Let #r# be the radius. Then, using the Figure given,
#IH=rcos theta, IA=rsin theta, "where, "theta=18.43^@#.
#:." Area of the Rectangle "IHBA=IH*IA=r^2sin thetacos theta#.
#=r^2/2*sin 2theta=r^2/2*(2tan theta)/(1+tan^2 theta)#
#=(r^2tan theta)/(1+tan^2 theta)#
For #theta=18.43^@, tan theta=1/3#
#"Hence, Area of the Rectangle"=(r^2/3)/(1+1/9)=3r^2/10#
#"Also, Area of the Quarter of the Circle"=1/4*pir^2#
#:. "Total Area of the Composite Shape"=3r^2/10+pir^2/4=r^2/20(6+5pi)#, which is, #570" sq.ft."#
#:. r^2/20(6+5pi)=570#
#:. r^2=(570*20)/(6+5pi)=11400/(6+15.7)=11400/21.7~~525.3456#
#:. r~~22.92 "ft."#
