2013 AMC 10A Problem 15

Let $h_1$ be the length of the altitude to the side of length $10$ and similarly define $h_2$ for the other given side. We have that $$10h_1 = 15h_2$$$$h_1 = \dfrac{3}{2}h_2.$$ The third altitude is the average of the other two, which makes its length $$\dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2.$$ Let the third side have length $x.$ Then $$\dfrac{5}{4}h_2x = 15h_2$$$$x = 12.$$ Thus, D is the correct answer.