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

 
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== Problem ==
 
== Problem ==
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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>?
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<math> \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 </math>
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== Solution ==
 
== Solution ==
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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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<math> m_1 = \frac{178-(-22)}{2006-6} = \frac{1}{10} </math>
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<math> m_2 = \frac{y-(-22)}{8-6} = \frac{y+22}{2} </math>
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Since <math>AB</math> and <math>AD</math> form a right angle:
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<math> m_2 = -\frac{1}{m_1} </math>
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<math> m_2 = -10 </math>
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<math> \frac{y+22}{2} = -10 </math>
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<math> y = -42 </math>
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Using the [[distance formula]]:
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<math> AB = \sqrt{ (2006-6)^2 + (178-(-22))^2 } = \sqrt{ (2000)^2 + (200)^2 } = 200\sqrt{101} </math>
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<math> AD = \sqrt{ (8-6)^2 + (-42-(-22))^2 } = \sqrt{ (2)^2 + (-20)^2 } = 2\sqrt{101} </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>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 13:43, 17 July 2006

Problem

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

$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$

Solution

Let the slope of $AB$ be $m_1$ and the slope of $AD$ be $m_2$.

$m_1 = \frac{178-(-22)}{2006-6} = \frac{1}{10}$

$m_2 = \frac{y-(-22)}{8-6} = \frac{y+22}{2}$

Since $AB$ and $AD$ form a right angle:

$m_2 = -\frac{1}{m_1}$

$m_2 = -10$

$\frac{y+22}{2} = -10$

$y = -42$

Using the distance formula:

$AB = \sqrt{ (2006-6)^2 + (178-(-22))^2 } = \sqrt{ (2000)^2 + (200)^2 } = 200\sqrt{101}$

$AD = \sqrt{ (8-6)^2 + (-42-(-22))^2 } = \sqrt{ (2)^2 + (-20)^2 } = 2\sqrt{101}$

Therefore the area of rectangle $ABCD$ is $200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E$

See Also

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