Question 106902
In other words, you have a 14" circle, what is the largest square that can fit inside of your circle. 
{{{drawing( 300, 300, -1.2, 1.2, -1.2, 1.2,grid( 1 ),circle( 0, 0, 1 ),
  red(line( 0, 0, .707, .707)),
locate(.1,.55,7 in),
locate(.5,-.2,s),
green(line( -.707, .707, .707, .707)),
green(line(  .707, .707, .707,-.707)),
green(line(  .707,-.707,-.707,-.707)),
green(line( -.707,-.707,-.707, .707)))}}}
The diagonal line is also the radius of the circle (7 inches). 
The diagonal line is also the hypotenuse of the right, isoceles triangle with sides s,s, and 7 inches. 
Since it's a right triangle, you can use the Pythagorean theorem. 
{{{s^2+s^2=7^2}}}
{{{2s^2=7^2}}}
{{{s=7/sqrt(2)}}}
The side of the square is 2s long, so the largest square that can be cut from the circle will be 
{{{t=2s=14/sqrt(2)}}}
t=9.9 inches
b) The area of the circle equals the area of the square plus the leftover(mulch). 
1.{{{A[c]=A[s]+A[m]}}}
Percentage of mulch area to total area is equal to 
{{{A[m]/A[c]}}}
Divide equation 1. by {{{A[c]}}}
1.{{{A[c]/A[c]=A[s]/A[c]+A[m]/A[c]}}}
{{{1=t^2/(pi)R^2+A[m]/A[c]}}}
{{{A[m]/A[c]=1-t^2/(pi)R^2}}}
{{{A[m]/A[c]=1-(14^2/2)/((pi)*7^2)}}}
{{{A[m]/A[c]=1-(128)/(153.9)}}}
{{{A[m]/A[c]=.168}}}
16.8% of the tree ends up as mulch. 
83.2% of the tree ends up as the square.