2006 AMC 10B Problems/Problem 20
Problem
In rectangle , we have , , , for some integer . What is the area of rectangle ?
Solution
Solution 1
Let the slope of be and the slope of be .
Since and form a right angle:
Using the distance formula:
Therefore the area of rectangle is
Solution 2
This solution is the same as Solution 1 up to the point where we find that .
We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse has legs and , while the triangle with hypotenuse has legs and . Aha! The two triangles are similar by SAS, with one triangle having side lengths times the other!
Let . Then from our reasoning above, we have . Finally, the area of the rectangle is .
Solution 3
We do not need to solve for y. We form a right triangle with as the hypotenuse and two adjacent sides lengths 200 and 2000, respectively. We form another right triangle with as the hypotenuse and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because and are perpendicular. , so the area
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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