Question 1204129
Solve for {{{R }}}:


{{{2piR^2 + 2piRH - S = 0 }}}


{{{2piR^2 + 2piRH = S }}}


{{{2pi(R^2 + RH) = S }}}


{{{R^2 + RH= S/(2pi) }}}....complete square{{{ R^2 + RH }}}=>{{{(R^2 + RH+(H/2)^2)-(H/2)^2 }}}=>{{{(R + H/2)^2-(H/2)^2 }}}


{{{(R + H/2)^2-(H/2)^2= S/(2pi) }}}


{{{(R + H/2)^2= S/(2pi)+(H/2)^2 }}}


{{{R + H/2= sqrt(S/(2pi)+(H/2)^2) }}}


{{{R  = sqrt(S/(2pi)+(H/2)^2)-H/2 }}}