Question

In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use `pi=22/7` and `sqrt5=2.236`)

Answer 1

The remaining solid is a frustum of the given cone

Total surface area of the frustum = πl(r₁+r₂)+πr₁²+πr₂²

Where

h = Height of the frustum = 12−4 = 8 cm

r₁ = Larger radius of the frustum = 6 cm

r₂ = Smaller radius of the frustum

l = Slant height of the frustum

In the given figure, ∆ABC ​~ ∆ADE by AA similarity criterion.

`:.(BC)/(DC)=(AB)/(AD)`

`=>r_2/6=4/12`

⇒r₂=2 cm

We know

`l = sqrt(h^2+(r_1-r_2)^2)`

`=>l=sqrt(8^2+(6-1)^2)`

`=>l = 4sqrt5 `

∴ Total surface area of the frustum = πl(r₁+r₂)+πr₁²+πr₂²

= π×4`sqrt5`(6+2)+π×6²+π×2²

`=pi(32sqrt5+40)`

`=22/7xx111.552`

= 350.592 cm²

Hence, the total surface area of the remaining solid is 350.592 cm2.