If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm,

Solution:

We have,

Height, $h=28 \mathrm{~cm}$,

Radius of the upper end, $R=28 \mathrm{~cm}$ and

Radius of the lower end, $r=7 \mathrm{~cm}$

Also,

The slant height, $l=\sqrt{(R-r)^{2}+h^{2}}$

$=\sqrt{(28-7)^{2}+28^{2}}$

$=\sqrt{21^{2}+28^{2}}$

$=\sqrt{441+784}$

$=\sqrt{1225}$

$=35 \mathrm{~cm}$

Now,

Capacity of the bucket $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 28 \times\left(28^{2}+7^{2}+28 \times 7\right)$

$=\frac{88}{3} \times(784+49+196)$

$=\frac{88}{3} \times 1029$

$=30184 \mathrm{~cm}^{3}$

Also,

Total surface area of the bucket $=\pi l(R+r)+\pi r^{2}$

$=\frac{22}{7} \times 35 \times(28+7)+\frac{22}{7} \times 7 \times 7$

$=110 \times(35)+154$

$=3850+154$

$=4004 \mathrm{~cm}^{2}$

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.