#### Question

Find the area of the following regular hexagon.

#### Solution

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)

\[=(\frac{1}{2}\times OM\times AN)+(RP\times PO)+(\frac{1}{2}\times RP\times BQ)\]

\[=(\frac{1}{2}\times24\times5)+(24\times13)+(\frac{1}{2}\times24\times5)\]

\[=60+312+60\]

\[ {=432 cm}^2\]

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)

\[=(\frac{1}{2}\times OM\times AN)+(RP\times PO)+(\frac{1}{2}\times RP\times BQ)\]

\[=(\frac{1}{2}\times24\times5)+(24\times13)+(\frac{1}{2}\times24\times5)\]

\[=60+312+60\]

\[ {=432 cm}^2\]

Is there an error in this question or solution?

Solution Find the Area of the Following Regular Hexagon. Concept: Area of a Polygon.