From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^{2 }[Use π = 22/7]

#### Solution

Given that,

Height (*h*) of the conical part = Height (*h*) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (*r*) of the cylindrical part = 0.7 cm

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

`= sqrt((0.7)^2+(2.4)^2) =sqrt(0.49+5.76)`

`=sqrt(6.25) = 2.5`

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2πrh + πrl + πr^{2}

`= 2xx 22/7xx 0.7xx2.4+22/7xx0.7xx2.5+22/7xx0.7xx0.7`

= 4.4x2.4+2.2x2.5+2.2x0.7

=10.56+5.50+1.54 = 17.60 cm^{2}

The total surface area of the remaining solid to the nearest cm^{2} is 18 cm^{2}.