Pythagorean Theorem Word Problems: 1. The bottom of a ladder must be placed 3 feet from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall... Pythagorean Theorem Word Problems: 1. The bottom of a ladder must be placed 3 feet from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall? 2. A soccer field is a rectangle 90 meters wide and 120 meters long. The coach asks players to run from one corner to the corner diagonally across. What is this distance? 3. How far from the base of the house do you need to place a 15-foot ladder so that it exactly reaches the top of a 12-foot tall wall? 4. What is the length of the diagonal of a 10 cm by 15 cm rectangle? 5. The diagonal of a rectangle is 25 in. The width is 15 inches. What is the length? 6. Two sides of a right triangle are 8 and 12. a. Find the missing side if 8 and 12 are legs. b. Find the missing side if 8 and 12 are a leg and hypotenuse. 7. The area of a square is 81 square centimeters. Find the length of a side. Find the length of the diagonal. 8. An isosceles triangle has congruent sides of 20 cm. The base is 10 cm. Find the height of the triangle. 9. A baseball diamond is a square that is 90 feet on each side. What is the distance from home to second base? 10. Jill's front door is 42" wide and 84" tall. She purchased a circular table that is 96 inches in diameter. Will the table fit through the front door? Explain using approximations. Read more

Steps to Solve

1. Problem 1: Ladder and Wall

We're given the distance of the ladder's base from the wall (3 feet) and the ladder's length (12 feet). We need to find the height the ladder reaches on the wall. Let's call this height $h$. This forms a right triangle, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$, where $a = 3$, $b = h$, and $c = 12$.

$3^2 + h^2 = 12^2$ $9 + h^2 = 144$ $h^2 = 144 - 9$ $h^2 = 135$ $h = \sqrt{135} \approx 11.62$ feet

2. Problem 2: Soccer Field Diagonal

A soccer field is a rectangle with width 90 meters and length 120 meters. The diagonal forms the hypotenuse of a right triangle. Let the diagonal be $d$. Using Pythagorean theorem: $90^2 + 120^2 = d^2$

$8100 + 14400 = d^2$ $22500 = d^2$ $d = \sqrt{22500} = 150$ meters

3. Problem 3: Ladder and Wall (again)

We have a 15-foot ladder and a 12-foot wall. We need to find the distance from the base of the house to the ladder's base. Let this distance be $x$. Using the Pythagorean theorem: $x^2 + 12^2 = 15^2$

$x^2 + 144 = 225$ $x^2 = 225 - 144$ $x^2 = 81$ $x = \sqrt{81} = 9$ feet

4. Problem 4: Rectangle Diagonal

We have a rectangle of 10 cm by 15 cm. The diagonal, $d$, can be found by: $10^2 + 15^2 = d^2$

$100 + 225 = d^2$ $325 = d^2$ $d = \sqrt{325} \approx 18.03$ cm

5. Problem 5: Rectangle Length

The diagonal is 25 inches and the width is 15 inches. Let the length be $l$. Then, $15^2 + l^2 = 25^2$

$225 + l^2 = 625$ $l^2 = 625 - 225$ $l^2 = 400$ $l = \sqrt{400} = 20$ inches

6. Problem 6: Right Triangle Sides a. 8 and 12 are legs. We need to find the hypotenuse, $c$. $8^2 + 12^2 = c^2$.

$64 + 144 = c^2$ $208 = c^2$ $c = \sqrt{208} \approx 14.42$

b. 8 and 12 are a leg and hypotenuse. Let's assume 12 is the hypotenuse. Then $8^2 + b^2 = 12^2$, where $b$ is the other leg.

$64 + b^2 = 144$ $b^2 = 144 - 64$ $b^2 = 80$ $b = \sqrt{80} \approx 8.94$

7. Problem 7: Square Area and Diagonal

Area of square = 81 $cm^2$. If $s$ is the side length, $s^2 = 81$, so $s = \sqrt{81} = 9$ cm. The diagonal, $d$, satisfies $s^2 + s^2 = d^2$. So, $9^2 + 9^2 = d^2$

$81 + 81 = d^2$ $162 = d^2$ $d = \sqrt{162} \approx 12.73$ cm

8. Problem 8: Isosceles Triangle Height

Isosceles triangle with congruent sides of 20 cm, and base = 10 cm. The height bisects the base, creating a right triangle with hypotenuse 20 and one leg 5 (half of 10). Let the height be $h$.

$h^2 + 5^2 = 20^2$ $h^2 + 25 = 400$ $h^2 = 375$ $h = \sqrt{375} \approx 19.36$ cm

9. Problem 9: Baseball Diamond Distance

Baseball diamond is a square, 90 feet on each side. Home to second base is the diagonal, $d$. $90^2 + 90^2 = d^2$

$8100 + 8100 = d^2$ $16200 = d^2$ $d = \sqrt{16200} \approx 127.28$ feet

10. Problem 10: Table and Door

Door is 42" wide and 84" tall. Table diameter is 96".

The table will not fit through the door because the diameter of the table (96 inches) is greater than both the width (42 inches) and the height (84 inches) of the door. Even if tilted, the table's diameter is still too large, so it won't fit.