2013 AMC 10A Problems/Problem 15

$Insert formula here$ ==Problem== Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution 1 (cheap way)

Credit to the Infuzion17 for this solution

The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, therefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\boxed{\textbf{(D) }12}$ .

Solution 2 (actually solving)

Let the height to the side of length 15 be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$ .

Because the area of a triangle is $\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$ . From the problem, we know that $2h_{3} = h_{1} + h_{2}$ . Substituting, we get tha

$h_{3} = 1.25h_{1}$ .

Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

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2013 AMC 10A Problems/Problem 15

Solution 1 (cheap way)

Solution 2 (actually solving)

See Also