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SSAT Upper Level Math : Geometry

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

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All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : How To Find The Length Of A Line With Distance Formula

A line segment has endpoints at $(-2, -1)\text{ and }(8, 1)$ . Find the length of this line.

Possible Answers:

$6\sqrt{3}$

$2\sqrt{26}$

Correct answer:

$2\sqrt{26}$

Explanation:

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

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

Where,

(x_1,y_1)=(-2,-1) and (x_2,y_2)=(8,1) .

$\text{Distance}=\sqrt{(8-(-2))^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$

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Example Question #4 : How To Find The Length Of A Line With Distance Formula

Find the length of the line segment that has endpoints $(0, 4)\text{ and }(12, 6)$ .

Possible Answers:

10.659

12.412

8.451

12.166

Correct answer:

12.166

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)=(0,4) and (x_2,y_2)=(12,6) .

$\text{Distance}=\sqrt{(12-0)^2+(6-4)^2}=\sqrt{144+4}=\sqrt{148}=12.166$

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Example Question #151 : Geometry

Find the length of the line segment that has the endpoints $(2, 4)\text{ and }(7, 8)$ .

Possible Answers:

$\sqrt{29}$

$\sqrt{31}$

$\sqrt{41}$

Correct answer:

$\sqrt{41}$

Explanation:

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

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

Where,

(x_1,y_1)=(2,4) and (x_2,y_2)=(7,8) .

$\text{Distance}=\sqrt{(7-2)^2+(8-4)^2}=\sqrt{25+16}=\sqrt{41}$

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

One leg of a triangle has endpoints at the coordinates $(-1, -2)\text{ and }(-4, -1)$ . Find the length of this leg.

Possible Answers:

$\sqrt{10}$

$2\sqrt2$

Correct answer:

$\sqrt{10}$

Explanation:

Use the distance formula to find the length of the leg.

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

Where,

(x_1,y_1)=(-1,-2) and (x_2,y_2)=(-4,-1) .

$\text{Length}=\sqrt{(-4-(-1))^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}$

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Example Question #5 : How To Find The Length Of A Line With Distance Formula

A side of a square is graphed onto a coordinate plane. The side has endpoints at $(-3, 1)\text{ and }(4, -1)$ . Find the length of the side of the square.

Possible Answers:

$3\sqrt5$

$\sqrt{53}$

Correct answer:

$\sqrt{53}$

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)=(4,-1)

$\text{Distance}=\sqrt{(4-(-3))^2+(-1-1)^2}=\sqrt{49+4}=\sqrt{53}$

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Example Question #6 : How To Find The Length Of A Line With Distance Formula

The side of a triangle is graphed onto a coordinate plane. The side has endpoints at $(2, 2)\text{ and }(1, -5)$ . Find the length of this side.

Possible Answers:

$5\sqrt2$

$4\sqrt{3}$

Correct answer:

$5\sqrt2$

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)=(2,2) and (x_2,y_2)=(1, -5) .

$\text{Distance}=\sqrt{(1-2)^2+(-5-2)^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt2$

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Example Question #151 : Geometry

Find the length of a line segment that has endpoints at $(1, 2)\text{ and }(5, 1)$ .

Possible Answers:

$\sqrt{15}$

$\sqrt{17}$

Correct answer:

$\sqrt{17}$

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)=(1,2) and (x_2,y_2)=(5,1)

$\text{Distance}=\sqrt{(5-1)^2+(1-2)^2}=\sqrt{16+1}=\sqrt{17}$

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Example Question #101 : Coordinate Geometry

Find the length of a line segment that has endpoints at $\left(\frac{5}{2}, \frac{5}{2}\right)$ and $\left(4, \frac{4}{5}\right)$ .

Possible Answers:

4.19

2.27

5.14

3.78

Correct answer:

2.27

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,

$(x_1,y_1)=\left(\frac{5}{2}, \frac{5}{2} \right )$ and $(x_2,y_2)=\left(4,\frac{4}{5} \right )$ .

$\text{Distance}=\sqrt{(4-\frac{5}{2})^2+(\frac{4}{5}-\frac{5}{2})^2}=\sqrt{\frac{9}{4}+\frac{289}{100}}=2.27$

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Example Question #102 : Coordinate Geometry

Find the distance of a line segment that has endpoints at $\left(\frac{1}{2}, 4\right)$ and $\left(\frac{3}{2}, 2\right)$ .

Possible Answers:

$\sqrt3$

$\sqrt5$

Correct answer:

$\sqrt5$

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,

$(x_1,y_1)=\left(\frac{1}{2}, 4 \right )$ and $(x_2,y_2)=\left(\frac{3}{2}, 2 \right )$ .

$\text{Distance}=\sqrt{(\frac{3}{2}-\frac{1}{2})^2+(2-4)^2}=\sqrt{1+4}=\sqrt{5}$

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Example Question #103 : Coordinate Geometry

Find the length of a line segment that has endpoints at $\left(\frac{1}{3}, 2\right)$ and $\left(\frac{4}{3}, 3\right)$ .

Possible Answers:

$2\sqrt2$

$\sqrt2$

Correct answer:

$\sqrt2$

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,

$(x_1,y_1)=\left(\frac{1}{3}, 2\right)$ and $(x_2,y_2)=\left(\frac{4}{3}, 3 \right )$ .

$\text{Distance}=\sqrt{(\frac{4}{3}-\frac{1}{3})^2+(3-2)^2}=\sqrt{1+1}=\sqrt2$

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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 #3 : How To Find The Length Of A Line With Distance Formula

Example Question #4 : How To Find The Length Of A Line With Distance Formula

Example Question #151 : Geometry

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #5 : How To Find The Length Of A Line With Distance Formula

Example Question #6 : How To Find The Length Of A Line With Distance Formula

Example Question #151 : Geometry

Example Question #101 : Coordinate Geometry

Example Question #102 : Coordinate Geometry

Example Question #103 : Coordinate Geometry

All SSAT Upper Level Math Resources

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