Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm,

Solution:

Let $\mathrm{AL}=\mathrm{DM}=\mathrm{x} \mathrm{cm}$

$\mathrm{LM}=\mathrm{BC}=13 \mathrm{~cm}$

$\therefore x+13+x=23$

$\Rightarrow 2 x+13=23$

$\Rightarrow 2 x=(23-13)$

$\Rightarrow 2 x=10$

$\Rightarrow x=5$

$\therefore \mathrm{AL}=5 \mathrm{~cm}$

From the right $\Delta \mathrm{AFL}$, we have $:$

$F L^{2}=A F^{2}-A L^{2}$

$\Rightarrow F L^{2}=\left\{\left(13^{2}\right)-(5)^{2}\right\}$

$\Rightarrow F L^{2}=(169-25)$

$\Rightarrow F L^{2}=144$

$\Rightarrow F L=\sqrt{144}$

$\Rightarrow F L=12 \mathrm{~cm}$

$\therefore F L=B L=12 \mathrm{~cm}$

Area of a regular hexagon $=(A$ rea of the trapezium ADEF $)+(A$ rea of the trapezium ABCD $)$

$=2($ Area of trapezium ADEF $)$

$=2\left\{\frac{1}{2} \times(A D+E F) \times F L\right\}$

$=2\left\{\frac{1}{2} \times(23+13) \times 12\right\} \mathrm{cm}^{2}$

$=2\left(\frac{1}{2} \times 36 \times 12\right) \mathrm{cm}^{2}$

$=432 \mathrm{~cm}^{2}$

Hence, the area of the given regular hexagon is $432 \mathrm{~cm}^{2}$.