What is the area of the given figure ?

Solution(By Examveda Team)

$$\eqalign{ & AD = \sqrt {{4^2} + {6^2}} cm \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {52} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {13} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times 3.6} \right)cm \cr & \,\,\,\,\,\,\,\,\,\,\, = 7.2\,cm \cr} $$

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