A frustum of a right circular cone has a diameter of base 20 cm,

Solution:

The radii of the bottom and top circles are r1 = 10 cm and r2 = 6 cm respectively. The height of the frustum cone is h = 3 cm. Therefore, the volume of the bucket is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(10^{2}+10 \times 6+6^{2}\right) \times 3$

$=\frac{1}{3} \times \frac{22}{7} \times 196 \times 3$

$=616 \mathrm{~cm}^{3}$

Hence volume $=616 \mathrm{~cm}^{3}$

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(10-6)^{2}+3^{2}}$

 

$=\sqrt{25}$

$=5 \mathrm{~cm}$

The total surface area of the frustum cone is

$=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{1}^{2}+\pi r_{2}^{2}$

$=\frac{22}{7} \times(10+6) \times 5+\frac{22}{7} \times 10^{2}+\frac{22}{7} \times 6^{2}$

$=\frac{4752}{7}$ Square $\mathrm{cm}$

$=678.85$ Square $\mathrm{cm}$

Hence Total surface area $=678.85$