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 13:05, 29 December 2013
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
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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
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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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