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SSAT Upper Level Math : Geometry

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Example Questions

Example Question #384 : Ssat Upper Level Quantitative (Math)

Find the length of a line segment that has endpoints at $(2, 1)\text{ and }(-1, 0)$ .

Possible Answers:

$2\sqrt2$

$\sqrt{10}$

Correct answer:

$\sqrt{10}$

Explanation:

Use the distance formula to find the length of the line segment.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

(x_1,y_1)=(2,1) and (x_2,y_2)=(-1,0) .

$\text{Distance}=\sqrt{(-1-2)^2+(0-1)^2}=\sqrt{9+1}=\sqrt{10}$

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Example Question #385 : Ssat Upper Level Quantitative (Math)

Find the length of a line segment that has endpoints at $(3, 1)\text{ and }(-2, -2)$ .

Possible Answers:

$4\sqrt3$

$\sqrt{34}$

Correct answer:

$\sqrt{34}$

Explanation:

Use the distance formula to find the length of the line segment.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

(x_1,y_1)=(3,1) and (x_2,y_2)=(-2,-2) .

$\text{Distance}=\sqrt{(-2-3)^2+(-2-1)^2}=\sqrt{25+9}=\sqrt{34}$

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Example Question #104 : Coordinate Geometry

If the two points on a line are $“\left(“\frac{1}{3}, \frac{1}{3}“\right)”$ and (2,4) , what is the approximate distance connecting the two points?

Possible Answers:

3.7

4.5

4.2

3.5

Correct answer:

Explanation:

Write the distance formula.

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Substitute the point values.

$d=\sqrt{\left( 4-\frac{1}{3}\right)^2+\left( 2-\frac{1}{3}\right)^2} = \sqrt{\left( \frac{11}{3}\right)^2+\left( \frac{5}{3}\right)^2}$

$d=\sqrt{13.\bar{4}+2.\bar{7}}= \sqrt{16.\bar{2}}= 4.027$

The approximated distance is 4.

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Example Question #105 : Coordinate Geometry

A segment on the coordinate plane whose endpoints are (A, A) and the origin has length 10. Calculate .

Possible Answers:

$A = \pm 5 \sqrt{3}$

$A = \pm 10 \sqrt{2}$

$A = \pm 5$

$A = \pm 5 \sqrt{2}$

$A = \pm 10$

Correct answer:

$A = \pm 5 \sqrt{2}$

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

If we set $x_{1}= y_{1} = 0, x_{2}= y_{2} = A, d=10$ , we can solve for :

$10= \sqrt{\left (A-0 \right )^{2}+\left (A-0 \right )^{2}}$

$10= \sqrt{ A ^{2}+A ^{2}}$

$10= \sqrt{ 2A ^{2}}$

$10^{2}=\left ( \sqrt{ 2A ^{2}} \right )^{2}$

$100 = 2A^{2 }$

$100 \div 2 = 2A^{2 } \div 2$

$50 = A^{2}$

$A = \pm \sqrt{50 } = \pm \sqrt{25} \cdot \sqrt{2} = \pm 5 \sqrt{2}$

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Example Question #11 : How To Find The Length Of A Line With Distance Formula

A segment on the coordinate plane whose endpoints are (A, 2A) and the origin has length 10. If is assumed to be positive, calculate .

Possible Answers:

A = 4

A = 2

$A = 2\sqrt{5}$

The correct answer is not given among the other responses.

$A = 4\sqrt{5}$

Correct answer:

$A = 2\sqrt{5}$

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

If we set $x_{1}= y_{1} = 0, x_{2}=2A, y_{2} = A, d=10$ , we can solve for :

$10= \sqrt{\left (2A-0 \right )^{2}+\left (A-0 \right )^{2}}$

$10= \sqrt{\left (2A \right )^{2}+A^{2}}$

$10= \sqrt{4A ^{2}+A^{2}}$

$10= \sqrt{5A ^{2} }$

$10^{2}=\left ( \sqrt{ 5A ^{2}} \right )^{2}$

$100 = 5A^{2}$

$100\div 5 = 5A^{2} \div 5$

$20 = A^{2}$

$A = \sqrt{20} = \sqrt{4} \cdot \sqrt{5} = 2\sqrt{5}$

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Example Question #111 : Coordinate Geometry

Give the length of a segment on the coordinate plane whose endpoints are $(\sqrt{11}, \sqrt{13})$ and the origin.

Possible Answers:

The correct answer is not among the other responses.

$\sqrt{290}$

$2 \sqrt{6}$

Correct answer:

$2 \sqrt{6}$

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Set $x_{1}= y_{1}=0, x_{2}= \sqrt{11}, y_{2} = \sqrt{13}$ :

$d = \sqrt{\left (\sqrt{11}-0 \right )^{2}+\left ( \sqrt{13}-0\right )^{2}}$

$= \sqrt{\left (\sqrt{11} \right )^{2}+\left ( \sqrt{13}\right )^{2}}$

$= \sqrt{11+13} = \sqrt{24} = \sqrt{4} \cdot \sqrt {6} = 2 \sqrt {6}$

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Example Question #390 : Ssat Upper Level Quantitative (Math)

Give the length of a segment on the coordinate plane whose endpoints are $\left (\frac{1}{5} , \frac{4}{5} \right )$ and $\left ( -\frac{3}{5}, \frac{1}{5} \right )$ .

Possible Answers:

$\frac{\sqrt{1}}{5}$

$\frac{1}{5}$

$\frac{\sqrt{18}}{5}$

$\frac{\sqrt{32}}{5}$

The correct answer is not among the other responses.

Correct answer:

The correct answer is not among the other responses.

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Set $x_{1}= \frac{1}{5}, y_{1}= \frac{4}{5}, x_{2}= -\frac{3}{5}, y_{2} = \frac{1}{5}$ :

$d = \sqrt{\left ( \frac{1}{5}- \left ( - \frac{3}{5}\right ) \right )^{2}+\left (\frac{4}{5}-\frac{1}{5} \right )^{2}}$

$= \sqrt{\left ( \frac{4}{5} \right )^{2}+\left (\frac{3}{5} \right )^{2}}$

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

$= \sqrt{ \frac{25}{25} }$

$= \sqrt{1}$

This is not given among the responses.

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Example Question #112 : Coordinate Geometry

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are (A+2, A-2) and the origin.

Possible Answers:

$\sqrt{ 2A^{2}- 8 }$

$2 \sqrt{2A}$

$\sqrt{ 2A^{2}+ 8A+8 }$

$\sqrt{ 2A^{2}+ 8 }$

$\sqrt{ 2A^{2}- 8A+8 }$

Correct answer:

$\sqrt{ 2A^{2}+ 8 }$

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Setting $x_{1}= y_{1} = 0, x_{2}=A+2, y_{2} = A-2$ , we can find our expression:

$d = \sqrt{\left ( \left [A+2 \right )-0\right ]^{2}+\left [\left (A-2 \right )-0 \right ] ^{2}}$ $= \sqrt{\left ( A+2 \right ) ^{2}+ \left (A-2 \right ) ^{2}}$

$= \sqrt{\left ( A^{2}+4A+4 \right )+\left ( A^{2}-4A+4 \right ) }$

$= \sqrt{ 2A^{2}+ 8 }$

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Example Question #113 : Coordinate Geometry

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are (A,A) and (3,3)

Possible Answers:

A - 3

$\sqrt{2A^{2}-12A+18 }$

$\sqrt{2A^{2} -18 }$

$\sqrt{2A^{2} +18 }$

$\sqrt{2A^{2}+12A+18 }$

Correct answer:

$\sqrt{2A^{2}-12A+18 }$

Explanation:

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Setting $x_{1}= y_{1} =3, x_{2}= y_{2} = A$ , we can find our expression:

$d = \sqrt{\left (A-3 \right )^{2}+\left (A-3\right )^{2}}$

$= \sqrt{\left (A^{2}-6A+9 \right ) + \left (A^{2}-6A+9 \right ) }$

$= \sqrt{2A^{2}-12A+18 }$

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Example Question #1477 : Concepts

A line segment on the coordinate plane has endpoints (a-5,-b) and (b+8, a-4) . In terms of and , as applicable, give the -coordinate of its midpoint.

Possible Answers:

a - b - 4

a + b + 3

$\frac{a + b + 3}{2}$

$\frac{a - b - 4}{2}$

$\frac{a - b - 13}{2}$

Correct answer:

$\frac{a + b + 3}{2}$

Explanation:

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$ .

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All SSAT Upper Level Math Resources

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SSAT Upper Level Math : Geometry

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #384 : Ssat Upper Level Quantitative (Math)

Example Question #385 : Ssat Upper Level Quantitative (Math)

Example Question #104 : Coordinate Geometry

Example Question #105 : Coordinate Geometry

Example Question #11 : How To Find The Length Of A Line With Distance Formula

Example Question #111 : Coordinate Geometry

Example Question #390 : Ssat Upper Level Quantitative (Math)

Example Question #112 : Coordinate Geometry

Example Question #113 : Coordinate Geometry

Example Question #1477 : Concepts

All SSAT Upper Level Math Resources

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