2014 AMC 10A Problem 9

Problem 9 of 25EasierGeometry

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?

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Solution

We get that the area of the triangle is

\dfrac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}.

The length of the hypotenuse is

\sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{48} = 4\sqrt{3}.

Dropping the altitude,

h,

from the vertex to the hypotenuse, we get that

\dfrac{1}{2} \cdot h \cdot 4\sqrt{3} = 6\sqrt{3}

h = 3.

Thus, C is the correct answer.

Tagged: altitude · triangle area · Pythagorean Theorem