3-Dimensional Geometry - SAT Subject Test in Math II

Card 0 of 20

Question

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 ?

Answer

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.

Compare your answer with the correct one above

Question

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 .

Answer

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 :

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

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

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

Compare your answer with the correct one above

Question

A line segment in three-dimensional space has endpoints with Cartesian coordinates and $\left (5, -2, 7 \right )$ . To the nearest tenth, give the length of the segment.

Answer

Use the three-dimensional version of the distance formula:

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

$d= \sqrt{(-4-5)^{2} +(1-(-2))^{2}+(8-7)^{2}}$

$d= \sqrt{(-9)^{2} +3^{2}+1^{2}}$

$d= \sqrt{81+9+1}$

$d= \sqrt{91}$

$d \approx 9.5$

Compare your answer with the correct one above

Question

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates , and the origin. Give the volume of this pyramid.

Answer

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length 6; the segment connecting the origin and is the other leg of the base and has length 9; the area of the base is therefore

$B = \frac{1}{2} \cdot 6 \cdot 9 = 27$

The segment connecting the origin and is the altitude; its length - the height of the pyramid - is 12.

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 27 \cdot 12 = 108$

Compare your answer with the correct one above

Question

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates , and the origin. Give the volume of this pyramid.

Answer

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length ; the segment connecting the origin and is the other leg of the base and has length ; the area of the base is therefore

$B = \frac{1}{2} XY$

The segment connecting the origin and is the altitude; its length - the height of the pyramid - is .

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot \frac{1}{2}XY \cdot Z= \frac{1}{6}XYZ$

Compare your answer with the correct one above

Question

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 .

Answer

The midpoint formula for the -coordinate

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

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

First, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -100$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}=-100$

$z_{1}+100 = -200$

$z_{1}=-300$

Now, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -300$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}= -300$

$z_{1}+100 = -600$

$z_{1}=-700$

Compare your answer with the correct one above

Question

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 .

Answer

The midpoint formula for the -coordinate

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

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

First, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 150$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 150$

$x_{1}+100 = 300$

$x_{1}=200$

Now, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 200$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 200$

$x_{1}+100 = 400$

$x_{1}=300$

Compare your answer with the correct one above

Question

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

Answer

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

Set and and solve for :

The polyhedron has eighteen edges.

Compare your answer with the correct one above

Question

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?

Answer

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 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.

Compare your answer with the correct one above

Question

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?

Answer

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 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.

Compare your answer with the correct one above

Question

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.

Answer

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 :

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

Compare your answer with the correct one above

Question

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

Answer

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

.

Set and and solve for :

The polyhedron has seventeen edges.

Compare your answer with the correct one above

Question

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

Answer

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

Set and and solve for :

The polyhedron has nine faces.

Compare your answer with the correct one above

Question

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

Answer

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

Set and and solve for :

The polyhedron has seven faces.

Compare your answer with the correct one above

Question

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

Answer

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

Set and and solve for :

The polyhedron has eight faces.

Compare your answer with the correct one above

Question

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

Answer

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

Therefore, set and solve for :

The polyhedron has twenty-four faces.

Compare your answer with the correct one above

Question

A convex polyhedron has twenty faces and thirty-six vertices. How many edges does it have?

Answer

The number of vertices , edges , and faces of any convex polyhedron are related by By Euler's Formula:

Setting and solving for :

The polyhedron has 54 edges.

Compare your answer with the correct one above

Question

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?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

Since the length of the pool is ten meters longer than twice its width , its length is .

The inside of the pool can be seen as a rectangular prism. The bottom, or the base, has dimensions and , 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 and therefore have area

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

The total area of the inside of the pool is

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Tap the card to reveal the answer