Question

A cylinder of base radius 7 cm and height 14 cm has been hollowed out from he top of a cube 20 cm find the surface area of the remaining solid

Answer 1

Sol:Radius of the cylinder = 6 cmHeight of the cylinder = 8 cmVolume of the solid cylinder = πr2h= 22/7 × 6 × 6 × 8= 905.14 cm2Radius of the base of the conical cavity = 6 cmHeight of the conical cavity = 8 cmVolume of the conical cavity = 1/3 πr2h= 1/3 × 22/7 × 6 × 6 × 8= 301.71 cm2Volume of the remaining solid = 905.14 - 301.71= 603.43 cm2Total surface area of the remaining solid =CSA of the cylinder + CSA of the conical cavity + Area of the base of the cylinderCSA of the cylinder = 2πrh = 2π × 6 x 8 = 96 πCSA of the conical cavity = πrl= πr√[(r)2+ (h)2]= π x (6)2x √[(6)2+ (8)2]= π x (6)2x √100= π x 36 x 10= 360 πArea of the base of the cylinder = π r2= π (6)2= 36 πTotal surface area of the remaining solid =96 π + 360 π + 36 π = 492 π = 492 x 3.1412 = 1545.4704 cm2.I hope this answer is helpful to you