Question

A hemispherical bowl is made of steel 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. [Use π = $\frac{22}{7}$]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Inner}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hemispherical}\mathrm{bowl},r=5\mathrm{cm}\mathrm{and} \\ \mathrm{Thickness}\mathrm{of}\mathrm{the}\mathrm{bowl}=0.25\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{outer}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{bowl},R=5+0.25=5.25\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{outer}\mathrm{curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bowl}=2\pi R^2 \\ =2\times \frac{22}{7}\times 5.25\times 5.25 \\ =173.25{\mathrm{cm}}^2 \end{gathered}$$

So, the outer curved surface area of the hemispherical bowl is 173.25 cm^2.