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

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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>.
 
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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===Solution 3===
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We do not need to solve for y. We form a right triangle with <math>AB</math> as the hypotenuse and two adjacent sides lengths 200 and 2000, respectively. We form another right triangle with <math>AD</math> as the hypotenuse and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because <math>AD</math> and <math>AB</math> are perpendicular. <math> \frac{AB}{AD} = \frac{200}{2} </math>, so the area <math> AB \cdot AD = \frac {AB^2}{100} = \frac {2000^2 + 200^2}{100} = \boxed{40400 \text{  (E)}} </math>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:43, 29 May 2023

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 by SAS, 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 $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400 \text{ (E)}}$.

Solution 3

We do not need to solve for y. We form a right triangle with $AB$ as the hypotenuse and two adjacent sides lengths 200 and 2000, respectively. We form another right triangle with $AD$ as the hypotenuse and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because $AD$ and $AB$ are perpendicular. $\frac{AB}{AD} = \frac{200}{2}$, so the area $AB \cdot AD = \frac {AB^2}{100} = \frac {2000^2 + 200^2}{100} = \boxed{40400 \text{ (E)}}$

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

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