$\frac{\text{AD}}{\text{DC}}=\frac32$
Let $AD = 3x$ and $DC = 2x$
Area of $\triangle\text{ABC}=\frac12\times\text{AC}\times\text{BC}$ $(BE = h)$
$\Rightarrow40=\frac12\times5\text{x}\times\text{h}$
$\Rightarrow 80 = 5xh$
$\Rightarrow xh =$ $16 \mathrm{~cm}^2$$ ....(1)$
Now, Area of $\triangle\text{ABD}=\frac12\times3\text{x}\times\text{h}=\frac{3\times\text{h}}{2}=\frac{3}{2}\times16=24\text{cm}^2$
Area of $\triangle\text{BDC}=$ Area of $\triangle\text{ABC}-$ Area of $\triangle\text{ABD}=40-24=16\text{cm}^2$
Hence, correct option is (a).