ABC is a right angled triangle in which ∠B = 90°, AB = 5 cm and BC = 12 cm. What is the approximate volume (in cmc: of the double cone formed by rotating the triangle about its hypotenuse?

Hypotenuse AC = $$\sqrt{5^2+12^2}=\sqrt{25+144}$$

= $$\sqrt{169}=13$$ cm

Area of $$\triangle$$ ABC = $$\frac{1}{2}\times(AB)\times(AC)$$ = $$\frac{1}{2}\times(AC)\times(OB)$$

=> $$OB\times13=5\times12$$

=> $$OB=\frac{60}{13}$$ cm

Volume of double cone = Volume of cone 1 + Volume of cone 2

= $$\frac{1}{3}\pi r^2h_1+\frac{1}{3}\pi r^2h_2$$

= $$\frac{1}{3}\pi r^2(h_1+h_1)$$

= $$\frac{1}{3} \times 3.14 \times (OB)^2 \times (OA+OC)$$

= $$\frac{1}{3} \times 3.14 \times (\frac{60}{13})^2\times13$$

= $$\frac{11304}{39}=289.85 \approx 290$$ $$cm^3$$

=> Ans - (B)