Question 881310
{{{drawing(300,300,-10,50,-10,50,blue(line(0,0,40,0)),
blue(line(40,0,40,10)),
blue(line(40,10,0,10)),
blue(line(0,10,0,0)),
blue(line(40,10,60,10)),
green(line(40,10,40,21)),
locate(42,5,r),
locate(42,17,h),
circle(40,0,21))}}}
You need to calculate the area of a circular segment which is the area above the blue line and bounded by the circle.
Do a search on circular segment and you'll see the derivation of that area.
http://mathworld.wolfram.com/CircularSegment.html
{{{R=r+h}}}
In this case, {{{r=10}}}, {{{h=11}}} since {{{R=21}}}
{{{A[cs]=R^2*cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2)}}}
{{{A[cs]=21^2*cos^(-1)((21-11)/21)-(21-11)sqrt(2(21)11-11^2)}}}
{{{A[cs]=441*1.074-(10)sqrt(341)}}}
{{{A[cs]=473.6-184.7}}}
{{{A[cs]=288.9}}}
This is the entire area above the blue line, we're only interested in half of it.
{{{A[cs2]=(1/2)(288.9)}}}
{{{A[cs2]=144.5}}}
The area of 1/4 of the entire circle is,
{{{A[c4]=(1/4)piR^2=(1/4)pi(21)^2=346.4}}}
So the area bounded by blue rectangle and the circle is the difference between these two areas and is the area on which the cow can graze.
{{{A[cow]=A[c4]-A[cs2]}}}
{{{A[cow]=346.4-144.5}}}
{{{A[cow]=201.9}}}{{{m^2}}}