2017 AMC 10B Problems/Problem 15
Problem
Rectangle has and . Point is the foot of the perpendicular from to diagonal . What is the area of ?
Solution 1
First, note that because is a right triangle. In addition, we have , so . Using similar triangles within , we get that and .
Let be the foot of the perpendicular from to . Since and are parallel, is similar to . Therefore, we have . Since , . Note that is an altitude of from , which has length . Therefore, the area of is
Solution 2
From similar triangles, we know that (see Solution 1). Furthermore, we also know that from the rectangle. Using the sine formula for area, we have But, note that . Thus, we see that ~coolwiz
Solution 3
Alternatively, we can use coordinates. Denote as the origin. We find the equation for as , and as . Solving for yields . Our final answer then becomes
Solution 4
We note that the area of must equal area of because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of to be
Solution 5
We know all right triangles are 5-4-3, so the areas are proportional to the square of like sides. Area of is of . Using similar logic in Solution 4, Area of is the same as .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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