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 wal... 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. Find the missing side if: a. 8 and 12 are legs. b. 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. Ladder Problem

   • The ladder, wall, and ground form a right triangle. The ladder is the hypotenuse (12 ft), the distance from the wall is one leg (3 ft), and the height the ladder reaches on the wall is the other leg which we want to find.
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $a = 3$, $c = 12$, and we want to find $b$.
   • Substitute and solve: $3^2 + b^2 = 12^2 \Rightarrow 9 + b^2 = 144 \Rightarrow b^2 = 144 - 9 = 135 \Rightarrow b = \sqrt{135} \approx 11.62$ feet.

2. Soccer Field Problem

   • The soccer field is a rectangle, and running diagonally across means finding the length of the diagonal. The diagonal, width, and length form a right triangle.
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $a = 90$, $b = 120$, and we want to find $c$.
   • Substitute and solve: $90^2 + 120^2 = c^2 \Rightarrow 8100 + 14400 = c^2 \Rightarrow c^2 = 22500 \Rightarrow c = \sqrt{22500} = 150$ meters.

3. Ladder and Wall Problem

   • Similar to problem 1, the ladder, wall, and ground form a right triangle. The ladder is the hypotenuse (15 ft), the wall is one leg (12 ft), and the distance from the base of the house to the ladder is the other leg which we want to find.
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $b = 12$, $c = 15$, and we want to find $a$.
   • Substitute and solve: $a^2 + 12^2 = 15^2 \Rightarrow a^2 + 144 = 225 \Rightarrow a^2 = 225 - 144 = 81 \Rightarrow a = \sqrt{81} = 9$ feet.

4. Rectangle Diagonal Problem

   • The diagonal of the rectangle divides it into two right triangles. The diagonal is the hypotenuse, and the sides of the rectangle are the legs.
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $a = 10$, $b = 15$, and we want to find $c$.
   • Substitute and solve: $10^2 + 15^2 = c^2 \Rightarrow 100 + 225 = c^2 \Rightarrow c^2 = 325 \Rightarrow c = \sqrt{325} \approx 18.03$ cm.

5. Rectangle Length Problem

   • Similar to problem 4, the diagonal of the rectangle divides it into two right triangles. We know the hypotenuse (diagonal) and one leg (width).
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $a = 15$, $c = 25$, and we want to find $b$.
   • Substitute and solve: $15^2 + b^2 = 25^2 \Rightarrow 225 + b^2 = 625 \Rightarrow b^2 = 625 - 225 = 400 \Rightarrow b = \sqrt{400} = 20$ inches.

6. Right Triangle Problems

   • a. Find the missing side if 8 and 12 are legs: $8^2 + 12^2 = c^2 \Rightarrow 64 + 144 = c^2 \Rightarrow c^2 = 208 \Rightarrow c = \sqrt{208} \approx 14.42$
   • b. Find the missing side if 8 and 12 are a leg and hypotenuse: $8^2 + b^2 = 12^2 \Rightarrow 64 + b^2 = 144 \Rightarrow b^2 = 144 - 64 = 80 \Rightarrow b = \sqrt{80} \approx 8.94$

7. Square Problem

   • The area of a square is $s^2$, where $s$ is the side length. If the area is 81, then $s^2 = 81 \Rightarrow s = \sqrt{81} = 9$ cm.
   • The diagonal of the square can be found using the Pythagorean Theorem: $9^2 + 9^2 = d^2 \Rightarrow 81 + 81 = d^2 \Rightarrow d^2 = 162 \Rightarrow d = \sqrt{162} \approx 12.73$ cm.

8. Isosceles Triangle Problem

   • The height of the isosceles triangle bisects the base, creating two right triangles. The congruent side is the hypotenuse (20 cm), half the base is one leg (5 cm), and the height is the other leg.
   • Use the Pythagorean Theorem: $a^2 + b^2 = c^2$, where $a = 5$, $c = 20$, and we want to find $b$.
   • Substitute and solve: $5^2 + b^2 = 20^2 \Rightarrow 25 + b^2 = 400 \Rightarrow b^2 = 400 - 25 = 375 \Rightarrow b = \sqrt{375} \approx 19.36$ cm.

9. Baseball Diamond Problem

   • The distance from home to second base is the diagonal of the square.
   • Use the Pythagorean Theorem: $90^2 + 90^2 = c^2 \Rightarrow 8100 + 8100 = c^2 \Rightarrow c^2 = 16200 \Rightarrow c = \sqrt{16200} \approx 127.28$ feet.

10. Table and Door Problem

• The door is 42" wide and 84" tall. The table is 96" in diameter. Since the diameter of the table (96") is larger than both the width (42") and the height (84") of the door, the table will not fit through the door.