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 r1 = 30 cm and r2 = 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 cm3
Hence, volume of the frustum cone is 5292π cm3
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 302 + π x 182
= π (48 x 15) + 900 π + 324 π
= 720π + 900π + 324π
= 1944π cm2
Hence, the total surface area of the freustum cone is 1944π cm2
