2009 AMC 10A Problems/Problem 10
Contents
Problem
Triangle has a right angle at . Point is the foot of the altitude from , , and . What is the area of ?
Solution 1
It is a well-known fact that in any right triangle with the right angle at and the foot of the altitude from onto we have . (See below for a proof.) Then , and the area of the triangle is .
Proof: Consider the Pythagorean theorem for each of the triangles , , and . We get:
- .
Substituting equations 2 and 3 into the left hand side of equation 1, we get .
Alternatively, note that .
Solution 2
For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas as well to solve the problem.
Assume the length of is equal to . Then, by Pythagoras, we have,
Then, by area formulas, we know that
Squaring and solving the subsequent equation yields our solution that Since the area of the triangle is half of this quantity multiplied by the base, we have
Solution 3 (Power of a point)
Draw the circumcircle of the . Because is a right angle triangle, AC is the diameter of the circumcircle. By applying Power of a Point Theorem, we can have and . Then we have
~Bran_Qin
Video Solution
https://youtu.be/4_x1sgcQCp4?t=1195
~ pi_is_3.14
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AMC 10 Problems and Solutions |
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