Question
Find the area of the following regular hexagon.
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Answer
In the regular hexagon MNOPQR There are two triangle and one rectangle.
Join MQ, MO and RP
NQ = 23cm,
NA = BQ $=\frac{10}{2}=5\text{cm}$
MR = OP = 13cm
In right $\triangle\text{BDQ,}$
$\text{PQ}^2=\text{BQ}^2+\text{BP}^2$
$\Rightarrow(13)^2=(5)^2+\text{BP}^2$
$\Rightarrow169=25+\text{BP}^2$
$\Rightarrow\text{BP}^2=169-25=144=(12)^2$
$\text{BP}=12\text{cm}$
$\text{PR}=\text{MO}=2\times12=24\text{cm}$
Now area of rectangle RPOM $=R P \times P O=24 \times 13=312 cm^2$
Area of $\triangle PRQ =\frac{1}{2} \times PR \times BQ$
$=\frac{1}{2} \times 24 \times 5=60 cm^2$
Similarly area $\triangle M O N=60 cm^2$
Area of the hexagon MNOPQR $=312+60+60=432 cm^2$
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