1966 AHSME Problems/Problem 2

Problem

When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is

$\text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease}$

Solution

Let the base of the original triangle be $b$ and the height be $h$ . We know the new base is $\frac{11b}{10}$ and the new height is $\frac{9h}{10}$ . Thus we see the area of the original triangle is $\frac{bh}{2}$ and the area of the new triangle is $\frac{99bh}{200}$ . It follows the percentage decrease is $1$ % because $\frac{\frac{bh}{2}-\frac{99bh}{200}}{\frac{bh}{2}}=\frac{1}{100}$ . So our answer is $1$ % $\text{decrease}$ $(E)$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1966 AHSME Problems/Problem 2

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