2002 AMC 12B Problems/Problem 20
- The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.
Problem
Let be a right-angled triangle with . Let and be the midpoints of legs and , respectively. Given that and , find .
Solution
Let , . By the Pythagorean Theorem on respectively,
Summing these gives .
By the Pythagorean Theorem again, we have
Alternatively, we could note that since we found , segment . Right triangles and are similar by Leg-Leg with a ratio of , so
There is the solution, folks! Overall, this problem's topic is associated with the Pythagorean theorem. If you do not understand this solution, you should take a look at everything about Pythagorean theorem. Please contact us if there are any questions, concerns, or doubts upon this problem,
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See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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