Question

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

Answer 1

We know that, volume of a frustum of a cone $=\frac{\pi h}{3}[R_1^2+R_2^2+R_1{R}_2]$

Here, $R_1=\frac{60}{2}=30cm$

$R_2=\frac{36}{2}=18cm$

$h=9cm$

Hence, volume $=\frac{\pi \times 9}{3}[{30}^2+{18}^2+30\times 18]$

$=3\pi [900+324+540]$

$=3\pi \left[1764\right]$

$=5292\pi {cm}^3$

We also know that, TSA of a frustum of a cone $=\pi (R_1+R_2)\left(\sqrt{{(R_1-R_2)}^2+h^2}\right)+\pi (R_1^2+R_2^2)$

$=\pi [(30+18)\left(\sqrt{{(30-18)}^2+9^2}\right)+({30}^2+{18}^2)]$

$=\pi [\left(48\right)\left(\sqrt{144+81}\right)+(900+324)]$

$=\pi [\left(48\right)\left(\sqrt{225}\right)+\left(1224\right)]$

$=\pi [\left(48\right)\left(15\right)+\left(1224\right)]$

$=\pi [720+1224]$

$=1944\pi {cm}^2$

Hence, the volume of the frustrum of the cone is $5929\pi c{m}^3$ and its total surface area is $1944\pi c{m}^2$.