2002 AMC 10A Problems/Problem 25
Problem
In trapezoid with bases and , we have , , , and . The area of is
Solution
Solution 1
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet at point :
Since we have , with the ratio of proportionality being . Thus So the sides of are , which we recognize to be a right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),
Solution 2
Draw altitudes from points and :
Translate the triangle so that coincides with . We get the following triangle:
The length of in this triangle is equal to the length of the original , minus the length of . Thus .
Therefore is a well-known right triangle. Its area is , and therefore its altitude is .
Now the area of the original trapezoid is
See also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last problem | |
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All AMC 10 Problems and Solutions |
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