ABC26GN4486 · Area

Let the length of side of the square be $a$ units. $$\begin{aligned} & \text { Then, } B E=E C=D F=F C=\frac{a}{2} \\ & \begin{aligned} A E & =\sqrt{(A B)^{2}+(B E)^{2}}=\sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}} \\ & =\sqrt{a^{2}+\frac{a^{2}}{4}}=\sqrt{\frac{5 a^{2}}{4}}=\frac{\sqrt{5} a}{2} . \end{aligned} \end{aligned}$$ Similarly, $A F=\frac{\sqrt{5} a}{2}$. $$\begin{aligned} & E F=\sqrt{(C E)^{2}+(C F)^{2}}=\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}}=\sqrt{\frac{2 a^{2}}{4}}=\frac{a}{\sqrt{2}} . \\ & E X=\frac{1}{2} E F=\frac{a}{2 \sqrt{2}} . \end{aligned}$$ $$\begin{aligned} A X=\sqrt{(A E)^{2}-(E X)^{2}} & =\sqrt{\left(\frac{\sqrt{5 a}}{2}\right)^{2}-\left(\frac{a}{2 \sqrt{2}}\right)^{2}} \\ & =\sqrt{\frac{5 a^{2}}{4}-\frac{a^{2}}{8}}=\sqrt{\frac{9 a^{2}}{8}}=\frac{3 a}{2 \sqrt{2}} . \end{aligned}$$ $$\therefore \text { Area }(\triangle \mathrm{AEF})=\frac{1}{2} \times E F \times A X=\frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{3 a}{2 \sqrt{2}}=\frac{3 a^{2}}{8} .$$$$\text { Required ratio }=\frac{3 a^{2}}{8}: a^{2}=3: 8 .$$