Find the area of the figure below which is made of two equal trapeziums. Two parallel sides of a trapezium are of lengths $57cm$ and $39cm$. The distance between the parallel sides $AB$ and $FC$ is $28cm$.

The correct option is A. $2688c{m}^2$

Area of the figure $ABCDEF=$ Area of trapezium $BCAF+$ Area of trapezium $FCDE$

Let ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ be the parallel sides of the trapezium and ${}^{\prime }{h}^{\prime }$ be the height of the trapezium.

Area of trapezium $=\frac{1}{2}\left(\text{sum of parallel sides}\right)\times height$

$=\frac{1}{2}\times (57+39)\times 28=1344c{m}^2$

Area of the figure $ABCDEF=2\times$ Area of trapezium $BCAF$ [As two trapezium are same]

$=2\times 1344$

$=2688c{m}^2$