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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===Solution 1===
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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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===Solution 2===
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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>.
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== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=19|num-a=21}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:18, 4 July 2013

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

Solution 1

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 $200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E$

Solution 2

This solution is the same as Solution 1 up to the point where we find that $y=-42$.

We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse $AB$ has legs $200$ and $2000$, while the triangle with hypotenuse $AD$ has legs $2$ and $20$. Aha! The two triangles are similar, with one triangle having side lengths $100$ times the other!

Let $AD=x$. Then from our reasoning above, we have $AB=100x$. Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400 \text{  (E)}}$.

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
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

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