Question

$ABCDEF$ is a hexagon in which $AB=BC=CD=DE=2$ and $EF=FA=1$. Its interior angle $C$ is between ${90}^{\circ }$ and ${180}^{\circ }$ and $F$ is greater than ${180}^{\circ }$. The rest of the angles are ${90}^{\circ }$ each. What is its area?

Answer 1

Since, $FE||DC$

$FEDC$ is trapezium

$Area\left(FEDC\right)=\frac{1}{2}(1+2)\times 2=3$ $...........\left(1\right)$

Since $FABC$ is trapezium

$Area\left(FABC\right)=\frac{1}{2}(1+2)\times 2=3$ $...........\left(2\right)$

On adding both equations, we get

$Area\left(hexagon\right)=3+3=6unit{s}^2$

Hence, this is the answer.