The curved surface are of a right circular cone is 65 $$\pi$$ $$cm^{2}$$ and the radius of its base in 5 cm. What is half of the volume of the cone, in $$cm^{3}$$ ?

Given, 

Radius = 5 cm

Curved surface area = $$\pi\ \times\ r\times\ l=65\pi\ $$

$$\pi\ \times\ 5\times\ l=65\pi\ $$

by solving , we get , $$l=13$$

as shown in the figure, l = slant height, r = radius of cone, h = height of cone 

We know, $$l^2\ =\ r^2+\ h^2$$

$$\therefore13^2\ =\ 5^2+\ h^2\ $$

By solving , we get h = 12 

According to question , 

Half of the volume = $$\frac{1}{2}\times\ \frac{1}{3}\times\ \pi\ \times\ r^2\times\ h$$

$$=\ \frac{1}{2}\times\ \frac{1}{3}\times\ \pi\ \times\ 5^2\times\ 12$$

$$=50\ \pi\ $$

Hence, Option A is correct.