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.) 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. Ladder Problem (Question 1)

   • Let $a$ be the distance from the wall to the base of the ladder, $b$ be the height the ladder reaches on the wall, and $c$ be the length of the ladder. We are given $a = 3$ feet and $c = 12$ feet. We need to find $b$.
   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • Substituting the given values: $3^2 + b^2 = 12^2$
   • Simplifying: $9 + b^2 = 144$
   • Solving for $b^2$: $b^2 = 144 - 9 = 135$
   • Solving for $b$: $b = \sqrt{135} = \sqrt{9 \cdot 15} = 3\sqrt{15} \approx 11.62$ feet

2. Soccer Field Problem (Question 2)

   • Let $a$ be the width of the rectangle (90 meters) and $b$ be the length (120 meters). Let $c$ be the diagonal distance.
   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • Substituting the given values: $90^2 + 120^2 = c^2$
   • Simplifying: $8100 + 14400 = c^2$
   • Solving for $c^2$: $c^2 = 22500$
   • Solving for $c$: $c = \sqrt{22500} = 150$ meters

3. Ladder and Wall Problem (Question 3)

   • Let $a$ be the distance from the base of the house to the ladder, $b$ be the height of the wall (12 feet), and $c$ be the length of the ladder (15 feet). We need to find $a$.
   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • Substituting the given values: $a^2 + 12^2 = 15^2$
   • Simplifying: $a^2 + 144 = 225$
   • Solving for $a^2$: $a^2 = 225 - 144 = 81$
   • Solving for $a$: $a = \sqrt{81} = 9$ feet

4. Rectangle Diagonal Problem (Question 4)

   • Let $a$ be the length of the rectangle (15 cm) and $b$ be the width (10 cm). We need to find the diagonal $c$.
   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • Substituting the given values: $15^2 + 10^2 = c^2$
   • Simplifying: $225 + 100 = c^2$
   • Solving for $c^2$: $c^2 = 325$
   • Solving for $c$: $c = \sqrt{325} = \sqrt{25 \cdot 13} = 5\sqrt{13} \approx 18.03$ cm

5. Rectangle Length Problem (Question 5)

   • Let $w$ be the width of the rectangle (15 inches), $l$ be the length, and $d$ be the diagonal (25 inches). We need to find $l$.
   • Using the Pythagorean theorem: $w^2 + l^2 = d^2$
   • Substituting the given values: $15^2 + l^2 = 25^2$
   • Simplifying: $225 + l^2 = 625$
   • Solving for $l^2$: $l^2 = 625 - 225 = 400$
   • Solving for $l$: $l = \sqrt{400} = 20$ inches

6. Right Triangle Problem (Question 6)

   • (a) If 8 and 12 are legs:
     • Let $a = 8$ and $b = 12$. We need to find the hypotenuse $c$.
     • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
     • Substituting: $8^2 + 12^2 = c^2$
     • Simplifying: $64 + 144 = c^2$
     • Solving for $c^2$: $c^2 = 208$
     • Solving for $c$: $c = \sqrt{208} = \sqrt{16 \cdot 13} = 4\sqrt{13} \approx 14.42$
   • (b) If 8 and 12 are a leg and hypotenuse:
     • Case 1: 12 is the hypotenuse. Let $a = 8$ and $c = 12$. We need to find $b$.
       • $a^2 + b^2 = c^2$
       • $8^2 + b^2 = 12^2$
       • $64 + b^2 = 144$
       • $b^2 = 144 - 64 = 80$
       • $b = \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5} \approx 8.94$
     • Case 2: 8 is the hypotenuse. This is impossible because the hypotenuse must be the longest side, so it cannot be smaller than the leg length 12 (Thanks user comment from answer with ID: tk0w7g8).

7. Square Problem (Question 7)

   • Side length:
     • Area of a square $= s^2 = 81$
     • $s = \sqrt{81} = 9$ cm
   • Diagonal:
     • Let $s = 9$ be the side length. The diagonal $d$ can be found using the Pythagorean theorem: $s^2 + s^2 = d^2$
     • $9^2 + 9^2 = d^2$
     • $81 + 81 = d^2$
     • $d^2 = 162$
     • $d = \sqrt{162} = \sqrt{81 \cdot 2} = 9\sqrt{2} \approx 12.73$ cm

8. Isosceles Triangle Problem (Question 8)

   • The congruent sides are 20 cm, and the base is 10 cm. We want to find the height. The height bisects the base, creating a right triangle with a hypotenuse of 20 cm and a base of 5 cm (half of 10 cm).
   • Let $a = 5$ and $c = 20$. We need to find the height $b$.
   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • Substituting: $5^2 + b^2 = 20^2$
   • Simplifying: $25 + b^2 = 400$
   • Solving for $b^2$: $b^2 = 400 - 25 = 375$
   • Solving for $b$: $b = \sqrt{375} = \sqrt{25 \cdot 15} = 5\sqrt{15} \approx 19.36$ cm

9. Baseball Diamond Problem (Question 9)

   • The baseball diamond is a square with sides of 90 feet. The distance from home to second base is the diagonal of the square.
   • Let $s = 90$ be the side length. We need to find the diagonal $d$.
   • Using the Pythagorean theorem: $s^2 + s^2 = d^2$
   • Substituting: $90^2 + 90^2 = d^2$
   • Simplifying: $8100 + 8100 = d^2$
   • Solving for $d^2$: $d^2 = 16200$
   • Solving for $d$: $d = \sqrt{16200} = \sqrt{8100 \cdot 2} = 90\sqrt{2} \approx 127.28$ feet

10. Table and Door Problem (Question 10)

    • The door is 42 inches wide and 84 inches tall. The table has a diameter of 96 inches.
    • Since the table is circular, the widest part is its diameter. The diameter is 96 inches, which is greater than both the width and height of the door.
    • Therefore, the table will not fit through the front door.