**Question:**

The area of the base of a right circular cone is 154 cm2 and its height is 14 cm. Its curved surface area is

(a) $154 \sqrt{5} \mathrm{~cm}^{2}$

(b) $154 \sqrt{7} \mathrm{~cm}^{2}$

(c) $77 \sqrt{7} \mathrm{~cm}^{2}$

(d) $77 \sqrt{5} \mathrm{~cm}^{2}$

**Solution:**

(a) $154 \sqrt{5} \mathrm{~cm}^{2}$

Area of the base of the of a right circular cone $=\pi r^{2}$

Therefore,

$\pi r^{2}=154$

$\Rightarrow \frac{22}{7} \times r^{2}=154$

$\Rightarrow r^{2}=\left(154 \times \frac{7}{22}\right)$

$\Rightarrow r^{2}=49$

$\Rightarrow r=7 \mathrm{~cm}$

Now, *r* = 7 cm and *h* = 14 cm

Then, slant height of the cone, $l=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(7)^{2}+(14)^{2}}$

$=\sqrt{49+196}$

$=\sqrt{245}$

$=7 \sqrt{5} \mathrm{~cm}$

Hence, the curved surface area of the cone $=\pi r l$

$=\left(\frac{22}{7} \times 7 \times 7 \sqrt{5}\right) \mathrm{cm}^{2}$

$=154 \sqrt{5} \mathrm{~cm}^{2}$