2006 AMC 10B Problems/Problem 20
Contents
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
Solution 4 (answer choices)
In order to find the area of rectangle , we need to find first. Using the distance formula, we can derive:
Now we can look at the answer choices. Because all of them are integers, then we know that has to also contain to ensure that is an integer. Once you know that, you can guarantee that the answer must be a multiple of 101. (If you're wondering why there can't be a fraction like , there can't because is an integer the number under the square root in , is an integer.) Having that information narrows down the answer choices to just , and .
Looking back at what we found for , which is , and what we know about , that it has to contain . we know that the answer is at least , which is already greater than answer choice , which is . From this, we can conclude that the answer is
~ @flyingkinder123
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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