Geometry: Finding Distance with the Pythagorean Theorem (CCSS.8.G.8)

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Common Core 8th Grade Math › Geometry: Finding Distance with the Pythagorean Theorem (CCSS.8.G.8)

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1

A ladder leans against a wall and reaches a window 12 feet above the ground. The ladder is 13 feet long. How far from the wall is the base of the ladder?

5 feet

CORRECT

12 feet

0

13 feet

0

25 feet

0

Explanation

The wall and ground form a right angle; the ladder is the hypotenuse. By the Pythagorean Theorem, base $=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5$ feet. This models a real ladder–wall right triangle.

2

A rectangular field is 24 meters long and 7 meters wide. What is the length of a straight path across the field from one corner to the opposite corner (the diagonal)?

31 meters

0

25 meters

CORRECT

17 meters

0

625 meters

0

Explanation

The diagonal is the hypotenuse of a right triangle with legs 24 and 7. By Pythagorean Theorem: $d=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25$ meters. This models the shortest straight path across the rectangle.

3

A $15$ ft ladder leans against a wall so its base is $9$ ft from the wall. How high up the wall does the ladder reach?

$15$ ft

0

$18$ ft

0

$6$ ft

0

$12$ ft

CORRECT

Explanation

The ground and wall are perpendicular, forming a right triangle with legs $9$ ft (base) and $h$ ft (height), and hypotenuse $15$ ft (ladder). By the Pythagorean Theorem, $h=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12$ ft.

4

A rectangular field is $50$ m wide and $120$ m long. What is the length of a straight path from one corner to the opposite corner (the diagonal)?

$170$ m

0

$70$ m

0

$130$ m

CORRECT

$110$ m

0

Explanation

The width and length are perpendicular, so the diagonal is the hypotenuse of a right triangle with legs $50$ m and $120$ m. By the Pythagorean Theorem, $d=\sqrt{50^2+120^2}=\sqrt{2500+14400}=\sqrt{16900}=130$ m.

5

What is the distance between the points (-3, 4) and (5, -2) on the coordinate plane?

8

0

10

CORRECT

6

0

12

0

Explanation

Use the distance formula, which comes from the Pythagorean Theorem: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Here, $\Delta x=5-(-3)=8$ and $\Delta y=-2-4=-6$. So $d=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10$. This applies $a^2+b^2=c^2$ to the horizontal and vertical legs between the two points.

6

A rectangular garden is 30 m by 40 m. What is the length of the diagonal path straight across the garden?

70 m

0

60 m

0

50 m

CORRECT

45 m

0

Explanation

Treat the garden as a right triangle with legs 30 m and 40 m; the diagonal is the hypotenuse. By the Pythagorean Theorem, $a^2+b^2=c^2$: $30^2+40^2=900+1600=2500$, so $c=\sqrt{2500}=50$ m. This is the classic real-world 3-4-5 right triangle scaled by 10.

7

A ladder is placed 5 ft from the base of a wall and reaches a window 12 ft above the ground. What is the length of the ladder?

7 ft

0

12 ft

0

17 ft

0

13 ft

CORRECT

Explanation

The wall and ground form a right angle, so the ladder is the hypotenuse. Apply the Pythagorean Theorem: $5^2+12^2=25+144=169$, so the ladder length is $\sqrt{169}=13$ ft. This models a real-world right triangle.

8

What is the distance between the points (-3, 4) and (5, -2) on a coordinate plane?

10

CORRECT

8

0

14

0

$\sqrt{28}$

0

Explanation

Use the distance formula from the Pythagorean Theorem: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Here, $d=\sqrt{(5-(-3))^2+(-2-4)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10$. The horizontal and vertical changes are the legs of a right triangle; the segment between the points is the hypotenuse.

9

A rectangular photo frame is 24 inches wide and 18 inches tall. What is the length of the diagonal?

24

0

18

0

30

CORRECT

42

0

Explanation

The diagonal of a rectangle is the hypotenuse of a right triangle with legs 24 and 18. By the Pythagorean Theorem, $d=\sqrt{24^2+18^2}=\sqrt{576+324}=\sqrt{900}=30$ inches.

10

What is the exact distance between the points (1, 2) and (7, 5)? Give your answer in simplest radical form.

$\sqrt{45}$

0

9

0

$\sqrt{9}$

0

$3\sqrt{5}$

CORRECT

Explanation

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Here, $d=\sqrt{(7-1)^2+(5-2)^2}=\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$. On the grid, the 6-by-3 right triangle's hypotenuse is the distance.