1959 AHSME Problems/Problem 22
Contents
[hide]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 1
Let be the length of the shorter base. Then:
Thus, our answer is .
Solution 2
Let the trapezoid be with , with as the midpoint of , and as the midpoint of , as in the diagram. As in the first solution, let the shorter base of the trapezoid () have length . Because , we can imagine shifting along by distance such that is at , at which point we get the following triangle:
Because is a midpoint connector of , , and so we have the equation . Solving for yields .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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