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`)
The remaining solid is a frustum of the given cone
Total surface area of the frustum = πl(r1+r2)+πr12+πr22
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.
`l = sqrt(h^2+(r_1-r_2)^2)`
`=>l = 4sqrt5 `
∴ Total surface area of the frustum = πl(r1+r2)+πr12+πr22
= 350.592 cm2
Hence, the total surface area of the remaining solid is 350.592 cm2.