What is the area of the given figure ?

Answer: Option B

$$\begin{array}{rl}& AD=\sqrt{4^2+6^2}cm\\ & =\sqrt{52}cm\\ & =2\sqrt{13}cm\\ & =\left(2\times 3.6\right)cm\\ & =7.2cm\end{array}$$

Area of the whole figure :
$$=\text{Area (}△\text{AED) + }$$   $$\text{ Area (}\square \text{ ABCD) + }$$   $$\text{Area (}△\text{BFC)}$$
$$=\left[\left(\frac{1}{2}\times 4\times 6\right)+\left(12\times 7.2\right)+\left(\frac{1}{2}\times 4\times 6\right)\right]c{m}^2$$
$$\begin{array}{rl}& =\left(24+86.4\right)c{m}^2\\ & =110.4c{m}^2\end{array}$$