# Solve this

Question:

Note Take $\pi=\frac{22}{7}$, unless stated otherwise.

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.

Solution:

Inner radius of the bowl, r = 5 cm

Let the outer radius of the bowl be R cm.

Thickness of the bowl = 0.25 cm         (Given)

∴ R − r = 0.25 cm

⇒ R = 0.25 + r = 0.25 + 5 = 5.25 cm

$\therefore$ Outer curved surface area of the bowl $=2 \pi r^{2}=2 \times \frac{22}{7} \times(5.25)^{2}=173.25 \mathrm{~cm}^{2}$

Thus, the outer curved surface area of the bowl is 173.25 cm2.