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 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 (or any other two positive real numbers that multiply to 70). We can find and by intersecting lines, and then we calculate the area of using shoelace formula. This yields .
Solution 4
Note that triangle is similar to triangle with ratio . Similarly, triangle is similar to triangle with ratio . Thus, if then we know that and meaning and thus the ratio of to is which equals the ratio of the areas of to . If , then we know that and since and we want to find of this, we get our answer is . -SuperJJ
Solution 5
, ,
, , ,
, ,
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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