# Difference between revisions of "2006 AMC 10B Problems/Problem 20"

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== Solution == | == Solution == | ||

+ | ===Solution 1=== | ||

Let the slope of <math>AB</math> be <math>m_1</math> and the slope of <math>AD</math> be <math>m_2</math>. | Let the slope of <math>AB</math> be <math>m_1</math> and the slope of <math>AD</math> be <math>m_2</math>. | ||

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Therefore the area of rectangle <math>ABCD</math> is <math> 200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E </math> | Therefore the area of rectangle <math>ABCD</math> is <math> 200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E </math> | ||

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+ | ===Solution 2=== | ||

+ | |||

+ | This solution is the same as Solution 1 up to the point where we find that <math>y=-42</math>. | ||

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+ | 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! | ||

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+ | 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>. | ||

== See Also == | == See Also == |

## Revision as of 10:48, 21 August 2011

## 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 .