Question 1186473
<pre>
I'm quite sure you mean</pre>
"Find the area of THE REGULAR octagon inscribed in the square", for there are
infinitely many non-regular octagons that could be inscribed in the square.<pre>
We want to find the dimensions of the square, given the radius 15m.
We draw a radius (in green) from the center of the semicircle to the upper
right corner of the square.  The shorter leg of the right triangle formed 
is one-half the side of the square and the vertical side is the whole side.
So we let half the side of the square be x and the whole side be 2x. The
hypotenuse is the radius of the circle which is 15.  We use the Pythagorean
theorem to find x.

{{{drawing(400,800/3,-1.5,1.5,-.5,1.5,line(-1,0,1.015,0),
arc(0,0,2.04,-2.04,0,180),green(line(0,0,cos(atan(2)),sin(atan(2))),
locate(.08,.5,15)),locate(.2,.02,x),locate(.46,.5,2x),
line(cos(atan(2)),sin(atan(2)),-cos(atan(2)),sin(atan(2))),
line(-cos(atan(2)),0,-cos(atan(2)),sin(atan(2))),
line(cos(atan(2)),0,cos(atan(2)),sin(atan(2))))}}}

{{{matrix(8,5,
x^2,""+"",(2x)^2,""="",15^2,
x^2,""+"",2^2*x^2,""="",225,
x^2,""+"",4x^2,""="",225,
"","",5x^2,""="",225,
"","",x^2,""="",45,
"","",x,""="",sqrt(45),
"","",x,""="",sqrt(9*5),
"","",x,""="",3sqrt(5)
)}}}

Now we know that each side of the square is twice that or 6√5 meters.
So the area of the square is 

{{{(6sqrt(5))^2}}}{{{""=""}}}{{{36*5}}}{{{""=""}}}{{{180}}} square meters.

We will draw in the regular octagon by cutting off the 4 corners of the
square, which are 4 45-45-90 right triangles, the appropriate size. We
let each side of the regular octagon be s.  We let the legs of the 4
45-45-90 right triangles be y.

{{{drawing(400,800/3,-1.5,1.5,-.5,1.5,line(-1,0,1.015,0),
arc(0,0,2.04,-2.04,0,180),
locate(0,0,s), locate(.23,.24,s),locate(-.28,.24,s),
locate(.28,0,y),locate(-.33,0,y),

red(line(.1122204246+.05,0,.4472135955,.3349931709-.05),
line(-.1122204246-.05,0,-.4472135955,.3349931709-.05),
line(-.1122204246-.05,.894427191,-.4472135955, .6094340201),
line(.1122204246+.05,.894427191,.4472135955, .6094340201)

),


line(cos(atan(2)),sin(atan(2)),-cos(atan(2)),sin(atan(2))),
line(-cos(atan(2)),0,-cos(atan(2)),sin(atan(2))),
line(cos(atan(2)),0,cos(atan(2)),sin(atan(2))))}}}

We use ratio and proportion to find y by comparing the right
triangle to the standard 45-45-90 right triangle with sides
1,1,√2.

{{{s/sqrt(2)=y/1}}}
{{{sqrt(2)*y=s}}}
{{{y=s/sqrt(2)}}}
rationalize the denominator
{{{y=s*sqrt(2)/2}}}

We know that y+s+y = side of the square, 
which we found = 6√5

{{{y+s+y=6sqrt(5)}}}
{{{2y+s=6sqrt(5)}}}
{{{2(s*sqrt(2)/2)+s=6sqrt(5)}}}
{{{s*sqrt(2)+s=6sqrt(5)}}}
{{{s(sqrt(2)+1)=6sqrt(5)}}}
{{{s=6sqrt(5)/(sqrt(2)+1)}}}
rationalize:
{{{s=6sqrt(5)(sqrt(2)-1)}}}

We find y

{{{y=s*sqrt(2)/2}}}
{{{y=(6sqrt(5)(sqrt(2)-1))*sqrt(2)/2}}}
{{{y=(3sqrt(5)(sqrt(2)-1))*sqrt(2)}}}
{{{y=3sqrt(5)(2-sqrt(2))}}}

The area of each of the right triangles is

{{{A=expr(1/2)base*height}}}

{{{A=expr(1/2)y*y}}}

{{{A=expr(1/2)y^2}}}

{{{A=expr(1/2)(3sqrt(5)(2-sqrt(2)))^2}}}}

{{{A=expr(3/2)sqrt(5)(4-4sqrt(2)+2)}}}}

{{{A=expr(3/2)sqrt(5)(6-4sqrt(2))}}}}
{{{A=expr(3/2)sqrt(5)*2(3-2sqrt(2))}}}}
{{{A=3sqrt(5)(3-2sqrt(2))}}}}

Since there are 4 of them, the amount we must subtract from the
area of the square is 4 times that or

{{{12sqrt(5)(3-2sqrt(2))}}}}

The area of the square is 180 square meters, so we subtract that
and get the area of the regular octagon as

{{{180-12sqrt(5)(3-2sqrt(2))}}}

That's about

{{{175.3962167}}} square meters.
 
Edwin</pre>