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

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

https://youtu.be/cd0yW4k4Fo8

~savannahsolver

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 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

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