1959 AHSME Problems/Problem 22

Problem

The line joining the midpoints of the diagonals of a trapezoid has length $3$ . If the longer base is $97,$ then the shorter base is: $\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89$

Solution

Let $x$ be the length of the shorter base. Then:

$3 = \frac{97-x}{2}$

$6 = 97-x$

$x = 91$

Thus, our answer is $\boxed{\textbf{(C) }91}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1959 AHSME Problems/Problem 22

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