It looks like you've uploaded a worksheet involving the
Pythagorean Theorem, which states:
> In a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
> That is:
> $$
> a^2 + b^2 = c^2
> $$
> where $ c $ is the hypotenuse, and $ a $ and $ b $ are the legs.
Each problem on the worksheet shows a right triangle with two sides labeled and asks you to find the missing side. Let's solve each one step by step.
---
Problem 1
- Given: Legs = 19 and 17
- Find: Hypotenuse ($ c $)
$$
c^2 = 19^2 + 17^2 = 361 + 289 = 650 \\
c = \sqrt{650} \approx 25.5
$$
✔ Answer: $ \boxed{25.5} $
---
Problem 2
- Given: Hypotenuse = 13, one leg = 5
- Find: Other leg ($ a $)
$$
a^2 + 5^2 = 13^2 \\
a^2 + 25 = 169 \\
a^2 = 144 \\
a = \sqrt{144} = 12
$$
✔ Answer: $ \boxed{12} $
---
Problem 3
- Given: Hypotenuse = 20, one leg = 10
- Find: Other leg ($ b $)
$$
b^2 + 10^2 = 20^2 \\
b^2 + 100 = 400 \\
b^2 = 300 \\
b = \sqrt{300} \approx 17.3
$$
✔ Answer: $ \boxed{17.3} $
---
Problem 4
- Given: Hypotenuse = 14, one leg = 7
- Find: Other leg ($ a $)
$$
a^2 + 7^2 = 14^2 \\
a^2 + 49 = 196 \\
a^2 = 147 \\
a = \sqrt{147} \approx 12.1
$$
✔ Answer: $ \boxed{12.1} $
---
Problem 5
- Given: Legs = 13 and 4
- Find: Hypotenuse ($ c $)
$$
c^2 = 13^2 + 4^2 = 169 + 16 = 185 \\
c = \sqrt{185} \approx 13.6
$$
✔ Answer: $ \boxed{13.6} $
---
Problem 6
- Given: Hypotenuse = 25, one leg = 4
- Find: Other leg ($ a $)
$$
a^2 + 4^2 = 25^2 \\
a^2 + 16 = 625 \\
a^2 = 609 \\
a = \sqrt{609} \approx 24.7
$$
✔ Answer: $ \boxed{24.7} $
---
Problem 7
- Given: Hypotenuse = 15, one leg = 8
- Find: Other leg ($ b $)
$$
b^2 + 8^2 = 15^2 \\
b^2 + 64 = 225 \\
b^2 = 161 \\
b = \sqrt{161} \approx 12.7
$$
✔ Answer: $ \boxed{12.7} $
---
Problem 8
- Given: Hypotenuse = 17, one leg = 9
- Find: Other leg ($ a $)
$$
a^2 + 9^2 = 17^2 \\
a^2 + 81 = 289 \\
a^2 = 208 \\
a = \sqrt{208} \approx 14.4
$$
✔ Answer: $ \boxed{14.4} $
---
Problem 9
- Given: Hypotenuse = 18, one leg = 8
- Find: Other leg ($ b $)
$$
b^2 + 8^2 = 18^2 \\
b^2 + 64 = 324 \\
b^2 = 260 \\
b = \sqrt{260} \approx 16.1
$$
✔ Answer: $ \boxed{16.1} $
---
Problem 10
- Given: Hypotenuse = 15, one leg = 7
- Find: Other leg ($ a $)
$$
a^2 + 7^2 = 15^2 \\
a^2 + 49 = 225 \\
a^2 = 176 \\
a = \sqrt{176} \approx 13.3
$$
✔ Answer: $ \boxed{13.3} $
---
✔ Final Answers Summary:
| Problem | Missing Side |
|--------|--------------|
| 1 | 25.5 |
| 2 | 12 |
| 3 | 17.3 |
| 4 | 12.1 |
| 5 | 13.6 |
| 6 | 24.7 |
| 7 | 12.7 |
| 8 | 14.4 |
| 9 | 16.1 |
| 10 | 13.3 |
> ⚠️ Note: All answers are rounded to the nearest tenth unless exact (like Problem 2).
Let me know if you'd like these in exact radical form or if you need help with any specific steps!
Parent Tip: Review the logic above to help your child master the concept of pythagorean triples worksheet.
