Question

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

Answer 1

AD = 23 cm, LM = 13 cm
$\therefore AL=MD=\frac{23-13}{2}=\frac{10}{2}$= 5 cm
In $\Delta$ ALB
$A{B}^2=A{L}^2+L{B}^2$
$(13{)}^2=(5{)}^2+L{B}^2$
$169=25+L{B}^2$
$\Rightarrow L{B}^2=169-25=144=(12{)}^2$
$\therefore LB=12cm$
$\therefore FB=EC=2\times 12=24cm$
AF =BC = LM = 13 cm
Area of ABCDEF = area of rect. ECBF + area $\Delta AFB+area\Delta DCE$
= $24\times 13+\frac{1}{2}BF\times AL+\frac{1}{2}EC\times DM$
= $312+\frac{1}{2}\times 24\times 5+\frac{1}{2}EC\times DM$
= $312+\frac{1}{2}\times 24\times 5+\frac{1}{2}\times 24\times 5$
= $312+60+60=432c{m}^2$