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 cm² 

[use `pi = 22/7`]

Answer 1

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 + πr²

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

= `4.4 xx 2.4 + 2.2 xx 2.5 + 2.2 xx 0.7`

= 10.56 + 5.50 +1.54

= 17.60 cm²

The total surface area of the remaining solid to the nearest cm² is 18 cm².