Diketahui jari jari kerucut 14 cm dan tinggi 5 cm berapakah luas selimut kerucut

s=[tex] \sqrt{r^{2}+t^{2}}[/tex]
  =[tex] \sqrt{(14cm)^{2}+(5cm)^{2}} [/tex]
  =[tex] \sqrt{196cm^{2}+25cm^{2}} [/tex]
  =[tex] \sqrt{221cm^{2}} [/tex]
  =√221cm

L selimut kerucut=πrs
                           =22/7(14cm)(√221cm)
                           =44√221cm²