Question 35494
First draw the picture of a circle with radius 1, and an octagon inside the circle.  Divide the octagon into a total of 8 triangles each with one vertex at the center of the circle and the other vertices on the edge of the circle.  The measure of each angle at the center of the circle will be 360 degrees divided by 8 or 45 degrees.  This makes the problem easier.  Now, look at the first triangle, whose base is on the x axis. This triangle will have a height of {{{(sqrt(2))/2 }}} and a base of 1, so area of this triangle will be 
{{{A = (1/2)*b*h= (1/2)* 1* ((sqrt(2) )/2)}}} ={{{(sqrt(2))/4}}}


There are 8 such triangles in the octagon, so the area of the octagon will be 
{{{A= 8*(sqrt(2))/4= 2*sqrt(2)}}} or approximately 2.828


As a check find the area of the unit circle, which would have been {{{A=pi*r^2= pi}}} or approximately 3.14.  The inscribed octagon should be (and it is!!) just less than the area of the circle.


R^2 at SCC