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SAT II Math II : 3-Dimensional Geometry

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

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Example Questions

Example Question #101 : Geometry

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:

$23.8 \textrm{ cm}$

$14.1 \textrm{ cm}$

$21.5 \textrm{ cm}$

$10.7 \textrm{ cm}$

$11.9 \textrm{ cm}$

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 #3 : 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:

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

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

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

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

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

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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Example Question #31 : 3 Dimensional Geometry

How many faces does a polyhedron with nine vertices and sixteen edges have?

Possible Answers:

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set V = 9 and E = 16 and solve for :

9 + F - 16 = 2

F - 7= 2

F = 9

The polyhedron has nine faces.

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

How many edges does a polyhedron with eight vertices and twelve faces have?

Possible Answers:

Insufficient information is given to answer the question.

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set V= 8 and F = 12 and solve for :

8 + 12 - E = 2

20 - E = 2

E = 18

The polyhedron has eighteen edges.

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

How many faces does a polyhedron with ten vertices and fifteen edges have?

Possible Answers:

Insufficient information is given to answer the question.

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set V = 10 and E = 15 and solve for :

10+ F - 15= 2

F - 5 = 2

F = 7

The polyhedron has seven faces.

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Example Question #7 : Faces, Face Area, And Vertices

How many faces does a polyhedron with ten vertices and sixteen edges have?

Possible Answers:

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2

Set V = 10 and E = 16 and solve for :

10+ F - 16= 2

F - 6 = 2

F = 8

The polyhedron has eight faces.

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Example Question #8 : Faces, Face Area, And Vertices

A convex polyhedron with eighteen faces and forty edges has how many vertices?

Possible Answers:

Correct answer:

Explanation:

The number of vertices, edges, and faces of a convex polygon— V, E, F —are related by the Euler's formula:

V-E + F = 2

Therefore, set F = 18, E = 40 and solve for :

V-40 + 18 = 2

V-22= 2

V= 24

The polyhedron has twenty-four faces.

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Example Question #9 : Faces, Face Area, And Vertices

How many edges does a polyhedron with fourteen vertices and five faces have?

Possible Answers:

Correct answer:

Explanation:

By Euler's Formula, the relationship between the number of vertices , the number of faces , and the number of edges of a polyhedron is

V + F - E = 2 .

Set V = 14 and F = 5 and solve for :

14+5 - E = 2

19 - E = 2

E = 17

The polyhedron has seventeen edges.

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Example Question #32 : 3 Dimensional Geometry

Which of the following numbers comes closest to the length of line segment in three-dimensional coordinate space whose endpoints are the origin and the point (3,4,5) ?

Possible Answers:

6.5

7.5

8.5

Correct answer:

Explanation:

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(3-0)^{2} +(4-0)^{2}+(5-0)^{2}}$

$d= \sqrt{3^{2} +4^{2}+5^{2}}$

$d= \sqrt{9+16+25}$

$d= \sqrt{50}$

$d \approx 7.1$

The closest of the five choices is 7.

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Example Question #1 : 3 Dimensional Axes And Coordinates

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Possible Answers:

$-1\frac{5}{6}$

$-\frac{5}{6}$

$2\frac{1}{6}$

$- 2\frac{5}{6}$

Correct answer:

$- 2\frac{5}{6}$

Explanation:

The midpoint formula for the -coordinate

$\frac{y_{1}+y_{2}}{2}=y_{M}$

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}= - \frac{1}{3}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}=- \frac{1}{3}$

$y_{1}+\frac{ 1 }{2}=- \frac{2}{3}$

$y_{1} =- 1\frac{1}{6}$

Now, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}=- 1\frac{1}{6}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}= - 1\frac{1}{6}$

$y_{1}+\frac{ 1 }{2} = - 2\frac{1}{3}$

$y_{1} = - 2\frac{5}{6}$

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SAT II Math II : 3-Dimensional Geometry

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

All SAT II Math II Resources

Example Questions

Example Question #101 : Geometry

Example Question #3 : Faces, Face Area, And Vertices

Example Question #31 : 3 Dimensional Geometry

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

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

Example Question #7 : Faces, Face Area, And Vertices

Example Question #8 : Faces, Face Area, And Vertices

Example Question #9 : Faces, Face Area, And Vertices

Example Question #32 : 3 Dimensional Geometry

Example Question #1 : 3 Dimensional Axes And Coordinates

All SAT II Math II Resources

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