Question 1191686
<pre>
We start with a rectangle which has area {{{xy}}}

{{{drawing(200,175,-8,8,-7,7,
locate(0,6,x), locate(7.1,.5,y),
line(-7,6,7,6),line(7,6,7,-6),line(7,6,7,-6),line(-7,-6,-7,6),line(-7,-6,7,-6)


  )}}}

Then we round off the corners, which means that we
subtract the 4 little corners that we round off from
the original area which is {{{xy}}}.

{{{drawing(200,175,-8,8,-7,7,

line(-7,6,7,6),line(7,6,7,-6),line(7,6,7,-6),line(-7,-6,-7,6),line(-7,-6,7,-6),

arc(-3,2,8,-8,90,180),
arc(3,2,8,-8,0,90),
arc(-3,-2,8,-8,180,270),
arc(3,-2,8,-8,270,360)

  )}}}

Next we draw in the quadrants of a circle of radius r to help us
see how much was subtracted from the corners in the rounding 
process.

{{{drawing(200,175,-8,8,-7,7,

line(-7,6,7,6),line(7,6,7,-6),line(7,6,7,-6),line(-7,-6,-7,6),line(-7,-6,7,-6),

arc(-3,2,8,-8,90,180),
arc(3,2,8,-8,0,90),
arc(-3,-2,8,-8,180,270),
arc(3,-2,8,-8,270,360),

line(-3,2,-7,2),line(3,2,7,2),line(3,-2,7,-2),line(-3,-2,-7,-2),
line(-3,2,-3,6),line(3,2,3,6),line(3,-2,3,-6),line(-3,-2,-3,-6)

  )}}}

Now we look at just those quadrants of a circle:

{{{drawing(200,175,-8,8,-7,7,

line(-7,6,-3,6),line(3,6,7,6),line(-7,-6,-3,-6),line(3,-6,7,-6),

line(-7,2,-7,6),line(7,2,7,6),line(-7,-2,-7,-6),line(7,-2,7,-6),

arc(-3,2,8,-8,90,180),
arc(3,2,8,-8,0,90),
arc(-3,-2,8,-8,180,270),
arc(3,-2,8,-8,270,360),

line(-3,2,-7,2),line(3,2,7,2),line(3,-2,7,-2),line(-3,-2,-7,-2),
line(-3,2,-3,6),line(3,2,3,6),line(3,-2,3,-6),line(-3,-2,-3,-6)

  )}}}

Now we put the four together and have a circle inscribed in a square

{{{drawing(200,175,-8,8,-7,7,

line(-4,0,4,0),line(0,-4,0,4),
line(-4,4,4,4),line(4,4,4,-4),
line(4,-4,-4,-4),
line(-4,-4,-4,4),
circle(0,0,4)  )}}}

Now we can see how much we have to subtract off from the
area of the original rectangle whose area is {{{xy}}}

We have the area of a square whose side is twice the radius r.
That's {{{(2r)^2}}}.  From that we subtract the area of the
circle {{{pi*r^2}}}, and get {{{(2r)^2-pi*r^2}}}.

That means the area of the rounded metal plate is xy minus
the area of those corners.

{{{A}}}{{{""=""}}}{{{xy-((2r)^2-pi*r^2)}}}

{{{A}}}{{{""=""}}}{{{xy-(4r^2-pi*r^2)}}}

{{{A}}}{{{""=""}}}{{{xy-4r^2+pi*r^2)}}}

We can factor out the r<sup>2</sup> in the last two terms:

{{{A}}}{{{""=""}}}{{{xy-r^2(4+pi))}}}  <--expression for A

---------------------------------

To get your answer for r, go back to the previous step:

{{{A}}}{{{""=""}}}{{{xy-4r^2+pi*r^2)}}}

{{{4r^2-pi*r^2}}}{{{""=""}}}{{{xy-A}}}

{{{r^2(4-pi)}}}{{{""=""}}}{{{xy-A}}}

{{{r^2}}}{{{""=""}}}{{{(xy-A)/(4-pi)}}}

{{{r}}}{{{""=""}}}{{{sqrt((xy-A)/(4-pi))}}}

Edwin</pre>