It is obvious from the above figure that each side of the regular octagon subtends equal angle of magnitude #360/8=45^@# at the center#O# 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
