Difference between revisions of "2002 AMC 10A Problems/Problem 25"
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Revision as of 16:16, 12 April 2013
Contents
[hide]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 | |
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 10 Problems and Solutions |