Practice problems for applying the Pythagorean Theorem to find missing side lengths in right triangles.

Let’s solve each problem one by one 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 use this formula for every triangle. If we’re solving for the hypotenuse, we add the squares of the two shorter sides.
If we’re solving for one of the shorter sides, we subtract the square of the known short side from the square of the hypotenuse.

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Problem 1:

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

AC² = AB² + BC² = 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

*(Check: 51² = 2601? Yes. 50²=2500, 51²=2601 → correct)*

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Problem 13:

AB = 7, BC = 9, right angle at B → so AC is hypotenuse

AC² = 7² + 9² = 49 + 81 = 130
→ AC = √130 ≈ 11.4 *(but since others are integers, maybe leave as √130? Wait — let me check if I misread.)*

Wait — in diagram 13, right angle is at B, sides AB=7, BC=9 → yes, AC is hypotenuse.

But 7²+9²=49+81=130 → not a perfect square. So answer is √130. But perhaps they expect decimal? Or did I mislabel?

Looking back: In problem 13, labels are A, B, C with right angle at B, AB=7, BC=9 → AC is hypotenuse → √(49+81)=√130.

But let’s keep exact value unless told otherwise. However, looking at other problems, all answers are integers. Maybe I made a mistake?

Wait — maybe the right angle is NOT at B? Let me recheck the image description.

Actually, in your original image description, for problem 13: “A triangle with points A, B, C. Right angle at B. Side AB = 7, side BC = 9. Find AC.”

Yes — so AC² = 7² + 9² = 130 → AC = √130.

But perhaps the problem expects us to write it as is? Or maybe I should compute approximate? Since other problems have integer answers, maybe there's an error? Let me double-check calculation:

7² = 49
9² = 81
49 + 81 = 130 → correct.

So unless the diagram has different labeling, we go with √130. But wait — maybe the right angle is at C or A? No, you said right angle at B.

Alternatively, perhaps in some versions, numbers are different. But based on given info, we proceed.

However, looking ahead — problem 14 also might not be integer. Let’s continue and see.

Actually, let me assume that for now, and if needed, adjust later. But for consistency, I'll calculate exactly.

But wait — perhaps I misread the sides. Let me think again.

Problem 13: Triangle ABC, right angle at B. So legs are AB and BC. AB=7, BC=9 → hypotenuse AC = √(49+81)=√130.

Similarly, problem 14: Right angle at B, AB=?, BC=12, AC=19 → so AB² = 19² - 12² = 361 - 144 = 217 → AB=√217.

Problem 15: Right angle at B, AB=15, BC=11, find AC → AC²=225+121=346 → √346.

These are not integers. But problems 1-12 were integers. Perhaps the last few are meant to be calculated as decimals? Or maybe I have a labeling error.

Wait — let me check problem 16: Right angle at B, AB=?, BC=5, AC=12 → AB² = 144 - 25 = 119 → √119.

Problem 17: Right angle at B, AB=?, BC=19, AC=23 → AB² = 529 - 361 = 168 → √168.

Problem 18: Right angle at B, AB=?, BC=21, AC=34 → AB² = 1156 - 441 = 715 → √715.

None of these are perfect squares. But perhaps the student is expected to simplify radicals or leave as is? Or maybe I made a mistake in identifying which side is which.

Alternatively, perhaps in some diagrams, the right angle is not where I think. But according to standard notation and your description, I think my approach is correct.

To avoid confusion, I will provide exact values (simplified radical if possible) or decimal approximations only if necessary. But since the first 12 are integers, and the rest aren't, perhaps the problem intends for us to compute numerically? Or maybe there's a typo in the problem set.

But for accuracy, I'll compute exactly.

However, let me verify problem 13 again. Is it possible that AB and BC are not both legs? No, because right angle is at B, so AB and BC are perpendicular → legs.

Perhaps the question is to find AC, and we leave it as √130. But let's see what the expected format is.

Since the user didn't specify, and to match the style of the first 12, I suspect that for problems 13-18, the answers are not integers, so we should report them as simplified radicals or decimals.

But to be precise, I'll calculate the numerical value rounded to one decimal place, as is common in such worksheets.

Let me do that for consistency.

For problem 13: √130 ≈ 11.4 (since 11.4² = 129.96)

Similarly, problem 14: √217 ≈ 14.7 (14.7²=216.09, 14.8²=219.04 → so 14.7)

But let's calculate properly.

Actually, to save time and since this is a homework helper, I'll provide exact values where possible, and for non-integers, I'll give the simplified radical or decimal as appropriate.

But looking at the context, perhaps the worksheet expects integer answers, so maybe I misidentified the right angle or sides.

Let me re-examine problem 13: "Triangle with points A, B, C. Right angle at B. Side AB = 7, side BC = 9. Find AC."

Yes, AC is hypotenuse.

Perhaps in some interpretations, the side lengths are assigned differently. Another possibility: maybe "AC = ?" means something else, but no.

I think I have to accept that not all answers are integers. So I'll proceed with calculations.

For efficiency, I'll list all answers now, having verified each.

Let me compile the final answers:

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 ≈ 11.4 (but let's keep exact for now)
Wait — perhaps the problem has a typo, or I need to output as is.

To resolve this, I recall that in many textbooks, they include non-integer answers too. So I'll provide the exact value.

But for the sake of this response, since the user likely expects numerical answers, and to match the format, I'll calculate the square roots for 13-18.

Let me do quick calculations:

13. AC = √(7² + 9²) = √(49+81) = √130 ≈ 11.4018 → 11.4
14. AB = √(19² - 12²) = √(361 - 144) = √217 ≈ 14.7309 → 14.7
15. AC = √(15² + 11²) = √(225 + 121) = √346 ≈ 18.6011 → 18.6
16. AB = √(12² - 5²) = √(144 - 25) = √119 ≈ 10.9087 → 10.9
17. AB = √(23² - 19²) = √(529 - 361) = √168 = √(4*42) = 2√42 ≈ 12.9615 → 13.0 (rounded)
18. AB = √(34² - 21²) = √(1156 - 441) = √715 ≈ 26.7395 → 26.7

But rounding might not be desired. Perhaps leave as radicals.

Given that the first 12 are exact integers, and the rest are not, I think the best approach is to provide the exact value for all, and for non-perfect squares, write the radical.

But for a student, decimal might be more helpful.

I think for this context, since it's a worksheet, and to be consistent, I'll provide the exact answer as per calculation, and if it's not integer, state it as sqrt(number).

But let's see the instruction: "Solve the problem accurately." So accuracy is key.

So for problem 13: AC = √130
Problem 14: AB = √217
etc.

But that might not be satisfactory. Perhaps I can simplify where possible.

For example, problem 17: √168 = √(4*42) = 2√42
Problem 18: √715 = √(5*143) = no simplification.

This is getting messy.

Another idea: perhaps in the original image, the numbers are different, but based on your text description, I have to go with what's given.

To move forward, I'll provide the answers as calculated, and for non-integers, I'll give the decimal approximation to one decimal place, as is common in such exercises.

So final decision:

For problems 1-12: integer answers as calculated.

For 13-18: decimal answers rounded to one decimal place.

Let me confirm the calculations once more for 13-18:

13. √130 = ? 11.4^2 = 129.96, close to 130 → 11.4
14. √217: 14.7^2 = 216.09, 14.8^2=219.04, so 14.7 is closer (217-216.09=0.91, 219.04-217=2.04) → 14.7
15. √346: 18.6^2 = 345.96, very close → 18.6
16. √119: 10.9^2 = 118.81, 11.0^2=121, so 10.9 (119-118.81=0.19) → 10.9
17. √168: 12.96^2 = ? 13^2=169, so √168≈12.961, round to 13.0
18. √715: 26.7^2 = 712.89, 26.8^2=718.24, 715-712.89=2.11, so 26.7 is better → 26.7

Okay, so I'll use these.

Now, compiling all answers:

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. 11.4
14. 14.7
15. 18.6
16. 10.9
17. 13.0
18. 26.7

Note: For problems 13-18, answers are approximate decimals rounded to one decimal place.

But to be precise, let's make sure about problem 17: √168 = 2√42, and 2*6.4807=12.9614, so 13.0 when rounded to one decimal.

Similarly, others are fine.

I think this is acceptable.

So the final answer is the list above.