Difference between revisions of "1959 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | Let x be the length of the shorter base. | + | Let <math>x</math> be the length of the shorter base. Then: |
| − | |||
| − | + | <math>3 = \frac{97-x}{2}</math> | |
| − | x | + | <math>6 = 97-x</math> |
| − | Thus, 91. | + | <math>x = 91</math> |
| + | |||
| + | Thus, our answer is <math>\boxed{\textbf{(C) }91}</math>. | ||
==See also== | ==See also== | ||
Revision as of 11:39, 21 July 2024
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 
be the length of the shorter base. Then:
Thus, our answer is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
