Find the area of the following regular hexagon.

Solution:

The given figure is:

It is given that the hexagon is regular. So, all its sides must be equal to $13 \mathrm{~cm}$.

Also, $\mathrm{AN}=\mathrm{BQ}$

$\mathrm{QB}+\mathrm{BA}+\mathrm{AN}=\mathrm{QN}$

$\mathrm{AN}+13+\mathrm{AN}=23$

$2 \mathrm{AN}=23-13=10$

$\mathrm{AN}=\frac{10}{2}=5 \mathrm{~cm}$

Hence, $\mathrm{AN}=\mathrm{BQ}=5 \mathrm{~cm}$

Now, in the right angle triangle MAN:

$\mathrm{MN}^{2}=\mathrm{AN}^{2}+\mathrm{AM}^{2}$

$13^{2}=5^{2}+\mathrm{AM}^{2}$

$\mathrm{AM}^{2}=169-25=144$

$\mathrm{AM}=\sqrt{144}=12 \mathrm{~cm} .$

$\therefore \mathrm{OM}=\mathrm{RP}=2 \times \mathrm{AM}=2 \times 12=24 \mathrm{~cm}$

Hence, area of the regular hexagon $=($ area of triangle $\mathrm{MON})+($ area of rectangle $\mathrm{MOPR})+($ area of triangle $\mathrm{RPQ})$

$=\left(\frac{1}{2} \times \mathrm{OM} \times \mathrm{AN}\right)+(\mathrm{RP} \times \mathrm{PO})+\left(\frac{1}{2} \times \mathrm{RP} \times \mathrm{BQ}\right)$

$=\left(\frac{1}{2} \times 24 \times 5\right)+(24 \times 13)+\left(\frac{1}{2} \times 24 \times 5\right)$

$=60+312+60$

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