### Find the area of the hexagon $ABCDEF$ in which each side measures $17 \space cm,$ width $BD = 30 \space cm,$ and height $CF = 33 \space cm.$ C A B E D D 33 cm F 30 cm 17 cm

$750 \space cm^2$
1. Let us join $CF$ and draw $AL \perp CF.$
Now, $AL = \dfrac { BD }{ 2 } = \dfrac { 30 }{ 2 } = 15 \space cm$
2. Area of hexagon $ABCDEF =$ Area of trapezium $ABCF$ + Area of trapezium $CDEF$ \begin{align} &= 2 \times \text { Area of trapezium ABCF } && \text{[Area of both the trapeziums will be same.]}\\ &= 2 \times \dfrac {1}{2}(AB + CF) \times AL \\ &= (17 + 33) \times 15 \space cm^2 \\ &= 750 \space cm^2 \end{align}