Question

From a solid right circular cone, whose height is 6 cm and radius of base is 12 cm, a right circular cylindrical cavity of height 3 cm and radius 4 cm is hollowed out such that bases of cone and cylinder form concentric circles. Find the surface area of the remaining solid in terms of π.

Answer 1

Given:

Cone radius (R) = 12 cm

Cone height (H) = 6 cm

Cylindrical cavity radius (r) = 4 cm

Cylindrical cavity height (h) = 3 cm

Slant height of the cone

l = `sqrt(R^2 + h^2)`

`= sqrt(12^2+6^2)`

`= sqrt(144+36)`

`=sqrt180`

`6sqrt5`

Slant height of the cone = πRl

`= pixx12xx6sqrt5`

`= 72sqrt5 pi`

CSA of cylinder = 2πrh₁

= 2π × 4 × 3

= 24π

Area of the remaining base

Area = π(R² − r²)

= π(12² − 4²)

= π(144 − 16)

= 128π

Total surface area of the remaining solid

`=75sqrt5 pi + 24pi + 128pi`

`= (72sqrt5 + 152)pi  cm^2`