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3D Applications of Pythagorean Theorem Warmup

3D Applications of Pythagorean Theorem Warmup

Assessment

Flashcard

•

Mathematics

•

8th Grade

•

Practice Problem

•

Hard

•
CCSS
8.G.B.8, HSG.GPE.B.7, 8.G.B.7

+2

Standards-aligned

Created by

Wayground Content

FREE Resource

Student preview

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14 questions

Show all answers

1.

FLASHCARD QUESTION

Front

What is the Pythagorean Theorem?

Back

The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b). It can be expressed as: c² = a² + b².

Tags

CCSS.8.G.B.8

2.

FLASHCARD QUESTION

Front

How do you calculate the length of a ladder leaning against a wall?

Back

Use the Pythagorean Theorem. If the height from the ground to the point where the ladder touches the wall is 'a' and the distance from the wall to the base of the ladder is 'b', then the length of the ladder 'c' can be calculated as: c = √(a² + b²).

Tags

CCSS.8.G.B.8

3.

FLASHCARD QUESTION

Front

If a ladder reaches 12 feet high and is placed 9 feet from the wall, what is the length of the ladder?

Back

Using the Pythagorean Theorem: c = √(12² + 9²) = √(144 + 81) = √225 = 15 feet.

Tags

CCSS.8.G.B.8

4.

FLASHCARD QUESTION

Front

What is the diagonal length of a rectangle with sides 6 inches and 8 inches?

Back

Using the Pythagorean Theorem: c = √(6² + 8²) = √(36 + 64) = √100 = 10 inches.

Tags

CCSS.8.G.B.8

5.

FLASHCARD QUESTION

Front

In 3D modeling, why is the Pythagorean Theorem important?

Back

It is used to measure distances between objects, calculate texture dimensions, and determine the correct scale of models, ensuring accuracy in virtual environments.

Tags

CCSS.8.G.B.8

6.

FLASHCARD QUESTION

Front

What is the distance from the base of a house to place a 15' ladder to reach a 12' wall?

Back

Using the Pythagorean Theorem: b = √(15² - 12²) = √(225 - 144) = √81 = 9 feet.

Tags

CCSS.8.G.B.8

7.

FLASHCARD QUESTION

Front

What is the formula for finding the distance between two points in 3D space?

Back

The distance formula in 3D is: d = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²).

Tags

CCSS.HSG.GPE.B.7

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