Difference between revisions of "2006 AMC 10B Problems/Problem 20"
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== 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>? | ||
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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 == | ||
| + | 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 
, 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 
