Difference between revisions of "1959 AHSME Problems/Problem 22"
(Created page with "Let x be the length of the shorter base. 3 = (97 - x)/2 6 = 97 - x x = 91 Thus, 91.") |
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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 == | ||
Let x be the length of the shorter base. | Let x be the length of the shorter base. | ||
3 = (97 - x)/2 | 3 = (97 - x)/2 | ||
+ | |||
6 = 97 - x | 6 = 97 - x | ||
+ | |||
x = 91 | x = 91 | ||
+ | |||
Thus, 91. | Thus, 91. |
Latest revision as of 19:23, 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.