Question

A hemispherical bowl is made up of stone whose thickness is 5 cm. If the inner radius is 35 cm, find the total surface area of the bowl.

Answer 1

Let R be outer radius and ${}^{\prime }{r}^{\prime }$ be inner radius Thickness of ring $=5cm$

$\therefore R=(r+5)cm=(35+5)cm=40cm$

Total Surface Area = CSA of outer hemisphere + CSA of inner hemisphere + area of the ring.

$=2\pi R^2+2\pi r^2+\pi (R^2-r^2)$

$=\pi (2R^2+2r^2+R^2-r^2)$

$=\frac{22}{7}(3R^2+r^2)=\frac{22}{7}(3\times {40}^2+{35}^2)c{m}^2$

$=\frac{6025\times 22}{7}c{m}^2$

$=18935.71c{m}^2$