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2006 AMC 10B Problem 20

Problem 20 of 25HarderGeometry

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

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Solution

The slope of AB‾\overline{AB} is 178−(−22)2006−6=2002000=110.\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}. Since AD‾⊥AB‾,\overline{AD}\perp\overline{AB}, its slope is −10,-10, so y+228−6=−10\tfrac{y+22}{8-6}=-10 gives y=−42.y=-42. Then AB=20002+2002AB=\sqrt{2000^2+200^2} =200101=200\sqrt{101} and AD=22+202=2101.AD=\sqrt{2^2+20^2}=2\sqrt{101}. The area is 200101⋅2101200\sqrt{101}\cdot2\sqrt{101} =400⋅101=400\cdot101 =40,400.=40{,}400. Thus, the correct answer is E.

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Concepts: coordinate geometry · slope · distance formula

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