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`)

#### Solution

The remaining solid is a frustum of the given cone

Total surface area of the frustum = πl(r_{1}+r_{2})+πr_{1}^{2}+πr_{2}^{2}

Where

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

r_{1} = Larger radius of the frustum = 6 cm

r_{2} = 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`

⇒r_{2}=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(r_{1}+r_{2})+πr_{1}^{2}+πr_{2}^{2}

= π×4`sqrt5`(6+2)+π×6^{2}+π×2^{2}

`=pi(32sqrt5+40)`

`=22/7xx111.552`

= 350.592 cm^{2}

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