2008 AMC 12B Problems/Problem 25
Problem
Let be a trapezoid with and . Bisectors of and meet at , and bisectors of and meet at . What is the area of hexagon ?
Solution
Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving.
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
Note: Once is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, and . . The height is one half of which is . So the area is
Solution 2
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 .
Solution 3
is the intersection of the angle bisectors of and . By definition, angle bisectors are always equidistant from the sides of the angle, so is equidistant from , , and . Likewise, point is equidistant from , , and . Because both points and are equidistant from and and the distance between and is constant, the common distances from each of the points to the mentioned segments is equal for and . Call this distance .
The distance between a point and a line is the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from to is , so the area of is equal to . Similarly, the area of is . The altitude of the trapezoid is , because it is the sum of the distances from either or to and . This means the area of trapezoid is . Now, the area of hexagon is the area of trapezoid , minus the areas of triangles and . This is . Now it remains to find .
We let and be the feet of the altitudes of and , respectively, to . We define and . We know that , so and . By the Pythagorean Theorem on and , we get and , respectively. Subtracting the second equation from the first gives us . The left hand side of this equation is a difference of squares and factors to . We know that , so . Now we can solve for by adding the two equations we just got to see that , or .
We now solve for . We know that , so and . We multiply both sides of this equation by to get . However, the area of hexagon is , so the answer is , or answer choice .
Solution 4
Let respectively. Since we have similarly we get Thus, is both an angle bisector and altitude of so Using the same logic on gives is a rhombus; similarly, is a rhombus. Then, where is the height of trapezoid Finding is the same as finding the altitude to the side of length in a triangle, and using Heron's, the area of such a triangle is Multiply to get our answer is
Solution 5
Like above solutions, find out the height of is Let be the intersection of the line through and parallel to and the intersection of the line through and parallel to Angle chasing shows that is the midpoint of and midpoint of Then from midline theorem, are collinear, and likewise for Thus, the line through is in fact the midline of
Let Then, angle chasing shows that not only bisects but is also perpendicular to it. This makes it a perpendicular bisector. The same is true for and Thus, and This means We can now find as the midline of Thus,
Now, the answer is simply finding the area of plus area of This is
~ MelonGirl
See Also
Video Solution:
https://www.youtube.com/watch?v=pwDV9p9eFQQ
2008 AMC 12B (Problems • Answer Key • Resources) | |
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