1963 AHSME Problems/Problem 25

Problem

Point $F$ is taken in side $AD$ of square $ABCD$ . At $C$ a perpendicular is drawn to $CF$ , meeting $AB$ extended at $E$ . The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:

$[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote[/asy]$

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

Solution

Because $ABCD$ is a square, $DC = CB = 16$ , $DC \perp DA$ , and $CB \perp BA$ . Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$ , $\angle DCF = \angle BCE$ . Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$ .

From the congruency, $CF = CE$ . Using the area formula for a triangle, $CE = 20$ . Finally, by the Pythagorean Theorem, $BE = \sqrt{20^2 - 16^2} = 12$ , which is answer choice $\boxed{\textbf{(A)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1963 AHSME Problems/Problem 25

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