Difference between revisions of "2008 AMC 12B Problems/Problem 25"
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Let <math>AP</math> and <math>BQ</math> meet <math>CD</math> at <math>X</math> and <math>Y</math>, respectively. | Let <math>AP</math> and <math>BQ</math> meet <math>CD</math> at <math>X</math> and <math>Y</math>, respectively. | ||
Revision as of 23:55, 25 September 2014
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
![2008 AMC 12B 25.jpg](https://wiki-images.artofproblemsolving.com//2/28/2008_AMC_12B_25.jpg)
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
Alternate Solution
Let and
meet
at
and
, respectively.
Since ,
, and they share
, triangles
and
are congruent.
By the same reasoning, we also have that triangles and
are congruent.
Hence, we have .
If we let the height of the trapezoid be , we have
.
Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.
Let the projections of and
to
be
and
, respectively.
We have ,
, and
.
Therefore, . Solving this, we easily get that
.
Multiplying this by 12, we find that the area of hexagon is
, which corresponds to answer choice
.
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 | |
All AMC 12 Problems and Solutions |
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