1961 AHSME Problems/Problem 16

Problem

An altitude $h$ of a triangle is increased by a length $m$. How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle?

$\textbf{(A)}\ \frac{bm}{h+m}\qquad \textbf{(B)}\ \frac{bh}{2h+2m}\qquad \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad \textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}$

Solution

Let $x$ be the value reduced from the base. Using the variables from the problem, the equation to solve is \[\frac{1}{2} (h + m)(b - x) = \frac{1}{2} \cdot \frac{bh}{2}\] Dividing both sides by $\frac{1}{2}$ and $h+m$ results in \[b - x = \frac{bh}{2(h+m)}\] \[-x = \frac{bh}{2(h+m)} - b\] \[x = b - \frac{bh}{2(h+m)}\] The right side can be written as a single fraction by giving $b$ a common denominator. \[x = \frac{2bh + 2bm}{2(h+m)} - \frac{bh}{2(h+m)}\] \[x = \frac{bh + 2bm}{2(h+m)}\] \[x = \frac{b(2m+h)}{2(h+m)}\] The answer is $\boxed{\textbf{(E)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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