Question

Note Take $\mathrm{\pi }=\frac{22}{7}$, 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.

Answer 1

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

∴ Outer curved surface area of the bowl = $2\mathrm{\pi }{r}^2=2\times \frac{22}{7}\times {\left(5.25\right)}^2$ = 173.25 cm²

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