#### Question

A solid is in the shape of a frustum of a cone. The diameter of two circular ends are 60cm and 36cm and height is 9cm. find area of its whole surface and volume?

#### Solution

Given that,

The radii of the upper and lower circle of the frustum of the cone are r_{1} = 30 cm and r_{2} = 18 cm respectively

the height of the frustum of a cone h =9 cm.

The slant height of the cone

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

`=sqrt(9^2+(30-18)^2`

`=sqrt(81+144)`

`=sqrt(225)`

= 15 cm

The volume of the frustum cone

`V=1/3pi(r_1^2+r_1r_2+r_2^2)xxh`

`=1/3pi(30^2+30xx18+18^2)xx9`

`=1/3pi(900+540+324)xx9`

`=1/3pixx1764xx9`

= π x 1764 x3

= 5292π cm^{3}

Hence, volume of the frustum cone is 5292π cm^{3}

The total surface narea of the frustum cone

`S=pi(r_1+r_2)xxl+pi r_1^2+pir_2^2`

= π (30+18) x 15 + π x 30^{2} + π x 18^{2}

= π (48 x 15) + 900 π + 324 π

= 720π + 900π + 324π

= 1944π cm^{2}

Hence, the total surface area of the freustum cone is 1944π cm^{2 }