Difference between revisions of "2009 AMC 10A Problems/Problem 10"
Arcticturn (talk | contribs) (→Solution 4) |
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== Solution 2 == | == Solution 2 == | ||
− | For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas | + | For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas to solve the problem. |
Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have, | Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have, | ||
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<cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath> | <cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath> | ||
− | Then, by area formulas, we know | + | Then, by area formulas, we know: |
<cmath>\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)</cmath> | <cmath>\frac{1}{2}\sqrt{(h^2+9)(h^2+16)} = \frac{1}{2}(7)(h)</cmath> | ||
− | Squaring and solving the | + | Squaring and solving the above equation yields our solution that <math>h^2 = 12 \Rightarrow h = 2\sqrt{3}.</math> Since the area of the triangle is half of this quantity multiplied by the base, we have |
<cmath>\text{area} = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath> | <cmath>\text{area} = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath> | ||
Revision as of 01:16, 29 October 2022
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
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.