1981 AHSME Problems/Problem 2

Problem

Point $E$ is on side $AB$ of square $ABCD$ . If $EB$ has length one and $EC$ has length two, then the area of the square is

$[asy] unitsize(2cm); size(200); pair A=(0,0), B=(1.732,0), C=(1.732,1.732), D=(0,1.732), E=(0.732,0); draw(A--B--C--D--cycle,black); draw(C--E,black); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,S); label("$2$", C--E, NW); label("$1$", B--E, S); [/asy]$

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$

Solution

Note that $\triangle BCE$ is a right triangle. By the Pythagorean theorem, $BC^2 = CE^2 - BE^2 = 2^2-1^2=3$ , so the area of $ABCD$ is $\boxed{\textbf{(C)}\ 3}$ .

~superagh, edited by j314andrews.

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 2

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