Question

Find the area of the following regular hexagon.

Answer 1

$\text{The given figure is:}$

Join QN.



$$\begin{gathered} \text{It is given that the hexagon is regular. So, all its sides must be equal to 13 cm.} \\ \text{Also, AN = BQ} \\ \text{QB+BA+AN = QN} \\ \text{AN+13+AN = 23} \\ \text{2AN = 23-13 = 10} \\ \text{AN =}\frac{10}{2}\text{= 5 cm} \\ \text{Hence, AN = BQ = 5 cm} \\ \text{Now, in the right angle triangle MAN:} \\ {\text{MN}}^2{\text{=AN}}^2{\text{+AM}}^2 \\ {\text{13}}^2{\text{=5}}^2{\text{+AM}}^2 \\ {\text{AM}}^2\text{=169-25=144} \\ \text{AM=}\sqrt{144}\text{=12cm.} \\ \text{∴ OM = RP = 2×AM = 2×12 = 24 cm} \\ \text{Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR)+(area of triangle RPQ)} \\ \text{=(}\frac{1}{2}\text{×OM×AN)+(RP×PO)+(}\frac{1}{2}\text{×RP×BQ)} \\ \text{=(}\frac{1}{2}\text{×24×5)+(24×13)+(}\frac{1}{2}\text{×24×5)} \\ \text{=60+312+60} \\ {\text{=432 cm}}^2 \end{gathered}$$