2018 AMC 8 第 22 题

Let the square have side length $s$.

Find the small triangle cut from $\triangle ACD$, then subtract it from half the square.

Let the square have side length $s,$ and let $H$ be the foot of the perpendicular from $F$ to $\overline{BC}.$ The right triangles $CAB$ and $CFH$ are similar, so $CH=FH.$ Also, triangles $BFH$ and $BEC$ are similar, giving $$\frac{FH}{EC}=\frac{BH}{BC}.$$ Since $EC=\frac{s}{2},$ $BC=s,$ and $BH=s-CH=s-FH,$ this becomes $\frac{2FH}{s}=1-\frac{FH}{s},$ so $FH=CH=\frac{s}{3}.$

Thus triangle $EFC$ has base $EC=\frac{s}{2}$ and height $CH=\frac{s}{3},$ so its area is $\frac{s^2}{12}.$ Therefore $$\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12}. \end{aligned}$$ Since this area is $45$, we get $s^2=108$, the area of the square.