Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{AL}=\mathrm{DM}=\mathrm{x}\mathrm{cm} \\ \mathrm{LM}=\mathrm{BC}=13\mathrm{cm} \\ \therefore x+13+x=23 \\ \Rightarrow 2x+13=23 \\ \Rightarrow 2x=\left(23-13\right) \\ \Rightarrow 2x=10 \\ \Rightarrow x=5 \\ \therefore \mathrm{AL}=5\mathrm{cm} \\ \mathrm{From}\mathrm{the}\mathrm{right}∆\mathrm{AFL},\mathrm{we}\mathrm{have}: \\ F{L}^2=A{F}^2-A{L}^2 \\ \Rightarrow F{L}^2=\left\{\left({13}^2\right)-{\left(5\right)}^2\right\} \\ \Rightarrow F{L}^2=\left(169-25\right) \\ \Rightarrow F{L}^2=144 \\ \Rightarrow FL=\sqrt{144} \\ \Rightarrow FL=12cm \\ \therefore \mathrm{FL}=\mathrm{BL}=12\mathrm{cm} \\ \mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{regular}\mathrm{hexagon}=\left(A\mathrm{rea}\mathrm{of}the\mathrm{trapezium}\mathrm{ADEF}\right)+\left(A\mathrm{rea}\mathrm{of}the\mathrm{trapezium}\mathrm{ABCD}\right) \end{gathered}$$
$$\begin{gathered} =2\left(AreaoftrapeziumADEF\right) \\ =2\left\{\frac{1}{2}\times \left(AD+EF\right)\times FL\right\} \\ =2\left\{\frac{1}{2}\times \left(23+13\right)\times 12\right\}c{m}^2 \\ =2\left(\frac{1}{2}\times 36\times 12\right)c{m}^2 \\ =432c{m}^2 \end{gathered}$$
$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{regular}\mathrm{hexagon}\mathrm{is}432{\mathrm{cm}}^2.$