# 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 $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$. 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 $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{\textbf{(C)}\ 3}$

## Solution 2

By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$. Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$. Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is $\boxed{\textbf{(C)}\ 3}$ (We can also check from the other side).

## Video Solution (CREATIVE THINKING)

https://youtu.be/ZvoMN5tJHBk


~Education, the Study of Everything

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 