Find the area of the following regular hexagon. Answer correctly with solution and get your answers as brainliest.
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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:
MN2 = AN2 + AM2
132 = 52 + AM2
AM2 = 169 – 25 = 144
AM = \(\sqrt{144}\) = 12cm.
Therefore, OM = RP = 2 x AM = 2 x 12 = 24 cm
Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR) + (area of triangle RPQ)
= (\(\frac{1}{2}\) x OM x AN) + (RP x PO) + (\(\frac{1}{2}\) x RP x BQ)
= (\(\frac{1}{2}\) x 24 x 5) + (24 x 13) + (\(\frac{1}{2}\) x 24 x 5)
= 60 + 312 + 60 = 432 cm2.
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:
MN2 = AN2 + AM2
132 = 52 + AM2
AM2 = 169 – 25 = 144
AM = \(\sqrt{144}\) = 12cm.
Therefore, OM = RP = 2 x AM = 2 x 12 = 24 cm
Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR) + (area of triangle RPQ)
= (\(\frac{1}{2}\) x OM x AN) + (RP x PO) + (\(\frac{1}{2}\) x RP x BQ)
= (\(\frac{1}{2}\) x 24 x 5) + (24 x 13) + (\(\frac{1}{2}\) x 24 x 5)
= 60 + 312 + 60 = 432 cm2.
Romancho:
plz tell where u found A and B
ok i understood
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