1988 AHSME Problems/Problem 20
Problem
In one of the adjoining figures a square of side is dissected into four pieces so that
and
are the midpoints
of opposite sides and
is perpendicular to
. These four pieces can then be reassembled into a rectangle as shown
in the second figure. The ratio of height to base,
, in this rectangle is
Solution
Within , the parallelogram piece has vertical side
, and diagonal side
. Thus the triangle in the bottom-right hand corner (the one with horizontal side
) must have hypotenuse
, and the only such triangle in the original figure is
, so we deduce
Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is
, and thus the vertical side
is
, so the required ratio is
, which is
.
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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