Question

A solid cylinder of height 15 cm and diameter 7cm . Two equal conical holes of radius 3 cm and height 4 cm are cut off , one from each circular end . Find the surface area of remaining solid

Answer 1

Given Diameter of cylinder = 7 cm So

Radius of solid cylinder = 3.5 cm

Height of cylinder = 15 cm

And

Radius of cone = 3 cm

Height of cone = 4 cm

So ,

Surface area of remaining solid cylinder = Total surface area of cylinder - Area of 2 bases of cones + curved surface area of cones

We know

Total surface area of cylinder = 2πr ( r + h ) , So

Total surface area of this solid cylinder = 2× 22/7 × 3.5 ( 3.5 + 15) ( As we know π = 22/7 )

Total surface area of this solid cylinder = 22 × 18.5 = 407 cm2

And

Area of base of cone = πr2 , So

Area of base of both cones = 2×πr2

Area of base of both cones = 2×2277 × 3 × 3

Area of base of both cones = 396/7 = 56.57 cm2

And

Slant height of cone l = (h^2 + r^2)^1/2=( 4^2 + 3^2)^1/2= (16 + 9)^1/2= 25^1/2= 5 cm

we know curved surface area of cone = πrl , So

Curved surface area of both cones = 2 × πrl

Curved surface area of both cones = 2 × 22/7 × 3 × 5

Curved surface area of both cones = 660/7 = 94.28 cm2

Then

Surface area of remaining solid cylinder = 407 cm2 - 56.57 cm2 + 94.28 cm2 = 444.71 cm2