Question

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 `c m^2`

Answer 1

Given,Diameter of cylinder =Diameter of conical cavity =1.4 cm
Radius of cylinder =Radius of conical cavity
`=("Diameter")/(2)=(14)/(2)=0.7 cm`
and Height of the cylinder =Height of the conical cavity =2.4 cm
`therefore` slant height of the conical cavity(l) = `sqrt(h^(2))+r^(2)=sqrt(2.4^(2))+(0.7^(2))`
`=sqrt(5.76+0.49)=sqrt(6.25)=2.5 cm`
Hence total surface area of remaining solid
=surface area of conical cavity +Total surface area of cylinder
`pirl+(2pirh+pir^(2)=pir(l+2h+r)`
`=(22)/(7)xx0.7(2.5+4.8+0.7)=2.2xx8=176=18 cm^(2)`