I4.4 In Figure 2, there are a square $A B C D$ with side length $(R-1) \mathrm{cm}$ and an equilateral triangle $A E F$. ( $E$ and $F$ are points on $B C$ and $C D$ respectively). If the area of $\triangle A E F$ is $(S-3) \mathrm{cm}^{2}$, find the value of $S$.
Let $A F=x \mathrm{~cm}=F E=A E$
$$
\begin{array}{l}
\angle F A E=60^{\circ}, \angle D A F=\angle B A E=15^{\circ} \\
A D=1=A F \cos 15^{\circ}=x \cos 15^{\circ} \Rightarrow x=\sec 15^{\circ}
\end{array}
$$
$$
S-3=\frac{1}{2} x^{2} \sin 60^{\circ}=\frac{1}{2} \cdot \frac{1}{\cos ^{2} 15^{\circ}} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \cdot \frac{1}{1+\cos 30^{\circ}}=\frac{\sqrt{3}}{2} \cdot \frac{1}{1+\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{2} \cdot \frac{2}{2+\sqrt{3}}=2 \sqrt{3}-3 ; S=2 \sqrt{3}
$$

Method 2
Let $C E=C F=x \mathrm{~cm}$
Then $B E=D F=(1-x) \mathrm{cm}$
By Pythagoras' theorem,
$$
\begin{array}{l}
A F=F E \Rightarrow 1+(1-x)^{2}=x^{2}+x^{2} \\
2-2 x+x^{2}=2 x^{2} \\
x^{2}+2 x-2=0 \Rightarrow x^{2}=2-2 x \\
x=-1+\sqrt{3}
\end{array}
$$

Area of $\triangle A F E=$ Area of square - area of $\triangle C E F-2$ area of $\triangle A D F$
$$
\begin{array}{c}
=1-\frac{x^{2}}{2}-2 \times \frac{1 \times(1-x)}{2}=x-\frac{x^{2}}{2}=x-\frac{2-2 x}{2}=2 x-1 \\
=2(-1+\sqrt{3})-1=2 \sqrt{3}-3 \\
S-3=2 \sqrt{3}-3 ; S=2 \sqrt{3}
\end{array}
$$