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Distance Between Points

Distance Between Points

Assessment

Flashcard

•

Mathematics

•

8th Grade

•

Practice Problem

•

Hard

•
CCSS
HSG.GPE.B.7, 5.NBT.A.4, 8.G.B.8

+3

Standards-aligned

Created by

Wayground Content

FREE Resource

Student preview

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15 questions

Show all answers

1.

FLASHCARD QUESTION

Front

What is the distance formula used to find the distance between two points (x1, y1) and (x2, y2)?

Back

The distance formula is given by: d=sqrt((x2−x1)2+(y2−y1)2)d = \text{sqrt}((x2 - x1)^2 + (y2 - y1)^2) .

Tags

CCSS.HSG.GPE.B.7

2.

FLASHCARD QUESTION

Front

Calculate the distance between the points (0, 0) and (3, 4).

Back

Using the distance formula: d=sqrt((3−0)2+(4−0)2)=sqrt(9+16)=sqrt(25)=5d = \text{sqrt}((3 - 0)^2 + (4 - 0)^2) = \text{sqrt}(9 + 16) = \text{sqrt}(25) = 5 .

Tags

CCSS.HSG.GPE.B.7

3.

FLASHCARD QUESTION

Front

What is the distance between the points (-3, 4) and (4, -4)?

Back

Using the distance formula: d=sqrt((4−(−3))2+(−4−4)2)=sqrt(72+(−8)2)=sqrt(49+64)=sqrt(113)d = \text{sqrt}((4 - (-3))^2 + (-4 - 4)^2) = \text{sqrt}(7^2 + (-8)^2) = \text{sqrt}(49 + 64) = \text{sqrt}(113) .

Tags

CCSS.HSG.GPE.B.7

4.

FLASHCARD QUESTION

Front

How do you round a number to the nearest hundredth?

Back

To round to the nearest hundredth, look at the third decimal place. If it's 5 or more, round up the second decimal place by one. If it's less than 5, keep the second decimal place the same.

Tags

CCSS.5.NBT.A.4

5.

FLASHCARD QUESTION

Front

What is the distance between the points (-8.2, -6.4) and (-3.7, -4.8)?

Back

Using the distance formula: d=sqrt((−3.7−(−8.2))2+(−4.8−(−6.4))2)=sqrt(4.52+1.62)=sqrt(20.25+2.56)=sqrt(22.81) which is approximately 4.78d = \text{sqrt}((-3.7 - (-8.2))^2 + (-4.8 - (-6.4))^2) = \text{sqrt}(4.5^2 + 1.6^2) = \text{sqrt}(20.25 + 2.56) = \text{sqrt}(22.81) \text{ which is approximately } 4.78 .

Tags

CCSS.HSG.GPE.B.7

6.

FLASHCARD QUESTION

Front

What is the distance between the points (11, 15) and (24, 21)?

Back

Using the distance formula: d=sqrt((24−11)2+(21−15)2)=sqrt(132+62)=sqrt(169+36)=sqrt(205)d = \text{sqrt}((24 - 11)^2 + (21 - 15)^2) = \text{sqrt}(13^2 + 6^2) = \text{sqrt}(169 + 36) = \text{sqrt}(205) .

Tags

CCSS.HSG.GPE.B.7

7.

FLASHCARD QUESTION

Front

What is the significance of the Pythagorean theorem in finding distances?

Back

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This is the basis for the distance formula.

Tags

CCSS.8.G.B.8

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