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 to solve the problem.
Assume the length of is equal to . Then, by Pythagoras, we have,
Then, by area formulas, we know:
Squaring and solving the above 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
Solution 4 (Fakesolve)/Answer Choices
The area of the triangle is . We want to fine what is. Now, we try each answer choice.
. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being . This does not work. . This is promising. This means that . Now, applying Pythagorean Theorem, we have the vertical sides is and the horizontal side is . Multiplying these and dividing by indeed gives us as desired. Therefore, the answer is
~Arcticturn
Video Solution by OmegaLearn
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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.