Difference between revisions of "1959 AHSME Problems/Problem 22"
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| + | == Problem == | ||
| + | The line joining the midpoints of the diagonals of a trapezoid has length <math>3</math>. If the longer base is <math>97,</math> then the shorter base is: <math>\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89</math> | ||
| + | |||
== Solution == | == Solution == | ||
Let x be the length of the shorter base. | Let x be the length of the shorter base. | ||
Revision as of 18:23, 7 April 2023
Problem
The line joining the midpoints of the diagonals of a trapezoid has length 
. If the longer base is 
then the shorter base is: 
Solution
Let x be the length of the shorter base. 3 = (97 - x)/2
6 = 97 - x
x = 91
Thus, 91.
