2014 AMC 10A Problem 9

In a right triangle, each leg stands straight up from the other leg, so each leg is already an altitude. The one altitude still missing is the one dropped onto the longest side, the hypotenuse.

1. Find the hypotenuse with the Pythagorean theorem:
$$(2\sqrt3)^2 + 6^2 = 12 + 36 = 48, \qquad \text{hypotenuse} = \sqrt{48} = 4\sqrt3.$$
2. Find the area using the two legs as base and height:
$$\text{area} = \frac12 \cdot 2\sqrt3 \cdot 6 = 6\sqrt3.$$
3. The area is the same no matter which side you call the base. Now use the hypotenuse as the base, with the unknown altitude $h$ as the height:
$$\frac12 \cdot 4\sqrt3 \cdot h = 6\sqrt3.$$
4. The left side is $2\sqrt3 \cdot h$. Divide both sides by $2\sqrt3$ to get $h = 3$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

A triangle has only one area, so every base-and-height pair must give the same number. In a right triangle the legs give the area straight away. Setting that equal to half the hypotenuse times its altitude gives a handy rule: the altitude to the hypotenuse equals (leg times leg) divided by the hypotenuse. Here that is $\frac{2\sqrt3 \cdot 6}{4\sqrt3} = 3$.

The legs $2\sqrt3$ and $6$ are in the ratio $1 : \sqrt3$, so this is a $30$-$60$-$90$ triangle, and the angle opposite the leg $2\sqrt3$ is $30^\circ$. That $30^\circ$ angle sits at the end of the leg of length $6$. The altitude from the right angle, together with the leg of length $6$ and part of the hypotenuse, makes a smaller $30$-$60$-$90$ triangle whose hypotenuse is $6$. The altitude is opposite the $30^\circ$ angle, so it is half of $6$, which is $3$.