Difference between revisions of "2013 AMC 10A Problems/Problem 15"

m (Solution 1 (Using the Answer Choices))
m (Added the section title)
Line 1: Line 1:
 +
==Problem==
 
Two sides of a triangle have lengths <math>10</math> and <math>15</math>.  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?
 
Two sides of a triangle have lengths <math>10</math> and <math>15</math>.  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?
  

Revision as of 22:32, 15 February 2019

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 (Process of Elimination)

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

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 that \[h_{3} = 1.25h_{1}.\] Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}$.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS