2017 AMC 10B Problems/Problem 15
Contents
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 the 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 corresponding sides. Area of is of . Using similar logic in Solution 4, Area of is the same as .
Solution 6
Drop an altitude from to and call its foot . We have that since both are equal to two times the area of . Since , , and , we can calculate that . If is extended to meet at point , . Therefore, .
- Fasolinka
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=369
~ pi_is_3.14
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |
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