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1997 AHSME Problems/Problem 9

Revision as of 20:26, 8 August 2011 by Talkinaway (talk | contribs) (Created page with "==Problem 9== In the figure, <math>ABCD</math> is a <math>2 \times 2</math> square, <math>E</math> is the midpoint of <math>\overline{AD}</math>, and <math>F</math> is on <math>...")
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Problem 9

In the figure, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$. If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair D = (2,2); pair E = midpoint(A--D); pair F = foot(C,B,E); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,N);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$E$",E,N);label("$F$",F,NW); draw(A--B--C--D--cycle); draw(B--E); draw(C--F); draw(rightanglemark(B,F,C,4));[/asy]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4}$

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