Pythagorean Theorem Worksheets - Free Printable

You’ve uploaded an answer key for a Pythagorean Theorem word problem worksheet (Math Worksheet 17). Since you asked to “solve the problem and explain the solution,” I’ll walk through each of the four problems shown, explaining how the Pythagorean Theorem is applied and why each answer is correct.

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## 🔷 Problem 1: Gavin’s Television

> Problem: Gavin bought a new television for his apartment. It is 50 inches wide and 30 inches tall. Find the diagonal length of the television.

✔ Solution:

We treat the TV as a rectangle. The diagonal forms the hypotenuse of a right triangle with legs = width and height.

Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Where:
- \( a = 50 \) inches (width)
- \( b = 30 \) inches (height)
- \( c = ? \) (diagonal)

\[
50^2 + 30^2 = c^2 \\
2500 + 900 = c^2 \\
3400 = c^2 \\
c = \sqrt{3400} ≈ 58.3095... ≈ \boxed{58.31} \text{ inches (rounded to nearest tenth)}
\]

✔ Answer: 58.31 inches

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## 🔷 Problem 2: Heather’s Walk Home

> Problem: Heather walks 3 miles North and then 4 miles East. What is the shortest distance she can cover to reach home?

✔ Solution:

This forms a right triangle:
- One leg = 3 miles (North)
- Other leg = 4 miles (East)
- Hypotenuse = straight-line distance home

\[
3^2 + 4^2 = c^2 \\
9 + 16 = c^2 \\
25 = c^2 \\
c = \sqrt{25} = \boxed{5} \text{ miles}
\]

✔ Answer: 5 miles

*(Note: This is a classic 3-4-5 right triangle — very common in Pythagorean problems!)*

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## 🔷 Problem 3: Mr. Cooper’s Right Triangle

> Problem: A right triangle has a hypotenuse of 200 mm and one leg of 172 mm. Find the length of the other leg.

✔ Solution:

Here, we know the hypotenuse (\(c = 200\)) and one leg (\(b = 172\)), and we need to find the other leg (\(a\)).

Use rearranged Pythagorean Theorem:
\[
a^2 = c^2 - b^2
\]

\[
a^2 = 200^2 - 172^2 \\
a^2 = 40000 - 29584 \\
a^2 = 10416 \\
a = \sqrt{10416} ≈ 102.058... ≈ \boxed{102.06} \text{ mm (rounded to nearest hundredth)}
\]

✔ Answer: 102.06 mm

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## 🔷 Problem 4: Ladder Against the Wall

> Problem: How far from the base of the house should an 18-foot ladder be placed so that it exactly reaches the top of a 15-foot wall?

✔ Solution:

The ladder is the hypotenuse (18 ft), the wall is one leg (15 ft), and we need the distance from the base (other leg).

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

\[
a^2 = 18^2 - 15^2 \\
a^2 = 324 - 225 \\
a^2 = 99 \\
a = \sqrt{99} ≈ 9.949... ≈ \boxed{9.95} \text{ feet (rounded to nearest hundredth)}
\]

✔ Answer: 9.95 feet

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## 📌 Summary of All Answers:

| Problem | Question | Answer |
|--------|----------|--------|
| 1 | Diagonal of TV (50" x 30") | 58.31 inches |
| 2 | Shortest distance home (3 mi N, 4 mi E) | 5 miles |
| 3 | Missing leg (hyp = 200 mm, leg = 172 mm) | 102.06 mm |
| 4 | Distance from wall for 18-ft ladder reaching 15-ft wall | 9.95 feet |

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## 💡 Key Takeaway:

The Pythagorean Theorem (\(a^2 + b^2 = c^2\)) applies whenever you have a right triangle and need to find a missing side. Always identify:
- Which sides are the legs (adjacent to the right angle),
- Which is the hypotenuse (opposite the right angle, longest side).

If you’re given two sides, plug into the formula. If you’re solving for a leg, rearrange to \(a^2 = c^2 - b^2\) or similar.

Let me know if you’d like to see visual diagrams or practice more problems!