G2.4 In the figure below, there are 4 identical circles placed inside an equilateral triangle. If the radii of the circles are 1 unit, what is the value of the area of the triangle?
Let the triangle be $A B C, O$ is the centre of the middle circle, $D, E, F$ are the centres of the other 3 circles respectively. Let $P, Q, R, S, T, U$ be the points of contact as shown.
$$
\begin{array}{l}
D P \perp A B, E Q \perp A B, E R \perp B C, F S \perp B C, F T \perp A C, \\
D U \perp A C \text { (tangent } \perp \text { radius) } \\
D P=E Q=E R=F S=F T=D U=1 \text { (radii) } \\
O D=O E=O F=2 \text { (radii } 1 \text { r radii } 1 \text { ) } \\
\triangle O D E \cong \triangle O E F \cong \triangle O F D \text { (S.S.S.) } \\
\angle D O E=\angle E O F=\angle F O D \text { (corr. } \angle \mathrm{s} \cong \triangle \mathrm{s}) \\
\angle D O E+\angle E O F+\angle F O D=360^{\circ}(\angle \mathrm{s} \text { at a point) } \\
\therefore \angle D O E=\angle E O F=\angle F O D=120^{\circ}
\end{array}
$$
$D P Q E, E R S F, F T U D$ are rectangles (opp. sides are eq. and //)
$$
D E=E F=F D=2 \times 2 \sin 60^{\circ}=2 \sqrt{3}=P Q=R S=T U
$$

In $\triangle A D U, \angle D A U=30^{\circ}, D U=1, D U \perp A U$,
$$
\begin{array}{l}
A U=1 \tan 60^{\circ}=\sqrt{3} \\
\therefore A B=B C=C A=2 \sqrt{3}+2 \sqrt{3}=4 \sqrt{3}
\end{array}
$$

Area of $\triangle A B C=\frac{1}{2} \cdot(4 \sqrt{3})^{2} \sin 60^{\circ}=12 \sqrt{3}$