Difference between revisions of "2006 AMC 10B Problems/Problem 20"
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This solution is the same as Solution 1 up to the point where we find that <math>y=-42</math>. | This solution is the same as Solution 1 up to the point where we find that <math>y=-42</math>. | ||
− | We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse <math>AB</math> has legs <math>200</math> and <math>2000</math>, while the triangle with hypotenuse <math>AD</math> has legs <math>2</math> and <math>20</math>. Aha! The two triangles are similar, with one triangle having side lengths <math>100</math> times the other! | + | We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse <math>AB</math> has legs <math>200</math> and <math>2000</math>, while the triangle with hypotenuse <math>AD</math> has legs <math>2</math> and <math>20</math>. Aha! The two triangles are similar, with one triangle having side lengths <math>100</math> times the other!! |
Let <math>AD=x</math>. Then from our reasoning above, we have <math>AB=100x</math>. Finally, the area of the rectangle is <math>100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400 \text{ (E)}}</math>. | Let <math>AD=x</math>. Then from our reasoning above, we have <math>AB=100x</math>. Finally, the area of the rectangle is <math>100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400 \text{ (E)}}</math>. |
Revision as of 14:27, 5 May 2012
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, with one triangle having side lengths times the other!!
Let . Then from our reasoning above, we have . Finally, the area of the rectangle is .
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |