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2006 AMC 10B Problems/Problem 20

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Revision as of 01:36, 4 August 2006 by Xantos C. Guin (Talk | contribs)

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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 $ABCD$ ?

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

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:

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

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

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

2006 AMC 10B Problems

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