A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12 \sqrt{2} cm$, then area of the triangle is
A$24 \sqrt{2} cm^2$
B$24 \sqrt{3} cm^2$
C$48 \sqrt{3} cm^2$
D$64 \sqrt{3} cm^2$
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D$64 \sqrt{3} cm^2$
D
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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 $\triangle\text{BDC}?$