Solve the Pythagorean Theorem word problems.

Steps to Solve

1. Ladder Problem

Let $a$ be the distance the ladder touches the wall above the ground. The ladder forms the hypotenuse of a right triangle, with the distance from the wall as one leg and the height on the wall as the other leg. Using the Pythagorean theorem:

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

2. Soccer Field Problem

Let $d$ be the diagonal distance across the soccer field. The diagonal forms the hypotenuse of a right triangle with the width and length as legs. $d^2 = 90^2 + 120^2$ $d^2 = 8100 + 14400$ $d^2 = 22500$ $d = \sqrt{22500} = 150$ meters

3. Ladder and Wall Problem

Let $b$ be the distance from the base of the house to the ladder. $b^2 + 12^2 = 15^2$ $b^2 + 144 = 225$ $b^2 = 81$ $b = \sqrt{81} = 9$ feet

4. Rectangle Diagonal Problem

Let $d$ be the length of the diagonal. $d^2 = 10^2 + 15^2$ $d^2 = 100 + 225$ $d^2 = 325$ $d = \sqrt{325} \approx 18.03$ cm

5. Rectangle Length Problem

Let $l$ be the length of the rectangle. $15^2 + l^2 = 25^2$ $225 + l^2 = 625$ $l^2 = 400$ $l = \sqrt{400} = 20$ inches

6. Right Triangle Problem

a. If 8 and 12 are legs, let $c$ be the hypotenuse. $c^2 = 8^2 + 12^2$ $c^2 = 64 + 144$ $c^2 = 208$ $c = \sqrt{208} \approx 14.42$

b. If 8 is a leg and 12 is the hypotenuse, let $b$ be the other leg. $8^2 + b^2 = 12^2$ $64 + b^2 = 144$ $b^2 = 80$ $b = \sqrt{80} \approx 8.94$

7. Square Problem

Side length $s = \sqrt{81} = 9$ cm. Diagonal $d$: $d^2 = 9^2 + 9^2$ $d^2 = 81 + 81$ $d^2 = 162$ $d = \sqrt{162} \approx 12.73$ cm

8. Isosceles Triangle Problem

Let $h$ be the height. The height bisects the base, creating a right triangle with legs $h$ and $10/2 = 5$, and hypotenuse 20. $h^2 + 5^2 = 20^2$ $h^2 + 25 = 400$ $h^2 = 375$ $h = \sqrt{375} \approx 19.36$ cm

9. Baseball Diamond Problem

Let $d$ be the distance from home to second base. $d^2 = 90^2 + 90^2$ $d^2 = 8100 + 8100$ $d^2 = 16200$ $d = \sqrt{16200} \approx 127.28$ feet

10. Table and Doorway Problem

The table's diameter is 96 inches. Since the door is only 42 inches wide, the table won't fit through the door without tilting. Let's see if it fits diagonally. The doorway is a rectangle and the diagonal can be found using the Pythagorean theorem. $d^2 = 42^2 + 84^2$ $d^2 = 1764 + 7056$ $d^2 = 8820$ $d = \sqrt{8820} \approx 93.91$ inches. Since the diagonal (approximately 93.91 inches) is less than the diameter of the table, (96 inches) the table will not fit through the door even diagonally