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>? | ||
| + | |||
| + | <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 == | ||
| + | ===Solution 1=== | ||
| + | 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> | ||
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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=== | ||
| + | |||
| + | 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 == | ||
| − | + | {{AMC10 box|year=2006|ab=B|num-b=19|num-a=21}} | |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 12:18, 4 July 2013
Problem
In rectangle 
, we have 
, 
, 
, for some integer 
. What is the area of rectangle ?
Solution
Solution 1
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 
Solution 2
This solution is the same as Solution 1 up to the point where we find that 
.
We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse has legs 
and 
, while the triangle with hypotenuse has legs 
and 
. Aha! The two triangles are similar, with one triangle having side lengths 
times the other!
Let 
. Then from our reasoning above, we have 
. Finally, the area of the rectangle is 
.
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
