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SAT II Math II : Geometry

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Example Questions

Example Question #5 : 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:

$\frac{1}{2}XYZ$

$\frac{2}{3}XYZ$

XYZ

$\frac{1}{3}XYZ$

$\frac{1}{6}XYZ$

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 #121 : Geometry

Which of the following equations represent a parabola?

Possible Answers:

y=2^x

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

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

y^2 = x^2

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

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 #1 : Circles, Ellipses, And Hyperbolas

Circle

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

Possible Answers:

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

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

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

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

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

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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Example Question #2 : Circles, Ellipses, And Hyperbolas

Circle

Refer to the above diagram. The circle has its center at the origin; is the point (-9, -12) . What is the length of the arc $\widehat{AB}$ , to the nearest tenth?

Possible Answers:

37.5

33.2

18.8

56.8

16.6

Correct answer:

33.2

Explanation:

First, it is necessary to determine the radius of the circle. This is the distance between (0,0) and (-9, -12) , so we apply the distance formula:

$r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$r = \sqrt{\left ( -9-0\right )^{2} +\left ( -12 -0 \right )^{2} }$

$r=\sqrt{ (-9)^{2}+(-12)^{2}}$

$r= \sqrt{81+144 }= \sqrt{225} = 15$

The circumference of the circle is

$C = 2 \pi r = 2 \pi \cdot 15 = 30 \pi$

Now we need to find the degree measure of the arc. We can do this best by examining this diagram:

Circle

The degree measure of $\widehat{AB}$ is also the measure $\theta$ of the central angle formed by the green radii. This is found using the relationship

$\tan \theta = \frac{y}{x}$

$\tan \theta = \frac{-12}{-9} \approx 1.333$

Using a calculator, we find that $\tan ^{-1} 1.3333 \approx 53.12^{\circ }$ . We can adjust for the location of :

$\theta \approx \left (180 - 53.12 \right )^{\circ } \approx 126.87^{\circ }$

which is the degree measure of the arc.

Now we can calculate the length of the arc:

$L \approx \frac{126.87}{360} \cdot C$

$L \approx \frac{126.87}{360} \cdot 30 \pi \approx 33.2$

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Example Question #1 : Circles, Ellipses, And Hyperbolas

On the coordinate plane, the vertices of a square are at the points with coordinates (7, 0), (0, 7), (-7, 0) , (0, -7) . Give the equation of the circle.

Possible Answers:

$(x- 7) ^{2}+ (y- 7) ^{2}= \frac{49}{4}$

$x^{2}+ y^{2}= \frac{7}{2}$

$x^{2}+ y^{2}= \frac{49}{2}$

$x^{2}+ y^{2}= \frac{49}{4}$

$(x- 7) ^{2}+ (y- 7) ^{2}= \frac{7}{2}$

Correct answer:

$x^{2}+ y^{2}= \frac{49}{2}$

Explanation:

The figure in question is below.

Incircle 1

The center of the circle can be seen to be the origin, so, if the radius is , the equation will be

$x^{2}+ y^{2} = r^{2}$ .

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint $(x_{m}, y_{m})$ of the segment with endpoints (7,0) and (0, 7) can be found by using the midpoint equations, setting $x_{1} = y_{2}= 7, x_{2} = y_{1}= 0$ :

$x_{m} = \frac{x_{1}+x_{2}}{2} = \frac{7+0}{2} = \frac{7}{2}$

$y_{m} = \frac{y_{1}+y_{2}}{2} = \frac{0+7 }{2} = \frac{7}{2}$

The circle passed though this midpoint $\left ( \frac{7}{2}, \frac{7}{2} \right )$ . The segment from this point to the origin (0,0) is a radius, and its length is equal to . Using the following form of the distance formula, since we only need the square of the radius:

$r ^{2} = (x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}$ ,

set $x_{1} = y_{1} = 0, x_{2}= y_{2}= \frac{7}{2}$ :

$r ^{2} =\left (\frac{7}{2}-0\right ) ^{2}+\left (\frac{7}{2}-0\right ) ^{2}$

$=\left (\frac{7}{2} \right ) ^{2}+\left (\frac{7}{2}\right ) ^{2}$

$= \frac{49}{4} + \frac{49}{4}$

$= \frac{49}{2}$

Substituting in the circle equation for $r ^{2}$ , we get the correct response,

$x^{2}+ y^{2}= \frac{49}{2}$

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Example Question #1 : Circles, Ellipses, And Hyperbolas

Find the diameter of the circle with the equation x^2-10x+y^2+2y+10=0 .

Possible Answers:

Correct answer:

Explanation:

Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:

(x-a)^2+(y-b)^2=r^2 , where the center of the circle is at (a,b) and the radius is .

x^2-10x+25+y^2+2y+1+10=25+1

(x-5)^2+(y+1)^2+10=26

(x-5)^2+(y+1)^2=16

From the equation, we know that r^2=16 .

$r=\sqrt{16}=4$

Since the radius is , double its length to find the length of the diameter. The length of the diameter is .

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Example Question #1 : Circles, Ellipses, And Hyperbolas

A triangle has its vertices at the points with coordinates (0,0) , (4, 4) , and (7, 0 ) . Give the equation of the circle that circumscribes it.

Possible Answers:

$x^{2}+y^{2}-7x-y=0$

$x^{2}+ y^{2}+7x -15y = 0$

$x^{2}+y^{2}+7x-y=0$

$x^{2}+y^{2}-7x-15 y=0$

None of these

Correct answer:

$x^{2}+y^{2}-7x-y=0$

Explanation:

The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.

In general form, the equation of a circle is

$x^{2}+ y^{2}+ Ax + By + C = 0$ .

Since the circle passes through the origin, substitute x = 0, y = 0 ; the equation becomes

$0^{2}+0^{2}+ A \cdot 0 + B \cdot 0 + C = 0$

0+0+0+0 + C = 0

C= 0

Therefore, we know the equation of any circle passing through the origin takes the form

$x^{2}+ y^{2}+ Ax + By = 0$

for some A , B .

Since the circle passes through (7,0) , substitute x = 7, y = 0 ; the equation becomes

$7^{2}+ 0^{2}+ A \cdot 7 + B \cdot 0 = 0$

49+7A= 0

Solving for :

7A = -49

A = -7

Now we know that the equation takes the form

$x^{2}+ y^{2}-7x + By = 0$

for some .

Since the circle passes through (4, 4) , substitute x = 7, y = 0 ; the equation becomes

$4^{2}+ 4^{2}- 7 \cdot 4 + B \cdot 4 = 0$

16+16- 28 + 4B = 0

4 + 4B = 0

Solving for :

4B = -4

B = -1

The general form of the equation of the circle is therefore

$x^{2}+ y^{2}-7x -y = 0$

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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 #5 : 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 #121 : Geometry

Example Question #1 : Circles, Ellipses, And Hyperbolas

Example Question #2 : Circles, Ellipses, And Hyperbolas

Example Question #1 : Circles, Ellipses, And Hyperbolas

Example Question #1 : Circles, Ellipses, And Hyperbolas

Example Question #1 : Circles, Ellipses, And Hyperbolas

All SAT II Math II Resources

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