# Difference between revisions of "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 1

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 $4\sqrt{3}h=2\times 6\sqrt{3}=12\sqrt{3}\implies h=\boxed{\textbf{(C)}\ 3}$

## Solution 2

Directly following from the formula for the area of a triangle, $\frac{bh}{2}$, we have that, for any right triangle, the product of the two legs of the triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse. (This is always true because the two legs can be the base and the height of the triangle, and so can the hypotenuse and the altitude of the hypotenuse.)

To find the hypotenuse, we can apply Pythagoras to obtain $$\sqrt{(2\sqrt{3})^2+6^3}$$ $$\implies \sqrt{12+36}$$ $$\implies \sqrt{48}=4\sqrt{3}$$

Let $a$ be the length of the third altutude - the altitude to the hypotenuse. We have $$2\sqrt3 \times 6 = a\times 4\sqrt3$$ $$\implies a=\frac{2\sqrt3\times6}{4\sqrt{3}}$$ $$\implies \frac{6}{2}=\boxed{\textbf{(C)}\ 3}$$

(Solution by bestwillcui1)

 2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 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