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. 3 = (97 - x)/2

6 = 97 - x

x = 91

Thus, 91.

Page

Toolbox

Search

1959 AHSME Problems/Problem 22

Problem

Solution