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
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
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 |
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