Let’s solve each problem step by step using the
Pythagorean Theorem:
> In a right triangle, \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse (the longest side, opposite the right angle).
We’ll find missing sides first, then calculate area and perimeter for triangles 1–4. For problems 5–9, we’re only asked to find the missing side (since it says “c is the hypotenuse” and gives two other sides).
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Problem 1:
Triangle with legs 17 and ? , hypotenuse 19.
Use: \( 17^2 + x^2 = 19^2 \)
→ \( 289 + x^2 = 361 \)
→ \( x^2 = 361 - 289 = 72 \)
→ \( x = \sqrt{72} ≈ 8.5 \) (rounded to nearest tenth)
✔ Missing side =
8.5
Now, area = \( \frac{1}{2} \times base \times height = \frac{1}{2} \times 17 \times 8.5 = 72.25 \)
Perimeter = 17 + 8.5 + 19 =
44.5
---
Problem 2:
Legs 5 and 13, hypotenuse = ?
Wait — in a right triangle, hypotenuse must be the LONGEST side. But here, if legs are 5 and 13, then hypotenuse should be longer than 13. Let’s check:
\( 5^2 + 13^2 = 25 + 169 = 194 \)
→ Hypotenuse = \( \sqrt{194} ≈ 13.9 \)
But wait — looking at the diagram, the side labeled “?” is the hypotenuse? Actually, in the drawing, the side labeled “?” is opposite the right angle? Let me recheck.
Actually, in problem 2, the triangle has vertical leg = 5, horizontal leg = 13, so hypotenuse is the slanted side marked “?”.
So yes: \( c = \sqrt{5^2 + 13^2} = \sqrt{25 + 169} = \sqrt{194} ≈ 13.9 \)
✔ Missing side =
13.9
Area = \( \frac{1}{2} \times 5 \times 13 = 32.5 \)
Perimeter = 5 + 13 + 13.9 =
31.9
---
Problem 3:
Hypotenuse = 20, one leg = 10, find other leg.
\( x^2 + 10^2 = 20^2 \)
→ \( x^2 + 100 = 400 \)
→ \( x^2 = 300 \)
→ \( x = \sqrt{300} ≈ 17.3 \)
✔ Missing side =
17.3
Area = \( \frac{1}{2} \times 10 \times 17.3 = 86.5 \)
Perimeter = 10 + 17.3 + 20 =
47.3
---
Problem 4:
Legs 5 and 14, hypotenuse = ?
\( c = \sqrt{5^2 + 14^2} = \sqrt{25 + 196} = \sqrt{221} ≈ 14.9 \)
✔ Missing side =
14.9
Area = \( \frac{1}{2} \times 5 \times 14 = 35 \)
Perimeter = 5 + 14 + 14.9 =
33.9
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Problems 5–9: Find missing side (c is hypotenuse)
#### Problem 5:
a=12, b=5 → c = ?
\( c^2 = 12^2 + 5^2 = 144 + 25 = 169 \)
→ c = √169 =
13 ✔ (matches handwritten answer)
#### Problem 6:
a=8, c=10 → b = ?
\( 8^2 + b^2 = 10^2 \)
→ 64 + b² = 100
→ b² = 36 → b =
6 ✔
#### Problem 7:
a=15, c=17 → b = ?
\( 15^2 + b^2 = 17^2 \)
→ 225 + b² = 289
→ b² = 64 → b =
8 ✔
#### Problem 8:
b=40, c=50 → a = ?
\( a^2 + 40^2 = 50^2 \)
→ a² + 1600 = 2500
→ a² = 900 → a =
30 ✔
#### Problem 9:
b=2, c=4 → a = ?
\( a^2 + 2^2 = 4^2 \)
→ a² + 4 = 16
→ a² = 12 → a = √12 = 2√3 ≈ 3.464... but they want exact form? Handwritten says 2√3 — that’s correct.
✔ a =
2√3
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## Final Answers:
For triangles 1–4 (missing length, area, perimeter):
1. Missing:
8.5, Area:
72.3, Perimeter:
44.5
*(Note: Area rounded to nearest tenth: 72.25 → 72.3)*
2. Missing:
13.9, Area:
32.5, Perimeter:
31.9
3. Missing:
17.3, Area:
86.5, Perimeter:
47.3
4. Missing:
14.9, Area:
35.0, Perimeter:
33.9
For problems 5–9 (missing side only):
5.
13
6.
6
7.
8
8.
30
9.
2√3
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Final Answer:
1. Missing: 8.5, Area: 72.3, Perimeter: 44.5
2. Missing: 13.9, Area: 32.5, Perimeter: 31.9
3. Missing: 17.3, Area: 86.5, Perimeter: 47.3
4. Missing: 14.9, Area: 35.0, Perimeter: 33.9
5. 13
6. 6
7. 8
8. 30
9. 2√3
Parent Tip: Review the logic above to help your child master the concept of pythagorean theorem work sheet.
