2014 AMC 10A Problems/Problem 9

Problem

The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

We find that the area of the triangle is $6\times \sqrt{3}=6\sqrt{3}$ . By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.

Let $h$ be the third height of the triangle. We have $4\sqrt{3}h=2\times 6\sqrt{3}=12\sqrt{3}\implies h=\boxed{\textbf{(C)} 3}$

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2014 AMC 10A Problems/Problem 9

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