Question.
Diameter of the base of a cone is $10.5 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its curved surface area. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Radius $(r)$ of the base of cone $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$
Slant height (l) of cone $=10 \mathrm{~cm}$
CSA of cone $=\pi r l$
$=\left(\frac{22}{7} \times 5.25 \times 10\right) \mathrm{cm}^{2}=(22 \times 0.75 \times 10) \mathrm{cm}^{2}=165 \mathrm{~cm}^{2}$
Therefore, the curved surface area of the cone is $165 \mathrm{~cm}^{2}$.
Radius $(r)$ of the base of cone $=\left(\frac{10.5}{2}\right) \mathrm{cm}=5.25 \mathrm{~cm}$
Slant height (l) of cone $=10 \mathrm{~cm}$
CSA of cone $=\pi r l$
$=\left(\frac{22}{7} \times 5.25 \times 10\right) \mathrm{cm}^{2}=(22 \times 0.75 \times 10) \mathrm{cm}^{2}=165 \mathrm{~cm}^{2}$
Therefore, the curved surface area of the cone is $165 \mathrm{~cm}^{2}$.