2006 AMC 10B Problem 20

Attempt Problem 20 of the 2006 AMC 10B below, then check your answer against the professionally curated solution 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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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

Answer: E

Concepts:coordinate geometry slope distance formula

Difficulty rating: 1580

Small hint:

\overline{AB}\perp\overline{AD},

so their slopes multiply to

-1

Big hint:

The area is

AB\cdot AD

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.