The area of an equilateral triangle with perimeter 12 cm is
A$16 \sqrt{3} cm^2$
B$8 \sqrt{3} cm^2$
C$4 \sqrt{3} cm^2$
D$6 \sqrt{3} cm^2$
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C$4 \sqrt{3} cm^2$
(c) $4 \sqrt{3} cm^2$ Let the length of each side be $a cm$. Then, $\begin{aligned}& \text { Perimeter }=12 cm \Rightarrow 3 a=12 cm \Rightarrow a=4 cm \\ \therefore & \text { Area }=\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4} \times 4^2 cm^2=4 \sqrt{3} cm^2\end{aligned}$
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In the given figure, the ratio $AD$ to $DC$ is $3$ to $2$. If the area of $\triangle\text{ABC}$ is $40 \mathrm{~cm}^2$, what is the area of $\triangle\text{BDC}?$