2004 AMC 10B Problems/Problem 18
Contents
Problem
In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?
Solution 1
Let . Because is divided into four triangles, .
Because of triangle area, .
and , so .
, so .
Solution 2
First of all, note that , and therefore .
Draw the height from onto as in the picture below:
Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is , hence . Now the area of can be computed as = .
Similarly we can find that as well.
Hence , and the answer is .
Solution 3 (Coordinate Geometry)
We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42.
Solution 4
You can also place a point on such that is , creating trapezoid . Then, you can find the area of the trapezoid, subtract the area of the two right triangles and , divide by the area of , and get the ratio of .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |
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