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 $\boxed{\textbf{(E)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1961 AHSME Problems/Problem 16

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