1981 AHSME Problems/Problem 2

Revision as of 21:16, 7 August 2020 by Superagh (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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


Note that $\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\sqrt{3}$. Since this is the side length of the square, the area of $ABCD$ is $\boxed{\textbf{(C)}\ 3}$.