Question

The thickness of a hemispherical bowl is $0.25cm$. The inner radius of the bowl is$5cm$. Find the outer curved surface area of the bowl. (Take $\pi =\frac{22}{7}$)

Answer 1

Let $r,R$ and $w$ be the inner and outer radii and thickness of the hemispherical bowl respectively.
Given that $r=5cm$, $w=0.25cm$
$\therefore$ $R=r+w=5+0.25$ $=5.25cm$
Now, outer surface area of the bowl $=2\pi R^2=2\times \frac{22}{7}\times 5.25\times 5.25$
Thus, the outer surface area of the bowl $=173.25sq.cm$