Question.
Find the total surface area of a cone, if its slant height is $21 \mathrm{~m}$ and diameter of its base is $24 \mathrm{~m}$. [ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Radius $(r)$ of the base of cone $=\left(\frac{24}{2}\right) \mathrm{m}=12 \mathrm{~m}$
Slant height (l) of cone $=21 \mathrm{~m}$
Total surface area of cone $=\pi r(r+l)$
$=\left[\frac{22}{7} \times 12 \times(12+21)\right] \mathrm{m}^{2}$
$=\left(\frac{22}{7} \times 12 \times 33\right) \mathrm{m}^{2}$
$=1244.57 \mathrm{~m}^{2}$
Radius $(r)$ of the base of cone $=\left(\frac{24}{2}\right) \mathrm{m}=12 \mathrm{~m}$
Slant height (l) of cone $=21 \mathrm{~m}$
Total surface area of cone $=\pi r(r+l)$
$=\left[\frac{22}{7} \times 12 \times(12+21)\right] \mathrm{m}^{2}$
$=\left(\frac{22}{7} \times 12 \times 33\right) \mathrm{m}^{2}$
$=1244.57 \mathrm{~m}^{2}$
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