Practice problems for applying the Pythagorean Theorem to find missing sides of right triangles.
A worksheet with 18 problems on the Pythagorean Theorem, each featuring a right triangle with two side lengths given and one side to solve for, labeled with vertices A, B, and C.
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Show Answer Key & Explanations
Step-by-step solution for: Worksheet) Pythagoras Theorem PDF | PDF | Triangle Geometry ...
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Show Answer Key & Explanations
Step-by-step solution for: Worksheet) Pythagoras Theorem PDF | PDF | Triangle Geometry ...
Let’s solve each problem using the Pythagorean Theorem.
The theorem says: In a right 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.
---
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
---
Problem 2:
Right angle at B. AB = 4, BC = 3, find AC.
AC² = 4² + 3² = 16 + 9 = 25
AC = √25 = 5
---
Problem 3:
Right angle at B. AB = 6, BC = 8, find AC.
AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10
---
Problem 4:
Right angle at B. AB = 24, BC = 7, find AC.
AC² = 24² + 7² = 576 + 49 = 625
AC = √625 = 25
---
Problem 5:
Right angle at B. AB = 12, BC = 9, find AC.
AC² = 12² + 9² = 144 + 81 = 225
AC = √225 = 15
---
Problem 6:
Right angle at B. AB = 15, BC = 8, find AC.
AC² = 15² + 8² = 225 + 64 = 289
AC = √289 = 17
---
Problem 7:
Right angle at B. AB = 12, AC = 20 (hypotenuse), find BC.
BC² = AC² - AB² = 20² - 12² = 400 - 144 = 256
BC = √256 = 16
---
Problem 8:
Right angle at B. BC = 5, AC = 13 (hypotenuse), find AB.
AB² = AC² - BC² = 13² - 5² = 169 - 25 = 144
AB = √144 = 12
---
Problem 9:
Right angle at B. BC = 18, AC = 30 (hypotenuse), find AB.
AB² = 30² - 18² = 900 - 324 = 576
AB = √576 = 24
---
Problem 10:
Right angle at B. AB = 24, AC = 40 (hypotenuse), find BC.
BC² = 40² - 24² = 1600 - 576 = 1024
BC = √1024 = 32
---
Problem 11:
Right angle at B. BC = 80, AC = 100 (hypotenuse), find AB.
AB² = 100² - 80² = 10000 - 6400 = 3600
AB = √3600 = 60
---
Problem 12:
Right angle at B. AB = 68, AC = 85 (hypotenuse), find BC.
BC² = 85² - 68² = 7225 - 4624 = 2601
BC = √2601 = 51
---
Problem 13:
Right angle at B. AB = 9, BC = 7, find AC.
AC² = 9² + 7² = 81 + 49 = 130
AC = √130 ≈ 11.40 (but since all others are integers, maybe check? Wait — 9²=81, 7²=49, sum=130 → not perfect square. But problem doesn’t say to round. So leave as √130? But looking at pattern, maybe I misread? No — diagram shows legs 9 and 7, so yes. But let’s double-check: 9 and 7 are legs, AC is hypotenuse → correct. So answer is √130. But perhaps they want exact form? Or maybe I made mistake? Let me recalculate: 81+49=130 → yes. So AC = √130. But in context of other problems being integers, maybe it’s intentional? I’ll write √130.
Wait — actually, looking back, problem 13: points A, B, C — right angle at B. Sides: AB=9, BC=7, so AC is hypotenuse → yes. So AC = √(81+49)=√130. But maybe simplify? 130=2×5×13 → no perfect squares → so √130 is simplest. But perhaps the problem expects decimal? Since others are integers, maybe I should check if I misassigned sides. Diagram: A to B is 9, B to C is 7, right angle at B → yes. So AC = √130. I’ll keep it as √130 for now.
But wait — let me see problem 14 and 15 — they also have non-integers? Problem 14: legs 12 and ? hypotenuse 19 → AB = √(19² - 12²) = √(361-144)=√217 → not integer. Problem 15: legs 11 and 15? Wait no — right angle at B, so legs are AB and BC? Diagram: AC=?, AB=15, BC=11, right angle at B → so AC = √(15² + 11²) = √(225+121)=√346 → not integer. So some problems don’t give integer answers. Okay, so for 13, AC = √130.
But let me confirm calculation: 9²=81, 7²=49, 81+49=130 → yes. So AC = √130.
---
Problem 14:
Right angle at B. BC = 12, AC = 19 (hypotenuse), find AB.
AB² = 19² - 12² = 361 - 144 = 217
AB = √217 ≈ 14.73 (but again, exact form is √217)
---
Problem 15:
Right angle at B. AB = 15, BC = 11, find AC.
AC² = 15² + 11² = 225 + 121 = 346
AC = √346 ≈ 18.60 (exact: √346)
---
Problem 16:
Right angle at B. BC = 5, AC = 12 (hypotenuse), find AB.
AB² = 12² - 5² = 144 - 25 = 119
AB = √119 ≈ 10.91 (exact: √119)
---
Problem 17:
Right angle at B. AC = 23 (hypotenuse), BC = 19, find AB.
AB² = 23² - 19² = 529 - 361 = 168
AB = √168 = √(4×42) = 2√42 ≈ 12.96 (exact: 2√42 or √168)
---
Problem 18:
Right angle at B. BC = 21, AC = 34 (hypotenuse), find AB.
AB² = 34² - 21² = 1156 - 441 = 715
AB = √715 ≈ 26.74 (exact: √715)
---
Now, compiling all 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
14. √217
15. √346
16. √119
17. √168 (or 2√42)
18. √715
But let me double-check problem 17: 23²=529, 19²=361, 529-361=168 → yes. √168 = √(4*42) = 2√42. Similarly, others are simplified if possible.
For consistency, I'll write exact forms where not integer.
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
14. √217
15. √346
16. √119
17. 2√42
18. √715
I think that's all.
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
The theorem says: In a right 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.
---
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
---
Problem 2:
Right angle at B. AB = 4, BC = 3, find AC.
AC² = 4² + 3² = 16 + 9 = 25
AC = √25 = 5
---
Problem 3:
Right angle at B. AB = 6, BC = 8, find AC.
AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10
---
Problem 4:
Right angle at B. AB = 24, BC = 7, find AC.
AC² = 24² + 7² = 576 + 49 = 625
AC = √625 = 25
---
Problem 5:
Right angle at B. AB = 12, BC = 9, find AC.
AC² = 12² + 9² = 144 + 81 = 225
AC = √225 = 15
---
Problem 6:
Right angle at B. AB = 15, BC = 8, find AC.
AC² = 15² + 8² = 225 + 64 = 289
AC = √289 = 17
---
Problem 7:
Right angle at B. AB = 12, AC = 20 (hypotenuse), find BC.
BC² = AC² - AB² = 20² - 12² = 400 - 144 = 256
BC = √256 = 16
---
Problem 8:
Right angle at B. BC = 5, AC = 13 (hypotenuse), find AB.
AB² = AC² - BC² = 13² - 5² = 169 - 25 = 144
AB = √144 = 12
---
Problem 9:
Right angle at B. BC = 18, AC = 30 (hypotenuse), find AB.
AB² = 30² - 18² = 900 - 324 = 576
AB = √576 = 24
---
Problem 10:
Right angle at B. AB = 24, AC = 40 (hypotenuse), find BC.
BC² = 40² - 24² = 1600 - 576 = 1024
BC = √1024 = 32
---
Problem 11:
Right angle at B. BC = 80, AC = 100 (hypotenuse), find AB.
AB² = 100² - 80² = 10000 - 6400 = 3600
AB = √3600 = 60
---
Problem 12:
Right angle at B. AB = 68, AC = 85 (hypotenuse), find BC.
BC² = 85² - 68² = 7225 - 4624 = 2601
BC = √2601 = 51
---
Problem 13:
Right angle at B. AB = 9, BC = 7, find AC.
AC² = 9² + 7² = 81 + 49 = 130
AC = √130 ≈ 11.40 (but since all others are integers, maybe check? Wait — 9²=81, 7²=49, sum=130 → not perfect square. But problem doesn’t say to round. So leave as √130? But looking at pattern, maybe I misread? No — diagram shows legs 9 and 7, so yes. But let’s double-check: 9 and 7 are legs, AC is hypotenuse → correct. So answer is √130. But perhaps they want exact form? Or maybe I made mistake? Let me recalculate: 81+49=130 → yes. So AC = √130. But in context of other problems being integers, maybe it’s intentional? I’ll write √130.
Wait — actually, looking back, problem 13: points A, B, C — right angle at B. Sides: AB=9, BC=7, so AC is hypotenuse → yes. So AC = √(81+49)=√130. But maybe simplify? 130=2×5×13 → no perfect squares → so √130 is simplest. But perhaps the problem expects decimal? Since others are integers, maybe I should check if I misassigned sides. Diagram: A to B is 9, B to C is 7, right angle at B → yes. So AC = √130. I’ll keep it as √130 for now.
But wait — let me see problem 14 and 15 — they also have non-integers? Problem 14: legs 12 and ? hypotenuse 19 → AB = √(19² - 12²) = √(361-144)=√217 → not integer. Problem 15: legs 11 and 15? Wait no — right angle at B, so legs are AB and BC? Diagram: AC=?, AB=15, BC=11, right angle at B → so AC = √(15² + 11²) = √(225+121)=√346 → not integer. So some problems don’t give integer answers. Okay, so for 13, AC = √130.
But let me confirm calculation: 9²=81, 7²=49, 81+49=130 → yes. So AC = √130.
---
Problem 14:
Right angle at B. BC = 12, AC = 19 (hypotenuse), find AB.
AB² = 19² - 12² = 361 - 144 = 217
AB = √217 ≈ 14.73 (but again, exact form is √217)
---
Problem 15:
Right angle at B. AB = 15, BC = 11, find AC.
AC² = 15² + 11² = 225 + 121 = 346
AC = √346 ≈ 18.60 (exact: √346)
---
Problem 16:
Right angle at B. BC = 5, AC = 12 (hypotenuse), find AB.
AB² = 12² - 5² = 144 - 25 = 119
AB = √119 ≈ 10.91 (exact: √119)
---
Problem 17:
Right angle at B. AC = 23 (hypotenuse), BC = 19, find AB.
AB² = 23² - 19² = 529 - 361 = 168
AB = √168 = √(4×42) = 2√42 ≈ 12.96 (exact: 2√42 or √168)
---
Problem 18:
Right angle at B. BC = 21, AC = 34 (hypotenuse), find AB.
AB² = 34² - 21² = 1156 - 441 = 715
AB = √715 ≈ 26.74 (exact: √715)
---
Now, compiling all 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
14. √217
15. √346
16. √119
17. √168 (or 2√42)
18. √715
But let me double-check problem 17: 23²=529, 19²=361, 529-361=168 → yes. √168 = √(4*42) = 2√42. Similarly, others are simplified if possible.
For consistency, I'll write exact forms where not integer.
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
14. √217
15. √346
16. √119
17. 2√42
18. √715
I think that's all.
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
Parent Tip: Review the logic above to help your child master the concept of print worksheet on pythagorean theorem.
