with the formula:

A=6*s^2

In this case the side length is 5, so plugging that into the formula will get the answer.

A=6*(5)^2=6*25=150

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Example Question #3 : Surface Area

Find the surface area of a rectangular prism with length, width, and height dimensions of $\frac{1}{2}$ , $\frac{3}{4}$ , and $\frac{5}{2}$ , respectively.

Possible Answers:

$\frac{15}{8}$

$\frac{15}{16}$

Correct answer:

Explanation:

Be careful not to confuse surface area with volume!

There are 6 faces in a rectangular prism, and we will need the sum of all the areas of each face.

Write the formula.

A=2(LW+LH+WH)

Substitute the dimensions.

$A= 2[(\frac{1}{2})(\frac{3}{4})+(\frac{1}{2})(\frac{5}{2})+(\frac{3}{4})(\frac{5}{2})]$

Evaluate each product in the bracket.

$A= 2[\frac{3}{8}+\frac{5}{4}+\frac{15}{8}] = \frac{3}{4}+\frac{10}{4}+\frac{15}{4}$

Combine the terms of the numerator.

$A = \frac{28}{4} =7$

The answer is:

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Example Question #101 : Geometry

What is the surface area of a cube with a side length of ?

Possible Answers:

54x^2

54x^6

27x^3

27x^2

Correct answer:

54x^2

Explanation:

Write the formula for the surface area of a cube.

A= 6s^2

Substitute the side length.

A= 6(3x)^2 = 6(3x)(3x) = 54x^2

The answer is: 54x^2

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Example Question #1 : Surface Area

Find the surface area of a sphere with a radius of 3.

Possible Answers:

$18\pi$

$36\pi$

$72\pi$

$48\pi$

$27\pi$

Correct answer:

$36\pi$

Explanation:

Write the formula for the surface area of a sphere.

$A=4\pi r^2$

Substitute the radius into the equation.

$A=4\pi(3)^2 = 4\pi (9) = 36\pi$

The answer is: $36\pi$

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Example Question #2 : Surface Area

Find the surface area of a sphere with a radius of 2.

Possible Answers:

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

$8\pi$

$16\pi$

$4\pi$

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

Correct answer:

$16\pi$

Explanation:

Write the formula for the surface area of a sphere.

$A=4\pi r^2$

Substitute the radius.

$A=4\pi (2)^2 = 16\pi$

The answer is: $16\pi$

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Example Question #1 : Surface Area

Find the surface area of a cube if the area of one of the square faces is .

Possible Answers:

$\textup{The answer cannot be determined.}$

256

Correct answer:

Explanation:

Write the formula for the surface area of a cube.

A = 6s^2

The area of a square side is already given. Substitute that as the s^2 term.

A=6(16) = 96

The answer is:

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Example Question #21 : 3 Dimensional Geometry

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

The total surface area of a given regular tetrahedron is 600 square centimeters. To the nearest tenth of a centimeter, what is the length of each edge?

Possible Answers:

$12.2 \textrm{ cm}$

$11.9 \textrm{ cm}$

$24.5 \textrm{ cm}$

$37.2\textrm{ cm}$

$18.6 \textrm{ cm}$

Correct answer:

$18.6 \textrm{ cm}$

Explanation:

The total surface area of the tetrahedron is 600 square centimeters; since the tetrahedron comprises four congruent faces, each has area $600 \div 4 = 150$ square centimeters.

The area of an equilateral triangle is given by the formula

$A = \frac{s^{2}\sqrt{3}}{4}$

Set A = 150 and solve for :

$\frac{s^{2}\sqrt{3}}{4} = 150$

$\frac{s^{2}\sqrt{3}}{4} \cdot \frac{4}{\sqrt{3}}=1 50 \cdot \frac{4}{\sqrt{3}}$

$s^{2} = \frac{600}{ \sqrt{3}} \approx \frac{600}{1.732} \approx 346.41$

$s \approx \sqrt{346.41} \approx 18.6$ centimeters.

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Example Question #2 : Faces, Face Area, And Vertices

A regular octahedron has eight congruent faces, each of which is an equilateral triangle.

The total surface area of a given regular octahedron is 400 square centimeters. To the nearest tenth of a centimeter, what is the length of each edge?

Possible Answers:

$14.1 \textrm{ cm}$

$10.7 \textrm{ cm}$

$11.9 \textrm{ cm}$

$21.5 \textrm{ cm}$

$23.8 \textrm{ cm}$

Correct answer:

$10.7 \textrm{ cm}$

Explanation:

The total surface area of the octahedron is 400 square centimeters; since the octahedron comprises eight congruent faces, each has area $400 \div 8 = 50$ square centimeters.

The area of an equilateral triangle is given by the formula

$A = \frac{s^{2}\sqrt{3}}{4}$

Set A = 50 and solve for :

$\frac{s^{2}\sqrt{3}}{4} = 50$

$\frac{s^{2}\sqrt{3}}{4} \cdot \frac{4}{\sqrt{3}}= 50 \cdot \frac{4}{\sqrt{3}}$

$s^{2} = \frac{200}{ \sqrt{3}} \approx \frac{200}{1.732} \approx 115.47$

$s \approx \sqrt{115.47} \approx 10.7$ centimeters.

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Example Question #1 : Faces, Face Area, And Vertices

A regular icosahedron has twenty congruent faces, each of which is an equilateral triangle.

A given regular icosahedron has edges of length four inches. Give the total surface area of the icosahedron.

Possible Answers:

$80 \sqrt{2} \textrm{ in}^{2}$

$40 \sqrt{3} \textrm{ in}^{2}$

$80 \textrm{ in}^{2}$

$40 \sqrt{2} \textrm{ in}^{2}$

$80 \sqrt{3} \textrm{ in}^{2}$

Correct answer:

$80 \sqrt{3} \textrm{ in}^{2}$

Explanation:

The area of an equilateral triangle is given by the formula

$A = \frac{s^{2}\sqrt{3}}{4}$

Since there are twenty equilateral triangles that comprise the surface of the icosahedron, the total surface area is

$SA = 20 A = 20 \cdot \frac{s^{2}\sqrt{3}}{4} = 5 s^{2}\sqrt{3}$

Substitute s = 4 :

$SA = 5 \cdot 4^{2}\sqrt{3} = 5 \cdot 16 \sqrt{3} = 80 \sqrt{3}$ square inches.

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All SAT II Math II Resources

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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 #2 : Surface Area

Example Question #1 : Surface Area

Example Question #3 : Surface Area

Example Question #101 : Geometry

Example Question #1 : Surface Area

Example Question #2 : Surface Area

Example Question #1 : Surface Area

Example Question #21 : 3 Dimensional Geometry

Example Question #2 : Faces, Face Area, And Vertices

Example Question #1 : Faces, Face Area, And Vertices

All SAT II Math II Resources

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