Area ##= 2sqrt(2)r^2##

where ##r## is the radius of the octagon

Consider the diagram below with radius ##r##:

A regular octagon can be thought of as being composed of ##4## kite shaped areas.

The area of a kite with diagonals ##d## and ##w## is

##color(white)(XXX)Area_kite=(d*w)/2##.

(This is fairly easy to prove if it isn’t a formula you already know).

Consider the kite ##PQCW## in the diagram above.

##/_QCW=pi/2## and ##|QC|=|WC|=r##

##color(white)(XXX)rArr |QW|=sqrt(2)r## (Pythagorean)

Therefore (since ##|PC|=r##)

##color(white)(XXX)Area_PQCW = (|PC|*|QW|)/2 = (r*sqrt(2)r)/2 = (sqrt(2)r^2)/2##

The octagon is composed of ##4## such kites so

##color(white)(XXX)Area_octagon = 2sqrt(2)r^2##