2008 AMC 12B Problems/Problem 25
Problem 25
Let be a trapezoid with
and
. Bisectors of
and
meet at
, and bisectors of
and
meet at
. What is the area of hexagon
?
Solution
Drop perpendiculars to from
and
, and call the intersections
respectively. Now,
and
. Thus,
.
We conclude
and
.
To simplify things even more, notice that
, so
.
Also,
So the area of
is:
Over to the other side: is
, and is therefore congruent to
. So
.
The area of the hexagon is clearly
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
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