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SAT II Math II : SAT Subject Test in Math II

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

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Example Questions

Example Question #1 : 3 Dimensional Axes And Coordinates

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:

8.5

7.5

6.5

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 #2 : 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:

$-\frac{5}{6}$

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

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

$2\frac{1}{6}$

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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Example Question #3 : 3 Dimensional Axes And Coordinates

A line segment in three-dimensional space has endpoints with Cartesian coordinates (-4, 1, 8) and $\left (5, -2, 7 \right )$ . To the nearest tenth, give the length of the segment.

Possible Answers:

15.1

9.5

3.0

9.1

1.7

Correct answer:

9.5

Explanation:

Use the three-dimensional version of the distance formula:

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

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

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

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

$d= \sqrt{91}$

$d \approx 9.5$

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Example Question #4 : 3 Dimensional Axes And Coordinates

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates (6,0,0) ,(0,9,0), (0,0,12) , and the origin. Give the volume of this pyramid.

Possible Answers:

108

648

162

Correct answer:

108

Explanation:

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 (6,0,0) ,(0,9,0) , and the origin, and, as its altitude, the segment with the origin and (0,0, 12) as its endpoints.

The segment connecting the origin and (6,0,0) is one leg of the base and has length 6; the segment connecting the origin and (0,9,0) 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 (0,0, 12) is the altitude; its length - the height of the pyramid - is 12.

The volume of the pyramid is

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

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Example Question #3 : 3 Dimensional Axes And Coordinates

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates (X,0,0) ,(0,Y,0), (0,0,Z) , and the origin. Give the volume of this pyramid.

Possible Answers:

$\frac{2}{3}XYZ$

$\frac{1}{2}XYZ$

$\frac{1}{3}XYZ$

$\frac{1}{6}XYZ$

XYZ

Correct answer:

$\frac{1}{6}XYZ$

Explanation:

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 (X,0,0) ,(0,Y,0) , and the origin, and, as its altitude, the segment with the origin and (0,0, Z) as its endpoints.

The segment connecting the origin and (X,0,0) is one leg of the base and has length ; the segment connecting the origin and (0,Y,0) 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 (0,0, Z) is the altitude; its length - the height of the pyramid - is .

The volume of the pyramid is

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

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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:

Correct answer:

Explanation:

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 :

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

$z_{1}+100 = -600$

$z_{1}=-700$

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Example Question #3 : 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:

500

300

250

Correct answer:

300

Explanation:

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 :

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

$x_{1}+100 = 400$

$x_{1}=300$

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Example Question #1 : Other 3 Dimensional Geometry

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

Possible Answers:

Correct answer:

Explanation:

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

V- E + F = 2

Setting V= 36, F = 20 and solving for :

36- E + 20 = 2

56- E = 2

E = 54

The polyhedron has 54 edges.

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Example Question #1 : Coordinate Geometry

Which of the following equations represent a parabola?

Possible Answers:

$y=\frac{1}{2x^2}$

$y=\frac{1}{2}x^2$

$y=\frac{1}{2x}$

y=2^x

y^2 = x^2

Correct answer:

$y=\frac{1}{2}x^2$

Explanation:

The parabola is represented in the form y=ax^2+bx+c . If there is a variable in the denominator or as an exponent, it is not a parabola.

The only equation that has an order of two is: $y=\frac{1}{2}x^2$

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Example Question #2 : Coordinate Geometry

Circle

Refer to the above figure. The circle has its center at the origin. What is the equation of the circle?

Possible Answers:

$\frac{x^{2}}{36}- \frac{y^{2} }{64}= 1$

$x^{2}+ y^{2} = 100$

$\frac{x^{2}}{36}+ \frac{y^{2} }{64}= 1$

$\frac{y^{2} }{64 } - \frac{x^{2}}{36} = 1$

$x^{2}+ y^{2} = 1 0$

Correct answer:

$x^{2}+ y^{2} = 100$

Explanation:

The equation of a circle with center (h, k) and radius is

$(x-h)^{2}+ \left ( y - k\right )^{2} = r^{2}$

The center is at the origin, or (0,0) , so h = k = 0 . To find $r^{2}$ , use the distance formula as follows:

$r^{2} = \left ( 6-0\right )^{2} +\left ( 8 -0 \right )^{2} = 6^{2}+8^{2} = 36 + 64 = 100$

Note that we do not actually need to find .

We can now write the equation of the circle:

$(x-0)^{2}+ \left ( y - 0\right )^{2} = 100$

$x^{2}+y^{2} = 100$

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #1 : 3 Dimensional Axes And Coordinates

Example Question #2 : 3 Dimensional Axes And Coordinates

Example Question #3 : 3 Dimensional Axes And Coordinates

Example Question #4 : 3 Dimensional Axes And Coordinates

Example Question #3 : 3 Dimensional Axes And Coordinates

Example Question #1 : 3 Dimensional Axes And Coordinates

Example Question #3 : 3 Dimensional Axes And Coordinates

Example Question #1 : Other 3 Dimensional Geometry

Example Question #1 : Coordinate Geometry

Example Question #2 : Coordinate Geometry

All SAT II Math II Resources

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