$$
\begin{array}{l}
\angle E D F=360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}(\angle \mathrm{s} \text { sum of polygon) } \\
\angle B A D=\angle B C D=180^{\circ}-120^{\circ}=60^{\circ} \text { (int. } \angle \mathrm{s} / / \text {-lines) } \\
B C=\frac{2}{\cos 60^{\circ}} \mathrm{cm}=4 \mathrm{~cm}=A D \\
B E=2 \tan 60^{\circ}=2 \sqrt{3} \mathrm{~cm} \\
A F=(4-1) \mathrm{cm}=3 \mathrm{~cm} \\
A B=\frac{3}{\cos 60^{\circ}} \mathrm{cm}=6 \mathrm{~cm}
\end{array}
$$

Area of $A B C D=A B \times B E=6 \times 2 \sqrt{3} \mathrm{~cm}^{2}=12 \sqrt{3} \mathrm{~cm}^{2}$
G4.1 In Figure $1, A B C D$ is a parallelogram, $B E \perp C D, B F \perp A D$, $C E=2 \mathrm{~cm}, D F=1 \mathrm{~cm}$ and $\angle E B F=60^{\circ}$.
If the area of the parallelogram $A B C D$ is $R \mathrm{~cm}^{2}$, find the value of $R$.