Difference between revisions of "2008 AMC 12B Problems/Problem 25"
FantasyLover (talk | contribs) (→Solution) |
(→See Also) |
||
Line 40: | Line 40: | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}} | {{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] |
Revision as of 22:36, 10 April 2013
Contents
[hide]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
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 |