2006 AMC 10B Problem 20

Below is the professionally curated solution for Problem 20 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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Concepts:coordinate geometry slope distance formula

Difficulty rating: 1580

20.

In rectangle $ABCD,$ we have $A=(6,-22),$ $B=(2006,178),$ and $D=(8,y)$ for some integer $y.$ What is the area of rectangle $ABCD?$

$4000$

$4040$

$4400$

$40{,}000$

$40{,}400$

Solution:

The slope of $\overline{AB}$ is $\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}.$ Since $\overline{AD}\perp\overline{AB},$ its slope is $-10,$ so $\tfrac{y+22}{8-6}=-10$ gives $y=-42.$

Then $AB=\sqrt{2000^2+200^2}=200\sqrt{101}$ and $AD=\sqrt{2^2+20^2}=2\sqrt{101}.$

The area is $200\sqrt{101}\cdot2\sqrt{101}=400\cdot101=40{,}400.$

Thus, the correct answer is E.

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