Collection of 18 geometry problems designed to test your ability to apply the Pythagorean theorem to find unknown side lengths in right-angled triangles.

Let’s solve each problem using the Pythagoras Theorem.

The Pythagoras Theorem says:
In a right-angled triangle,
a² + b² = c²
where c is the hypotenuse (the side opposite the right angle — it’s always the longest side), and a and b are the other two sides.

We’ll go one by one. I’ll show you how to find the missing side for each triangle.

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Problem 1:

Right triangle ABC, right angle at B.
AB = 12, BC = 5, find AC (hypotenuse).

AC² = AB² + BC²
AC² = 12² + 5² = 144 + 25 = 169
AC = √169 = 13

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Problem 2:

AB = 4, BC = 3, find AC (hypotenuse)

AC² = 4² + 3² = 16 + 9 = 25
AC = √25 = 5

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Problem 3:

AB = 6, BC = 8, find AC (hypotenuse)

AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10

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Problem 4:

BC = 7, AB = 24, find AC (hypotenuse)

AC² = 7² + 24² = 49 + 576 = 625
AC = √625 = 25

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Problem 5:

AB = 12, BC = 9, find AC (hypotenuse)

AC² = 12² + 9² = 144 + 81 = 225
AC = √225 = 15

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Problem 6:

AB = 15, BC = 8, find AC (hypotenuse)

AC² = 15² + 8² = 225 + 64 = 289
AC = √289 = 17

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Problem 7:

AB = 12, AC = 20 (hypotenuse), find BC

BC² = AC² - AB² = 400 - 144 = 256
BC = √256 = 16

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Problem 8:

BC = 5, AC = 13 (hypotenuse), find AB

AB² = AC² - BC² = 169 - 25 = 144
AB = √144 = 12

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Problem 9:

BC = 18, AC = 30 (hypotenuse), find AB

AB² = 30² - 18² = 900 - 324 = 576
AB = √576 = 24

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Problem 10:

AB = 24, AC = 40 (hypotenuse), find BC

BC² = 40² - 24² = 1600 - 576 = 1024
BC = √1024 = 32

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Problem 11:

BC = 80, AC = 100 (hypotenuse), find AB

AB² = 100² - 80² = 10000 - 6400 = 3600
AB = √3600 = 60

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Problem 12:

AB = 68, AC = 85 (hypotenuse), find BC

BC² = 85² - 68² = 7225 - 4624 = 2601
BC = √2601 = 51

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Problem 13:

AB = 7, BC = 9, find AC (hypotenuse)

AC² = 7² + 9² = 49 + 81 = 130
AC = √130 → This doesn’t simplify nicely. Let’s leave as √130 or approx 11.4, but since others are integers, maybe check again? Wait — diagram shows right angle at B, so yes, AC is hypotenuse. But 7-9-? isn't a standard triple. So answer is √130.

Wait — let me double-check: 7²=49, 9²=81, sum=130. Yes. So AC = √130

But looking back at original image — perhaps I misread? No, in problem 13, legs are 7 and 9, hypotenuse unknown. So we keep it as √130. But maybe they expect decimal? Since all others are whole numbers, perhaps typo? Or maybe not. We'll go with exact value.

Actually — wait! In some curricula, they accept radical form. So we’ll write √130

But let me check if 130 can be simplified: 130 = 2×5×13 → no perfect squares → so √130 is simplest.

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Problem 14:

BC = 12, AC = 19 (hypotenuse), find AB

AB² = 19² - 12² = 361 - 144 = 217
AB = √217 → again, not a perfect square. 217 = 7×31 → no simplification. So √217

Hmm — this seems odd. Maybe I misidentified hypotenuse? Diagram says right angle at B, so AC is hypotenuse. Yes. So AB = √(19² - 12²) = √217

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Problem 15:

AB = 15, BC = 11, find AC (hypotenuse)

AC² = 15² + 11² = 225 + 121 = 346
AC = √346 → 346 = 2×173 → no simplification → √346

Again, not integer. But okay.

Wait — let me recheck problem 15: right angle at B, AB=15, BC=11, so AC is hypotenuse. Correct.

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Problem 16:

BC = 5, AC = 12, find AB (hypotenuse?)

Wait — diagram: right angle at B. So AB and BC are legs, AC is hypotenuse? But here AC=12, BC=5, find AB.

So AB² + BC² = AC² → AB² + 25 = 144 → AB² = 119 → AB = √119

But that’s not nice. Wait — maybe I got it wrong.

Looking at diagram: points A, B, C. Right angle at B. Side BC=5, side AC=12? But AC would be hypotenuse only if right angle is at B — yes. Then AB is leg.

So AB² = AC² - BC² = 144 - 25 = 119 → AB = √119

Still messy. Perhaps typo? Or maybe in some problems they allow radicals.

But let's continue.

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Problem 17:

BC = 19, AC = 23 (hypotenuse), find AB

AB² = 23² - 19² = 529 - 361 = 168
AB = √168 = √(4×42) = 2√42 → or leave as √168? Better simplify: 168 = 4×42 → so 2√42

But again, not integer.

Wait — 23²=529, 19²=361, difference 168. Yes.

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Problem 18:

BC = 21, AC = 34 (hypotenuse), find AB

AB² = 34² - 21² = 1156 - 441 = 715
AB = √715 → 715 = 5×11×13 → no simplification → √715

This is getting strange — many non-integers. But let’s assume the diagrams are correct and proceed.

Wait — perhaps I made a mistake in identifying which side is which.

Let me quickly review the pattern:

Problems 1-12 all gave nice integer answers. Problems 13-18 seem to have non-integers. Maybe that’s intentional? Or perhaps I misread the labels.

For example, in problem 13: triangle with right angle at B, sides AB=7, BC=9, find AC. That should be √(49+81)=√130 — correct.

Similarly, problem 14: right angle at B, AB=?, BC=12, AC=19 → AB=√(361-144)=√217 — correct.

Problem 15: AB=15, BC=11, right angle at B, AC=? → √(225+121)=√346 — correct.

Problem 16: right angle at B, BC=5, AC=12, find AB → AB=√(144-25)=√119 — correct.

Problem 17: right angle at B, BC=19, AC=23, find AB → AB=√(529-361)=√168=2√42 — correct.

Problem 18: right angle at B, BC=21, AC=34, find AB → AB=√(1156-441)=√715 — correct.

So even though they’re not integers, mathematically they’re correct.

But let me check if any of these can be simplified further or if I miscalculated.

For problem 16: √119 — 119=7×17 → no.

Problem 17: √168 = √(4*42) = 2√42 — yes.

Problem 18: √715 — 715÷5=143, 143=11×13 → so √715 = √(5×11×13) — no simplification.

Okay, so we’ll report them as is.

But wait — perhaps in the original worksheet, some values are different? For example, in problem 13, maybe it’s 9 and 12? But according to your image description, it’s 7 and 9.

Since I must work with what’s given, I’ll proceed.

Now, to present final answers clearly.

Let me list all answers:

1. AC = 13
2. AC = 5
3. AC = 10
4. AC = 25
5. AC = 15
6. AC = 17
7. BC = 16
8. AB = 12
9. AB = 24
10. BC = 32
11. AB = 60
12. BC = 51
13. AC = √130
14. AB = √217
15. AC = √346
16. AB = √119
17. AB = √168 or 2√42
18. AB = √715

But for consistency, perhaps simplify where possible.

For 17: √168 = √(4×42) = 2√42 — better to simplify.

Others don’t simplify.

Alternatively, maybe the problem expects decimal approximations? But since it’s a math worksheet, exact form is preferred.

I think we should provide exact values.

Final Answer:

1. 13
2. 5
3. 10
4. 25
5. 15
6. 17
7. 16
8. 12
9. 24
10. 32
11. 60
12. 51
13. √130
14. √217
15. √346
16. √119
17. 2√42
18. √715

But let me verify problem 17 again: AB² = 23² - 19² = (23-19)(23+19) = 4 × 42 = 168 → yes, so AB = √168 = 2√42 — correct.

All others checked.

So here are the final answers.