Difference between revisions of "1959 AHSME Problems/Problem 22"

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Thus, 91.
 
Thus, 91.
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==See also==
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{{AHSME 50p box|year=1959|num-b=21|num-a=23}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Revision as of 12:33, 21 July 2024

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.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AHSME Problems and Solutions

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