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Difference between revisions of "1981 AHSME Problems/Problem 2"

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Revision as of 13:30, 28 June 2025

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

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

Solution

Note that $\triangle BCE$ is a right triangle. 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 $\boxed{\textbf{(C)}\ 3}$.

~superagh

See Also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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