![drawn]()
It is obvious from the above figure that each side of the regular octagon subtends equal angle of magnitude 360/8=45^@3608=45∘ at the centerOO of the octagon .
One of the eight isosceles triangle formed by eight sides of the regular octagon is Delta OAB in which OA=OB
OP is drawn perpendicular from O to AB
OP bisects the Delta OAB and it will be radius of the inscribed circle of the octagon. Given OP= 12 unit
Now area of the octagon ABCDEFGH
=8xxDelta OAB
=8xx2xx "area of " Delta OAP
=8xx2xx1/2xxOPxxAP
=8xxOP^2xx(AP)/(OP)
=8xxOP^2xxtan/_AOP
=8xx12^2xxtan22.5^@
=1152xx(2sin22.5^@cos22.5^2)/(2cos^2 22.5^@)
=1152xx(sin45^@)/(1+cos45^@)
=1152xx(1/sqrt2)/(1+1/sqrt2)
=1152xx1/(sqrt2+1)
=1152xx(sqrt2-1) sq unit
