2006 SMT/Geometry Problems/Problem 2

Problem

Given square $ABCD$ of side length $1$, with $E$ on $\overline{CD}$ and $F$ in the interior of the square so that $\overline{EF}\perp\overline{CD}$ and $\overline{AF}\cong\overline{BF}\cong\overline{EF}$, find the area of quadrilateral $ADEF$.

Solution

[asy] draw(unitsquare); draw((1/2,0)--(1/2,5/8)--(0,1)); draw((1/2,5/8)--(1,1)); draw((0,5/8)--(1/2,5/8)); label("$A$",(0,1),NW); label("$B$",(1,1),NE); label("$C$",(1,0),SE); label("$D$",(0,0),SW); label("$F$",(1/2,5/8),SE); label("$Q$",(0,5/8),W); label("$x$",(1/2,5/16),E); label("$x$",(0,5/16),W); label("$1-x$",(0,13/16),W); label("$x$",(1/4,13/16),NE); label("$\frac{1}{2}$",(1/4,0),S); label("$\frac{1}{2}$",(1/4,5/8),S); label("$x$",(3/4,13/16),NW); label("$E$",(1/2,0),S); [/asy]

Since $AF=BF$, $F$ lies on the perpendicular bisector of $AB$, which is the same as the perpendicular bisector of $CD$. Therefore, since $E$ is the foot of the perpendicular from $F$ to $CD$, $E$ is the midpoint of $CD$, and $ED=\frac{1}{2}$.


Let $Q$ be the foot of the perpendicular from $F$ to $DA$. Since $DEFQ$ is a rectangle, $FQ=ED=\frac{1}{2}$. Let $AF=BF=EF=x$. We have $DQ=EF=x$, and $AQ=1-x$. Therefore, from right triangle $AQF$, $(1-x)^2+\left(\frac{1}{2}\right)^2=x^2$, and solving this gives $x=\frac{5}{8}$.


Finally, $ADEF$ is a trapezoid with bases of length $1$ and $x=\frac{5}{8}$ and height $\frac{1}{2}$, so its area is $\left(\frac{1}{2}\right)\left(1+\frac{5}{8}\right)\left(\frac{1}{2}\right)=\boxed{\frac{13}{32}}$.

See Also

2006 SMT/Geometry Problems