2001 AMC 12 Problems/Problem 22
Problem
In rectangle , points and lie on so that and is the midpoint of . Also, intersects at and at . The area of the rectangle is . Find the area of triangle .
Solution
Solution 1
Note that the triangles and are similar, as they have the same angles. Hence .
Also, triangles and are similar, hence .
We can now compute as . We have:
- .
- is of , as these two triangles have the same base , and is of , therefore also the height from onto is of the height from . Hence .
- is of , as the base is of the base , and the height from is of the height from . Hence .
- is of for similar reasons, hence .
Therefore .
Solution 2
As in the previous solution, we note the similar triangles and prove that is in and in of .
We can then compute that .
As is the midpoint of , the height from onto is of the height from onto . Therefore we have .
Solution 3
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are and . We can find and by intersecting lines, and then we calculate the area of using shoelace formula. This yields .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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