Converse of the Pythagorean Theorem worksheet with problems to identify triangle types based on side lengths.

Let’s solve each problem one by one using the Converse of the Pythagorean Theorem.

The rule is:
- If \( a^2 + b^2 = c^2 \) → it’s a right triangle
- If \( a^2 + b^2 > c^2 \) → it’s an acute triangle
- If \( a^2 + b^2 < c^2 \) → it’s an obtuse triangle

Where \( c \) is the longest side (hypotenuse in right triangles).

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Problem 1: Sides 9 m, 12 m, 15 m

Longest side = 15 → check if \( 9^2 + 12^2 = 15^2 \)

\( 81 + 144 = 225 \)
\( 225 = 225 \) → ✔ Right triangle

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Problem 2: Sides 10 ft, 16 ft, \( 2\sqrt{39} \) ft

First, find decimal or square of \( 2\sqrt{39} \):

\( (2\sqrt{39})^2 = 4 × 39 = 156 \)

Now check which is largest:
10² = 100
16² = 256 ← biggest
So c = 16

Check: \( 10^2 + (2\sqrt{39})^2 =? 16^2 \)
→ \( 100 + 156 = 256 \)
→ \( 256 = 256 \) → ✔ Right triangle

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Problem 3: Sides 9 yd, \( \sqrt{115} \) yd, 11 yd

Square all:

9² = 81
(√115)² = 115
11² = 121 ← biggest → c = 11

Check: \( 9^2 + (\sqrt{115})^2 =? 11^2 \)
→ \( 81 + 115 = 196 \)
But 11² = 121 → 196 ≠ 121 → NOT right

Wait — that can’t be. Let me double-check.

Actually, 81 + 115 = 196, and 11² = 121 → 196 > 121 → so this would mean acute? But wait — we must assign c as the LONGEST side.

Is 11 the longest? √115 ≈ 10.72 → yes, 11 is longest.

But 9² + (√115)² = 81 + 115 = 196
c² = 121 → 196 > 121 → so actually, this suggests the triangle is acute? Wait — no!

Hold on — I think I made a mistake here.

If \( a^2 + b^2 > c^2 \), then angle opposite c is acute → whole triangle is acute.

But let’s verify numbers again.

Sides: 9, √115 (~10.72), 11 → so 11 is longest.

Compute: 9² + (√115)² = 81 + 115 = 196
c² = 121 → 196 > 121 → so YES, acute triangle.

But wait — that seems off because 196 is much bigger than 121. That would mean the sum of squares of two smaller sides is greater than square of largest → acute.

Yes, that’s correct.

BUT — let me check if these sides even form a valid triangle.

Triangle inequality: 9 + √115 > 11?
√115 ≈ 10.72 → 9 + 10.72 = 19.72 > 11 → ok
9 + 11 > 10.72 → ok
10.72 + 11 > 9 → ok

So valid triangle.

And since \( a^2 + b^2 > c^2 \) → acute triangle.

Wait — but let me recalculate: maybe I misassigned sides.

Actually, perhaps the diagram shows 11 as hypotenuse? But mathematically, with given values, 9² + (√115)² = 196, and 11²=121 → not equal → not right.

But 196 > 121 → acute.

Hmm — but let me check if maybe the side labeled 11 is not the longest? No, √115 ≈ 10.72 < 11.

So conclusion: acute triangle.

Wait — I think I see the issue. Maybe the problem expects us to assume the triangle is drawn with right angle at bottom left? But the instruction says “state if each triangle is a right triangle” — meaning use converse theorem, not visual.

So based on calculation: not right → and since sum of squares of two smaller > square of largest → acute.

But let me confirm with actual values:

a=9, b≈10.72, c=11

a² + b² = 81 + 115 = 196
c² = 121
196 > 121 → acute → ✔️

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Problem 4: Sides 32.5 ft, 39 ft, 48.5 ft

Longest = 48.5 → c

Check: 32.5² + 39² =? 48.5²

Calculate:

32.5² = (30+2.5)² = 900 + 2×30×2.5 + 6.25 = 900 + 150 + 6.25 = 1056.25
Or better: 32.5 × 32.5

32.5 × 32.5:
= (65/2)² = 4225 / 4 = 1056.25

39² = 1521

Sum = 1056.25 + 1521 = 2577.25

Now 48.5² = (97/2)² = 9409 / 4 = 2352.25? Wait, that can't be — 48.5 should be larger.

Wait — 48.5 × 48.5:

48² = 2304
0.5² = 0.25
2×48×0.5 = 48
So total = 2304 + 48 + 0.25 = 2352.25

But earlier sum was 2577.25 → which is greater than 2352.25 → so a² + b² > c² → acute?

Wait — but 32.5 and 39 are legs? Then hypotenuse should be sqrt(32.5² + 39²) = sqrt(2577.25) ≈ ?

sqrt(2577.25) — let's compute:

50²=2500, 51²=2601 → so ~50.77

But given side is 48.5 — which is less than 50.77 → so indeed, the actual hypotenuse would be longer than 48.5 → meaning the triangle with sides 32.5, 39, 48.5 has the largest side 48.5, but 32.5² + 39² > 48.5² → so angle opposite 48.5 is acute → whole triangle acute.

But let me verify calculations again.

32.5² = 1056.25
39² = 1521
Sum = 2577.25

48.5² = let's calculate properly:

48.5 × 48.5
= (50 - 1.5)² = 2500 - 2×50×1.5 + (1.5)² = 2500 - 150 + 2.25 = 2352.25

Yes, 2577.25 > 2352.25 → so acute triangle.

But wait — is 48.5 really the longest? Yes, 48.5 > 39 > 32.5.

So answer: acute.

But I recall sometimes problems like this are designed to be right triangles. Let me check if I swapped sides.

Perhaps 48.5 is not the hypotenuse? But it's the largest number.

Another thought: maybe the triangle is labeled with 48.5 as hypotenuse, but mathematically it doesn't satisfy.

Unless I miscalculated 32.5² or 39².

32.5²: 32.5 * 32.5

30*30=900, 30*2.5=75, doubled is 150, 2.5*2.5=6.25 → 900+150+6.25=1056.25 ✓

39²=1521 ✓

Sum 2577.25

48.5²: 48.5 * 48.5

48*48=2304, 48*0.5=24, doubled is 48, 0.5*0.5=0.25 → 2304+48+0.25=2352.25 ✓

2577.25 vs 2352.25 → difference is big.

So not right, and since sum of squares of two smaller > square of largest → acute.

Okay.

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Problem 5: Sides 6m, \( 2\sqrt{55} \)m, 17m

First, find squares:

6² = 36

(2√55)² = 4 * 55 = 220

17² = 289

Longest side = 17 → c

Check: 6² + (2√55)² = 36 + 220 = 256

c² = 289

256 < 289 → so a² + b² < c² → obtuse triangle

Also, check triangle inequality: 6 + 2√55 > 17?

2√55 ≈ 2*7.416 = 14.832

6 + 14.832 = 20.832 > 17 → ok

Others obvious.

So obtuse.

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Problem 6: Sides 4.8km, 28.6km, 29km

Longest = 29 → c

Compute:

4.8² = 23.04

28.6² = ? 28.6 * 28.6

28² = 784

0.6² = 0.36

2*28*0.6 = 33.6

So 784 + 33.6 + 0.36 = 817.96

Sum = 23.04 + 817.96 = 841

29² = 841

Exactly! 841 = 841 → right triangle

Nice.

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Problem 7: Sides 9cm, 12cm, 17cm

Longest = 17 → c

9² = 81

12² = 144

Sum = 225

17² = 289

225 < 289 → obtuse triangle

Check triangle inequality: 9+12=21>17, etc. → valid.

Obtuse.

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Problem 8: Sides 9.6in, 18in, 20.1in

Longest = 20.1 → c

Compute squares:

9.6² = 92.16

18² = 324

Sum = 92.16 + 324 = 416.16

20.1² = ? 20.1 * 20.1

20² = 400

0.1² = 0.01

2*20*0.1 = 4

So 400 + 4 + 0.01 = 404.01

Now compare: 416.16 > 404.01 → so a² + b² > c² → acute triangle

Verify: 9.6 + 18 = 27.6 > 20.1 → ok

So acute.

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Now summarize answers:

1) Right
2) Right
3) Acute
4) Acute
5) Obtuse
6) Right
7) Obtuse
8) Acute

Wait — for problem 3, I had doubt. Let me recheck problem 3.

Problem 3: sides 9 yd, √115 yd, 11 yd

√115 ≈ 10.7238

So sides approx: 9, 10.72, 11

Squares: 81, 115, 121

81 + 115 = 196

196 > 121 → so acute — yes.

But is there a possibility that the triangle is right with different assignment? No, because 11 is largest, and 9² + (√115)² = 196 ≠ 121.

Unless the side labeled 11 is not the hypotenuse, but in the diagram it might be shown as such, but the problem says "state if each triangle is a right triangle" using converse theorem, so we go by calculation.

Similarly for problem 4, we got acute.

But let me double-check problem 4 with exact fractions.

32.5 = 65/2, 39=39, 48.5=97/2

(65/2)^2 = 4225/4

39^2 = 1521 = 6084/4

Sum = (4225 + 6084)/4 = 10309/4

(97/2)^2 = 9409/4

10309/4 > 9409/4 → yes, so acute.

Okay.

Final Answers:

1) Right
2) Right
3) Acute
4) Acute
5) Obtuse
6) Right
7) Obtuse
8) Acute

I think that's correct.

Final Answer:
1) Right
2) Right
3) Acute
4) Acute
5) Obtuse
6) Right
7) Obtuse
8) Acute