Pythagorean Theorem: Word Problems | Worksheet - Free Printable

Problem: Pythagorean Theorem Word Problems

The task involves solving six word problems using the Pythagorean Theorem. The theorem states:

\[
a^2 + b^2 = c^2
\]

where:
- \( a \) and \( b \) are the lengths of the two legs of a right triangle,
- \( c \) is the length of the hypotenuse.

Let's solve each problem step by step.

---

Problem 1:

A contractor is building a new room onto the back of Donna’s house. He needs to attach a support bracket to the frame of one of the walls. The wall is 15 feet long, and the bracket has two holes that are 10 feet apart. How long is the bracket?

#### Solution:
1. Identify the given information:
- The wall is 15 feet long (one leg of the right triangle).
- The distance between the holes on the bracket is 10 feet (the other leg of the right triangle).
- We need to find the length of the bracket, which is the hypotenuse (\( c \)).

2. Apply the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 15 \) feet and \( b = 10 \) feet.
\[
15^2 + 10^2 = c^2
\]

3. Calculate the squares:
\[
15^2 = 225, \quad 10^2 = 100
\]
\[
225 + 100 = c^2
\]

4. Solve for \( c^2 \):
\[
325 = c^2
\]

5. Take the square root of both sides:
\[
c = \sqrt{325}
\]
Simplify \( \sqrt{325} \):
\[
\sqrt{325} = \sqrt{25 \times 13} = 5\sqrt{13}
\]

6. Approximate the value (if needed):
\[
\sqrt{13} \approx 3.6056 \implies 5\sqrt{13} \approx 5 \times 3.6056 \approx 18.028
\]

#### Final Answer:
\[
\boxed{5\sqrt{13} \text{ or approximately } 18.03 \text{ feet}}
\]

---

Problem 2:

Pete’s stick flies, and he finds it in an opposite corner of his uncle’s vacant lot. If the lot is 5 meters long and 4 meters wide, what is the distance between the holes?

#### Solution:
1. Identify the given information:
- The lot is rectangular with dimensions 5 meters (length) and 4 meters (width).
- The distance between the holes is the diagonal of the rectangle, which forms the hypotenuse of a right triangle.

2. Apply the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 5 \) meters and \( b = 4 \) meters.
\[
5^2 + 4^2 = c^2
\]

3. Calculate the squares:
\[
5^2 = 25, \quad 4^2 = 16
\]
\[
25 + 16 = c^2
\]

4. Solve for \( c^2 \):
\[
41 = c^2
\]

5. Take the square root of both sides:
\[
c = \sqrt{41}
\]

#### Final Answer:
\[
\boxed{\sqrt{41} \text{ meters}}
\]

---

Problem 3:

Anna is setting up a sprinkler net in the beach. The sprinkler casts water from the edge of the net to the ground at a distance of 9 feet away from the net. If the cable is 10 feet long, how high off the ground is the net?

#### Solution:
1. Identify the given information:
- The distance from the sprinkler to the ground is 9 feet (one leg of the right triangle).
- The length of the cable is 10 feet (the hypotenuse).
- We need to find the height of the sprinkler above the ground (\( h \)), which is the other leg of the right triangle.

2. Apply the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = h \), \( b = 9 \) feet, and \( c = 10 \) feet.
\[
h^2 + 9^2 = 10^2
\]

3. Calculate the squares:
\[
9^2 = 81, \quad 10^2 = 100
\]
\[
h^2 + 81 = 100
\]

4. Solve for \( h^2 \):
\[
h^2 = 100 - 81
\]
\[
h^2 = 19
\]

5. Take the square root of both sides:
\[
h = \sqrt{19}
\]

#### Final Answer:
\[
\boxed{\sqrt{19} \text{ feet}}
\]

---

Problem 4:

Penny is designing the layout for her rectangular garden. She plans to include a pathway connecting the opposite corners of the garden. The path will be 20 feet long, and the garden pattern will be 16 feet long. How wide is the garden?

#### Solution:
1. Identify the given information:
- The diagonal of the garden (pathway) is 20 feet (hypotenuse).
- The length of the garden is 16 feet (one leg of the right triangle).
- We need to find the width of the garden (\( w \)), which is the other leg of the right triangle.

2. Apply the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 16 \) feet, \( b = w \), and \( c = 20 \) feet.
\[
16^2 + w^2 = 20^2
\]

3. Calculate the squares:
\[
16^2 = 256, \quad 20^2 = 400
\]
\[
256 + w^2 = 400
\]

4. Solve for \( w^2 \):
\[
w^2 = 400 - 256
\]
\[
w^2 = 144
\]

5. Take the square root of both sides:
\[
w = \sqrt{144}
\]
\[
w = 12
\]

#### Final Answer:
\[
\boxed{12 \text{ feet}}
\]

---

Problem 5:

Ramon wants to buy a new TV before he goes to the store. He measures his TV and notes that it is 19.6 inches high and 27.2 inches wide. When he gets to the store, he realizes that TVs are measured by their diagonal length. What is the diagonal length of his current TV?

#### Solution:
1. Identify the given information:
- The height of the TV is 19.6 inches (one leg of the right triangle).
- The width of the TV is 27.2 inches (the other leg of the right triangle).
- We need to find the diagonal length (\( d \)), which is the hypotenuse.

2. Apply the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 19.6 \) inches, \( b = 27.2 \) inches, and \( c = d \).
\[
19.6^2 + 27.2^2 = d^2
\]

3. Calculate the squares:
\[
19.6^2 = 384.16, \quad 27.2^2 = 739.84
\]
\[
384.16 + 739.84 = d^2
\]

4. Solve for \( d^2 \):
\[
d^2 = 1124
\]

5. Take the square root of both sides:
\[
d = \sqrt{1124}
\]
Approximate \( \sqrt{1124} \):
\[
\sqrt{1124} \approx 33.53
\]

#### Final Answer:
\[
\boxed{33.53 \text{ inches}}
\]

---

Problem 6:

At marching band practice, Jade marches from one corner of her school’s field to the opposite corner. The field is 100 yards long and 60 yards wide. If it takes her 200 steps to get across the field, how wide is each step?

#### Solution:
1. Identify the given information:
- The field is rectangular with dimensions 100 yards (length) and 60 yards (width).
- The diagonal distance Jade marches is the hypotenuse of a right triangle.
- It takes her 200 steps to cover this diagonal distance.
- We need to find the width of each step.

2. Calculate the diagonal distance using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 100 \) yards and \( b = 60 \) yards.
\[
100^2 + 60^2 = c^2
\]

3. Calculate the squares:
\[
100^2 = 10000, \quad 60^2 = 3600
\]
\[
10000 + 3600 = c^2
\]

4. Solve for \( c^2 \):
\[
c^2 = 13600
\]

5. Take the square root of both sides:
\[
c = \sqrt{13600}
\]
Simplify \( \sqrt{13600} \):
\[
\sqrt{13600} = \sqrt{16 \times 850} = 4\sqrt{850}
\]
Approximate \( \sqrt{850} \):
\[
\sqrt{850} \approx 29.15 \implies 4\sqrt{850} \approx 4 \times 29.15 \approx 116.6
\]

6. Calculate the width of each step:
- Total diagonal distance = 116.6 yards.
- Number of steps = 200.
- Width of each step:
\[
\text{Width of each step} = \frac{116.6}{200} = 0.583 \text{ yards}
\]
Convert yards to feet (1 yard = 3 feet):
\[
0.583 \text{ yards} \times 3 = 1.749 \text{ feet}
\]

#### Final Answer:
\[
\boxed{0.583 \text{ yards or approximately } 1.75 \text{ feet}}
\]

---

Final Answers:

1. \(\boxed{5\sqrt{13} \text{ or approximately } 18.03 \text{ feet}}\)
2. \(\boxed{\sqrt{41} \text{ meters}}\)
3. \(\boxed{\sqrt{19} \text{ feet}}\)
4. \(\boxed{12 \text{ feet}}\)
5. \(\boxed{33.53 \text{ inches}}\)
6. \(\boxed{0.583 \text{ yards or approximately } 1.75 \text{ feet}}\)