Geometry - SAT Subject Test in Math II

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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.

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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 :

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

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

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

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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:

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

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

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

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

$d= \sqrt{91}$

$d \approx 9.5$

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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

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

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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

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

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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 :

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

$z_{1}+100 = -600$

$z_{1}=-700$

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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 :

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

$x_{1}+100 = 400$

$x_{1}=300$

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Question

Refer to the above diagram. Which of the following is not a valid name for $\angle DBF$ ?

Answer

$\angle B$ is the correct choice. A single letter - the vertex - can be used for an angle if and only if that angle is the only one with that vertex. This is not the case here. The three-letter names in the other choices all follow the convention of the middle letter being vertex and each of the other two letters being points on a different side of the angle.

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Question

Solve for x and y using the rules of quadrilateral

Answer

By using the rules of quadrilaterals we know that oppisite sides are congruent on a rhombus. Therefore, we set up an equation to solve for x. Then we will use that number and substitute it in for x and solve for y.

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Question

Use the rules of triangles to solve for x and y.

Answer

Using the rules of triangles and lines we know that the degree of a straight line is 180. Knowing this we can find x by creating and solving the following equation:

Now using the fact that the interior angles of a triangle add to 180 we can create the following equation and solve for y:

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Question

Use the facts of circles to solve for x and y.

Answer

In this question we use the rule that oppisite angles are congruent and a line is 180 degrees. Knowing these two facts we can first solve for x then solve for y.

Then:

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Question

Chords $\overline{AB}$ and $\overline{CD}$ intersect at point . $\overline{BX}$ is twice as long as $\overline{AX}$ ; and .

Give the length of $\overline{AX}$ .

Answer

If we let , then .

The figure referenced is below (not drawn to scale):

If two chords intersect inside the circle, then the cut each other so that for each chord, the product of the lengths of the two parts is the same; in other words,

$AX \cdot BX = CX \cdot DX$

Setting , and solving for :

$t \cdot 2t = 2 \cdot 9$

$2t^{2} =18$

$2t^{2} \div 2 =18 \div 2$

$t^{2} = 9$

Taking the positive square root of both sides:

$t = \sqrt{9} = 3$ ,

the correct length of $\overline{AX}$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. , , and $\angle ABC$ and $\angle ADB$ are right angles. What percent of $\Delta ABC$ is colored red?

Answer

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The area of $\Delta ADB$ , the shaded region, is half the products of its legs:

$A= \frac{1}{2} (AD) \times (DB ) =\frac{1}{2} \times 6 \times 12 = 36$

The area of $\Delta ABC$ is half the product of its hypoteuse, which we can see as the base, and the length of corresponding altitude $\overline{BD}$ :

$A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180$

$\Delta ADB$ comprises

$\frac{36}{180} \times \100 = 20 %$

of $\Delta ADC$ .

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Question

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a square garden (in green) surrounded by a dirt path (in orange). The dirt path is seven feet wide throughout. What is the area of the dirt path in square feet?

Answer

The area of the dirt path is the area of the outer square minus that of the inner square.

The outer square has sidelength 75 feet and therefore has area

$75 \times 75 = 5,625$ square feet.

The inner square has sidelength $75 - 2 \times 7 = 61$ feet and therefore has area

$61 \times 61 = 3,721$ square feet.

Subtract to get the area of the dirt path:

square feet.

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Question

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in orange). The dirt path is six feet wide throughout. Which of the following polynomials gives the area of the garden in square feet?

Answer

The length of the garden is $2 \times 6= 12$ feet less than that of the entire lot, or

;

The width of the garden is $2 \times 6= 12$ less than that of the entire lot, or

;

The area of the garden is their product:

$=2x \cdot x - 2x \cdot 12 - 12 \cdot x + 12 \cdot 12$

$= 2x ^{2} - 24x - 12 x + 144$

$= 2x ^{2} - 36 x + 144$

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Question

The circle in the above diagram has its center at the origin. To the nearest tenth, what is the area of the pink region?

Answer

First, it is necessary to determine the radius of the circle. This is the distance between and , so we apply the distance formula:

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

$r = \sqrt{\left ( -8 -0 \right )^{2} +\left ( 6-0\right )^{2} }=\sqrt{ 8^{2}+6^{2}} = \sqrt{64+36 }= \sqrt{100} = 10$

Subsequently, the area of the circle is

$A = \pi r^{2} = \pi \cdot 10^{2} = 100 \pi$

Now, we need to find the central angle of the shaded sector. This is found using the relationship

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

$\tan \theta = \frac{6}{-8} = -0.75$

Using a calculator, we find that $\tan^{-1}(-0.75) \approx -36.87^{\circ }$ ; since we want a degree measure between $90^{\circ }$ and $180 ^{\circ }$ , we adjust by adding $180 ^{\circ }$ , so

$\theta \approx \left (-36.87+ 180 \right ) ^{\circ } \approx 143.13^{\circ }$

The area of the sector is calculated as follows:

$A \approx \frac{143.13 }{360} \cdot 100 \pi \approx 124.9$

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Question

The above figure is a regular decagon. If , then to the nearest whole number, what is ?

Answer

As an interior angle of a regular decagon, $\angle B$ measures

$m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$ .

.

can be found using the Law of Cosines:

$\left ( AC\right )^{2} = \left ( AB\right )^{2}+ \left ( BC\right )^{2} - 2 \left ( AB\right ) \left ( BC\right ) \cos m\angle B$

$\left ( AC\right )^{2} = 100^{2}+ 100^{2} - 2 \cdot 100 \cdot 100 \cos 144^{\circ }$

$\left ( AC\right )^{2} \approx 10,000 + 10,000 - 20,000 (-0.8090)$

$\left ( AC\right )^{2} \approx 20,000 - (-16,180) \approx 36,180$

$AC \approx \sqrt{36,180} \approx 190$

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Question

You own a mug with a circular bottom. If the distance around the outside of the mug's base is $20 \pi cm$ what is the area of the base?

Answer

You own a mug with a circular bottom. If the distance around the outside of the mug's base is $20 \pi cm$ what is the area of the base?

Begin by solving for the radius:

$C=2\pi r$

$20 \pi cm=2\pi r$

Next, plug the radius back into the area formula and solve:

$A_{circle}=\pi r^2$

$A_{circle}=\pi (10cm)^2=100 \pi cm^2$

So our answer is:

$100 \pi cm^2$

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Question

You have a right triangle with a hypotenuse of 13 inches and a leg of 5 inches, what is the area of the triangle?

Answer

You have a right triangle with a hypotenuse of 13 inches and a leg of 5 inches, what is the area of the triangle?

So find the area of a triangle, we need the following formula:

$A_{tri}=\frac{1}{2}bh$

However, we only know one leg, so we only know b or h.

To find the other leg, we can either use Pythagorean Theorem, or recognize that this is a 5-12-13 triangle. Meaning, our final leg is 12 inches long.

To prove this:

$x=\sqrt{144in^2}=12in$

Now, we know both legs, let's just plug in and solve for area:

$A_{tri}=\frac{1}{2}(5in)(12in)=6in*5in=30in^2$

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Question

You have a rectangular-shaped rug which you want to put in your living room. If the rug is 12.5 feet long and 18 inches wide, what is the area of the rug?

Answer

You have a rectangular-shaped rug which you want to put in your living room. If the rug is 12.5 feet long and 18 inches wide, what is the area of the rug?

To begin, we need to realize two things.

Our given measurements are not in equivalent units, so we need to convert one of them before doing any solving.

The area of a rectangle is given by:

Now, let's convert 18 inches to feet, because it seems easier than 12.5 feet to inches:

$18inches*\frac{1ft}{12in}=1.5ft$

Now, using what we know from 2) we can find our answer

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