2000 AMC 10 Problems/Problem 19
Problem
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
Solution
Let the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .
By similar triangles, we have
This is choice
(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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