Question

Find the area of the following regular hexagon.

Answer 1

The given figure is:
Join QN.

It is given that the hexagon is regular. So, all its sides must be equal to 13 cm.
Also, AN = BQ
QB+BA+AN = QN
AN+13+AN = 23
2AN = 23-13 = 10
\[AN =\frac{10}{2}= 5 cm\]
Hence, AN = BQ = 5 cm
Now, in the right angle triangle MAN:
\[ {MN}^2 {=AN}^2 {+AM}^2 \]
\[ {13}^2 {=5}^2 {+AM}^2 \]
\[ {AM}^2 =169-25=144\]
\[AM=\sqrt{144}=12cm.\]
\[\therefore OM = RP = 2\times AM = 2\times12 = 24 cm\]
Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR)+(area of triangle RPQ)
\[=(\frac{1}{2}\times OM\times AN)+(RP\times PO)+(\frac{1}{2}\times RP\times BQ)\]
\[=(\frac{1}{2}\times24\times5)+(24\times13)+(\frac{1}{2}\times24\times5)\]
\[=60+312+60\]
\[ {=432 cm}^2\]