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

Ragnarok23 (talk | contribs) |
(Added problem and solution) |
||

Line 1: | Line 1: | ||

== Problem == | == Problem == | ||

+ | In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some integer <math>y</math>. What is the area of rectangle <math>ABCD</math>? | ||

+ | |||

+ | <math> \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 </math> | ||

+ | |||

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

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

+ | |||

+ | <math> m_1 = \frac{178-(-22)}{2006-6} = \frac{1}{10} </math> | ||

+ | |||

+ | <math> m_2 = \frac{y-(-22)}{8-6} = \frac{y+22}{2} </math> | ||

+ | |||

+ | Since <math>AB</math> and <math>AD</math> form a right angle: | ||

+ | |||

+ | <math> m_2 = -\frac{1}{m_1} </math> | ||

+ | |||

+ | <math> m_2 = -10 </math> | ||

+ | |||

+ | <math> \frac{y+22}{2} = -10 </math> | ||

+ | |||

+ | <math> y = -42 </math> | ||

+ | |||

+ | Using the [[distance formula]]: | ||

+ | |||

+ | <math> AB = \sqrt{ (2006-6)^2 + (178-(-22))^2 } = \sqrt{ (2000)^2 + (200)^2 } = 200\sqrt{101} </math> | ||

+ | |||

+ | <math> AD = \sqrt{ (8-6)^2 + (-42-(-22))^2 } = \sqrt{ (2)^2 + (-20)^2 } = 2\sqrt{101} </math> | ||

+ | |||

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

+ | |||

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

*[[2006 AMC 10B Problems]] | *[[2006 AMC 10B Problems]] |

## Revision as of 12:43, 17 July 2006

## Problem

In rectangle , we have , , , for some integer . What is the area of rectangle ?

## Solution

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