The Pythagorean Theorem and Its Converse online exercise for ... - Free Printable

Let's solve each problem step by step using the Pythagorean Theorem and related triangle classification rules.

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🔷 Pythagorean Theorem Recap:

For a triangle with sides $ a $, $ b $, and $ c $ (where $ c $ is the longest side):

- If $ a^2 + b^2 = c^2 $ → Right triangle
- If $ a^2 + b^2 > c^2 $ → Acute triangle
- If $ a^2 + b^2 < c^2 $ → Obtuse triangle

We'll apply this to each question.

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## ✔ Part 1: State if each triangle is a right triangle.

9) Triangle with sides: 15 m, 9 m, 12 m

Let’s check:
- Longest side: 15 m → possible hypotenuse
- Check: $ 9^2 + 12^2 = 81 + 144 = 225 $
- $ 15^2 = 225 $

✔ $ 9^2 + 12^2 = 15^2 $ → Right triangle

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10) Triangle with sides: 10 ft, 16 ft, $ 2\sqrt{39} $ ft

First, simplify $ 2\sqrt{39} $:
- $ \sqrt{39} \approx 6.245 $, so $ 2\sqrt{39} \approx 12.49 $ ft

So sides: 10, ~12.49, 16 → longest side is 16 ft

Check: $ 10^2 + (2\sqrt{39})^2 = 100 + 4 \cdot 39 = 100 + 156 = 256 $

$ 16^2 = 256 $

✔ $ 10^2 + (2\sqrt{39})^2 = 16^2 $ → Right triangle

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11) Triangle with sides: 9 yd, 11 yd, $ \sqrt{115} $ yd

Simplify $ \sqrt{115} \approx 10.72 $

So sides: 9, ~10.72, 11 → longest is 11

Check: $ 9^2 + (\sqrt{115})^2 = 81 + 115 = 196 $

$ 11^2 = 121 $

Compare: $ 196 > 121 $ → but we need to see if it satisfies Pythagoras.

Wait — is 11 the hypotenuse? No, because $ \sqrt{115} \approx 10.72 < 11 $, so 11 is the longest.

But let's test:

Is $ 9^2 + (\sqrt{115})^2 = 11^2 $?
→ $ 81 + 115 = 196 $, $ 11^2 = 121 $ → 196 ≠ 121

Now try: Is $ 9^2 + 11^2 = (\sqrt{115})^2 $?
→ $ 81 + 121 = 202 $, $ \sqrt{115}^2 = 115 $ → no

Or $ 11^2 + (\sqrt{115})^2 = 9^2 $? → too big

So none of them satisfy $ a^2 + b^2 = c^2 $

But wait — maybe I misread: is $ \sqrt{115} $ one leg?

Try: $ 9^2 + (\text{other leg})^2 = 11^2 $ → but the third side is $ \sqrt{115} $, which is about 10.72

So: $ 9^2 + (\sqrt{115})^2 = 81 + 115 = 196 $, and $ 11^2 = 121 $ → not equal

But is $ 9^2 + 11^2 = (\sqrt{115})^2 $? → $ 81 + 121 = 202 $, $ 115 $ → no

So none of the combinations work → Not a right triangle?

Wait — perhaps $ \sqrt{115} $ is the hypotenuse?

Then: $ 9^2 + 11^2 = 81 + 121 = 202 $

$ (\sqrt{115})^2 = 115 $ → 202 ≠ 115 → no

So no combination works → ✘ Not a right triangle

But wait — maybe I made an error.

Let’s suppose the two legs are 9 and $ \sqrt{115} $, then hypotenuse should be $ \sqrt{9^2 + (\sqrt{115})^2} = \sqrt{81 + 115} = \sqrt{196} = 14 $

But given hypotenuse is 11 → not 14 → not matching.

Alternatively, if 9 and 11 are legs: $ \sqrt{81 + 121} = \sqrt{202} \approx 14.2 $, not $ \sqrt{115} \approx 10.7 $

So no, not a right triangle.

✘ Not a right triangle

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12) Triangle with sides: 48.5 ft, 32.5 ft, 39 ft

Longest side: 48.5 ft → possible hypotenuse

Check: $ 32.5^2 + 39^2 = ? $

Calculate:
- $ 32.5^2 = (65/2)^2 = 4225 / 4 = 1056.25 $
- $ 39^2 = 1521 $
- Sum: $ 1056.25 + 1521 = 2577.25 $

Now $ 48.5^2 = (97/2)^2 = 9409 / 4 = 2352.25 $

Compare: $ 2577.25 \neq 2352.25 $

Since $ a^2 + b^2 > c^2 $, but not equal → not a right triangle

Also, $ 2577.25 > 2352.25 $ → acute? Wait — but we're just checking for right triangle.

So since $ a^2 + b^2 \neq c^2 $, it's not a right triangle

✘ Not a right triangle

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## ✔ Part 2: State if three side lengths form a right triangle

13) 10 cm, 49.5 cm, 50.5 cm

Longest side: 50.5 cm → test if $ 10^2 + 49.5^2 = 50.5^2 $

Compute:
- $ 10^2 = 100 $
- $ 49.5^2 = (50 - 0.5)^2 = 2500 - 50 + 0.25 = 2450.25 $
- Sum: $ 100 + 2450.25 = 2550.25 $

Now $ 50.5^2 = (50 + 0.5)^2 = 2500 + 50 + 0.25 = 2550.25 $

✔ Equal! So yes → Right triangle

✔️ Yes, forms a right triangle

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14) 9 in, 12 in, 15 in

Check: $ 9^2 + 12^2 = 81 + 144 = 225 $

$ 15^2 = 225 $

✔ Yes → Right triangle

✔️ Yes, forms a right triangle

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## ✔ Part 3: State if each triangle is acute, obtuse, or right

15) Sides: 17 cm, 12 cm, 9 cm

Longest side: 17 cm → test $ 12^2 + 9^2 $ vs $ 17^2 $

- $ 12^2 = 144 $
- $ 9^2 = 81 $
- Sum: $ 144 + 81 = 225 $
- $ 17^2 = 289 $

Compare: $ 225 < 289 $ → $ a^2 + b^2 < c^2 $ → Obtuse triangle

✘ Obtuse

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16) Sides: 9.6 in, 18 in, 20.1 in

Longest side: 20.1 in

Check: $ 9.6^2 + 18^2 $ vs $ 20.1^2 $

- $ 9.6^2 = 92.16 $
- $ 18^2 = 324 $
- Sum: $ 92.16 + 324 = 416.16 $
- $ 20.1^2 = (20 + 0.1)^2 = 400 + 4 + 0.01 = 404.01 $

Compare: $ 416.16 > 404.01 $ → $ a^2 + b^2 > c^2 $ → Acute triangle

✔️ Acute

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## ✔ Part 4: State if three side lengths form acute, obtuse, or right triangle

17) 6 mi, $ 2\sqrt{55} $ mi, 17 mi

First, compute $ 2\sqrt{55} $:

- $ \sqrt{55} \approx 7.416 $, so $ 2\sqrt{55} \approx 14.832 $

So sides: 6, ~14.832, 17 → longest is 17

Now check: $ 6^2 + (2\sqrt{55})^2 = 36 + 4 \cdot 55 = 36 + 220 = 256 $

$ 17^2 = 289 $

Compare: $ 256 < 289 $ → $ a^2 + b^2 < c^2 $ → Obtuse triangle

✘ Obtuse

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18) 4.8 km, 28.6 km, 29 km

Longest side: 29 km

Check: $ 4.8^2 + 28.6^2 $ vs $ 29^2 $

- $ 4.8^2 = 23.04 $
- $ 28.6^2 = (28 + 0.6)^2 = 784 + 33.6 + 0.36 = 817.96 $
- Sum: $ 23.04 + 817.96 = 841 $
- $ 29^2 = 841 $

✔ Equal → Right triangle

✔️ Right triangle

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## ✔ Final Answers:

State if each triangle is a right triangle:

9) ✔ Yes
10) ✔ Yes
11) ✘ No
12) ✘ No

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State if the three side lengths form a right triangle:

13) ✔ Yes
14) ✔ Yes

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State if each triangle is acute, obtuse, or right:

15) ✘ Obtuse
16) ✔️ Acute

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State if the three side lengths form an acute, obtuse, or right triangle:

17) ✘ Obtuse
18) ✔ Right

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Let me know if you'd like these summarized in a table format!