Difference between revisions of "2009 AMC 10B Problems/Problem 20"
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+ | == Video Solution == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=4816 | ||
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+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Latest revision as of 20:42, 24 January 2021
Problem
Triangle has a right angle at , , and . The bisector of meets at . What is ?
Solution 1
By the Pythagorean Theorem, . Then, from the Angle Bisector Theorem, we have:
Solution 2
Let . Notice and . By the double angle identity,
Solution 3
Let .
Make a line so that it is perpendicular to . Since is an angle bisector, is congruent to . Using the Pythagorean Theorem:
We know that by the congruent triangles, so . We know that , , and . We now have right triangle and its 3 sides. Using the Pythagoreon Thereom, we get:
So,
~HelloWorld21
Video Solution
https://youtu.be/4_x1sgcQCp4?t=4816
~ pi_is_3.14
See Also
2009 AMC 10B (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 10 Problems and Solutions |
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