From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid [take π=22/7]

#### Solution

The following figure shows the required cylinder and the conical cavity.

Given:

Height (*h*) of the conical part = Height (*h*) of the cylindrical part = 2.8 cm

Diameter of the cylindrical part = Diameter of the conical part = 4.2 cm

∴ Radius (*r*) of the cylindrical part = Radius (*r*) of the conical part = 2.1 cm

Slant height (*l*) of the conical part `=sqrt(r^2+h^2)`

`=sqrt((2.1)^2+(2.8)^2)cm`

`=sqrt(4.41+7.84)cm`

`=sqrt(12.25)cm=3.5 cm`

Total surface area of the remaining solid = Curved surface area of the cylindrical part + Curved surface area of the conical part + Area of the cylindrical base

`=2pirh+pirl+pir^2`

`=(2xx22/7xx2.1xx2.8+22/7xx2.1xx3.5+22/7xx2.1xx2.1) cm^2`

`=(36.96+23.1+13.86)cm^2`

`=73.92 cm^2`

Thus, the total surface area of the remaining solid is 73.92 cm^{2}.