Question 1178914
<pre><b>
{{{drawing(400,400,-3,3,-3,3,circle(0,.375,.04),

circle(0,.375,2.576941016), line(-2,2,2,2), line(-2,0,2,0), line(-1,-2,1,-2),

line(-2,0,-2,2),line(0,0,0,2), line(2,0,2,2),locate(.4,-1.74,8),

locate(1,1.2,r),locate(.6,-.7,r),

locate(-.16,.3,h),locate(-.6,1.3,16-h),locate(-.25,-1,16),locate(1,2.3,16),

red(line(0,.375,2,2),line(0,.375,1,-2)), green(line(0,.375,0,-2)),

line(-1,-2,-1,0), line(1,-2,1,0)   )}}}

We let h equal the little distance from the top of the bottom square to the
center of the circle.

The two red lines marked r are radii.

There are two right triangles, and their hypotenuses are the
radii, of length r.

Applying the Pythagorean theorem to the upper right triangle:

{{{(16-h)^2+16^2=r^2}}}
{{{256-32h+h^2+256=r^2}}}
{{{512-32h+h^2=r^2}}}

Applying the Pythagorean theorem to the lower right triangle:

{{{(16+h)^2+8^2=r^2}}}
{{{256+32h+h^2+64=r^2}}}
{{{320+32h+h^2=r^2}}}

We set the two expressions for r<sup>2</sup> equal:

{{{512-32h+h^2}}}{{{""=""}}}{{{320+32h+h^2}}}

Subtract h<sup>2</sup> from both sides:

{{{512-32h}}}{{{""=""}}}{{{320+32h}}}

{{{-64h}}}{{{""=""}}}{{{-192}}}

{{{h=3}}}

We substitute h=3 in

{{{320+32h+h^2=r^2}}}
{{{320+32(3)+(3)^2=r^2}}}
{{{320+96+9=r^2}}}
{{{425=r^2}}}

The area of a circle is

{{{A=pi*r^2}}}

So the area of the circle is

{{{A=425pi}}}   <--answer

Edwin</pre></b>