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Common Core: 8th Grade Math : Grade 8

Study concepts, example questions & explanations for Common Core: 8th Grade Math

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All Common Core: 8th Grade Math Resources

7 Diagnostic Tests 75 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

In order to get to work, Jeff leaves home and drives 4 miles due north, then 3 miles due east, followed by 6 miles due north and, finally, 7 miles due east. What is the straight line distance from Jeff’s work to his home?

Possible Answers:

2√5

6√2

10√2

Correct answer:

10√2

Explanation:

Jeff drives a total of 10 miles north and 10 miles east. Using the Pythagorean theorem (a²+b²=c²), the direct route from Jeff’s home to his work can be calculated. 10²+10²=c². 200=c². √200=c. √100√2=c. 10√2=c

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Example Question #111 : Plane Geometry

Jim leaves his home and walks 10 minutes due west and 5 minutes due south. If Jim could walk a straight line from his current position back to his house, how far, in minutes, is Jim from home?

Possible Answers:

6√6

5√5

√10

√5

Correct answer:

5√5

Explanation:

By using Pythagorean Theorem, we can solve for the distance “as the crow flies” from Jim to his home:

10² + 5² = x²

100 + 25 = x²

√125 = x, but we still need to factor the square root

√125 = √25*5, and since the √25 = 5, we can move that outside of the radical, so

5√5= x

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Example Question #71 : Geometry

You leave on a road trip driving due North from Savannah, Georgia, at 8am. You drive for 5 hours at 60mph and then head due East for 2 hours at 50mph. After those 7 hours, how far are you Northeast from Savannah as the crow flies (in miles)?

Possible Answers:

$\sqrt{90000}$

$\sqrt{100000}$

$\sqrt{10000}$

$\sqrt{80000}$

$\sqrt{70000}$

Correct answer:

$\sqrt{100000}$

Explanation:

Distance = hours * mph

North Distance = 5 hours * 60 mph = 300 miles

East Distance = 2 hours * 50 mph = 100 miles

Use Pythagorean Theorem to determine Northeast Distance

300^{2 +}100²=NE²

90000 + 10000 = 100000 = NE²

NE = √100000

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Example Question #52 : Triangles

An airplane is 8 miles west and 15 miles south of its destination. Approximately how far is the plane from its destination, in miles?

Possible Answers:

$7\ miles$

$17\ miles$

$30\ miles$

$23\ miles$

Correct answer:

$17\ miles$

Explanation:

A right triangle can be drawn between the airplane and its destination.

Destination

15 miles Act_math_170_01 Airplane

8 miles

We can solve for the hypotenuse, x, of the triangle:

8² + 15² = x²

64 + 225 = x²

289 = x²

x = 17 miles

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Example Question #11 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

If James traveled north $30\textup { miles}$ and John traveled $40\textup { miles}$ west from the same town, how many miles away will they be from each other when they reach their destinations?

Possible Answers:

$40\textup { miles}$

$50\textup { miles}$

$10\textup { miles}$

$30\textup { miles}$

$60\textup { miles}$

Correct answer:

$50\textup { miles}$

Explanation:

The distances when put together create a right triangle.

The distance between them will be the hypotenuse or the diagonal side.

You use Pythagorean Theorem or $a^{2}+b^{2}=c^{2}$ to find the length.

So you plug and for and which gives you,

$900+1600=c^{2}$ or $2500=c^{2}$ .

Then you find the square root of each side and that gives you your answer of c=50 .

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Example Question #74 : Right Triangles

Justin travels $15\textup{ feet}$ to the east and $20\textup{ feet}$ to the north. How far away from his starting point is he now?

Possible Answers:

$45\textup{ ft}$

$35\textup{ ft}$

$25\textup{ ft}$

$22\textup{ ft}$

$30\textup{ ft}$

Correct answer:

$25\textup{ ft}$

Explanation:

This is solving for the hypotenuse of a triangle. Using the Pythagorean Theorem, which says that a^2+b^2=c^2

15^2 + 20^2 = c^2

225+400=c^2

625=c^2

25=c

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Example Question #11 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Use the Pythagorean Theorem to calculate the length of the line shown on the provided coordinate plane. Round the answer to the nearest tenth.

Possible Answers:

$7.7\textup{ units}$

$4.9\textup{ units}$

$5.7\textup{ units}$

$6\textup{ units}$

$3.4\textup{ units}$

Correct answer:

$5.7\textup{ units}$

Explanation:

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

1 2

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

a^2+b^2=c^2

In this equation:

$a \textup{ and }b=\textup{legs}$

$c=\textup{hypotenuse}$

We can count the number of units on the coordinate plane that were used to create the legs of our drawn triangle. Afterwards, we can use the Pythagorean Theorem to solve for the length of the hypotenuse, or the original diagonal line.

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

4^2+4^2=c^2

16+16=c^2

32=c^2

$\sqrt{32}=\sqrt{c^2}$

5.7=c

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Example Question #462 : Grade 8

Use the Pythagorean Theorem to calculate the length of the line shown on the provided coordinate plane. Round the answer to the nearest tenth.

Possible Answers:

$9.9\textup{ units}$

$11.2\textup{ units}$

$12.3\textup{ units}$

$8.7\textup{ units}$

$10\textup{ units}$

Correct answer:

$9.9\textup{ units}$

Explanation:

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

2 2

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

a^2+b^2=c^2

In this equation:

$a \textup{ and }b=\textup{legs}$

$c=\textup{hypotenuse}$

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

7^2+7^2=c^2

49+49=c^2

98=c^2

$\sqrt{98}=\sqrt{c^2}$

9.9=c

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Example Question #12 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Use the Pythagorean Theorem to calculate the length of the line shown on the provided coordinate plane. Round the answer to the nearest tenth.

Possible Answers:

$11.7\textup{ units}$

$14.5\textup{ units}$

$14.8\textup{ units}$

$10.1\textup{ units}$

$9.4\textup{ units}$

Correct answer:

$11.7\textup{ units}$

Explanation:

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

3 2

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

a^2+b^2=c^2

In this equation:

$a \textup{ and }b=\textup{legs}$

$c=\textup{hypotenuse}$

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

4^2+11^2=c^2

16+121=c^2

137=c^2

$\sqrt{137}=\sqrt{c^2}$

11.7=c

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Example Question #12 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Use the Pythagorean Theorem to calculate the length of the line shown on the provided coordinate plane. Round the answer to the nearest tenth.

Possible Answers:

$13\textup{ units}$

$12.7\textup{ units}$

$14.8\textup{ units}$

$15.1\textup{ units}$

$11.2\textup{ units}$

Correct answer:

$11.2\textup{ units}$

Explanation:

Notice that the diagonal line from the problem could be the hypotenuse of a right triangle. If we add two more lines, then we can create a closed figure in the shape of a triangle:

4 2

Let's use the Pythagorean Theorem to calculate the length of the line that represents the hypotenuse of a right triangle. The Pythagorean Theorem states that for right triangles:

a^2+b^2=c^2

In this equation:

$a \textup{ and }b=\textup{legs}$

$c=\textup{hypotenuse}$

In order to solve for this problem we want to substitute in the known side lengths for the triangle's legs:

5^2+10^2=c^2

25+100=c^2

125=c^2

$\sqrt{125}=\sqrt{c^2}$

11.2=c

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Common Core: 8th Grade Math : Grade 8

Study concepts, example questions & explanations for Common Core: 8th Grade Math

All Common Core: 8th Grade Math Resources

Example Questions

Example Question #22 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #111 : Plane Geometry

Example Question #71 : Geometry

Example Question #52 : Triangles

Example Question #11 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Example Question #74 : Right Triangles

Example Question #11 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Example Question #462 : Grade 8

Example Question #12 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

Example Question #12 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8

All Common Core: 8th Grade Math Resources

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