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Let’s assume triangle ABC has a right angle at ∠B, with AB = 15 cm and BC = 20 cm. ∴ Hypotenuse AC = \( \sqrt{15^2 + 20^2} \) cm = \( \sqrt{625} \) cm = 25 cm Using the area of triangle ABC: \( \frac{1}{2} \times AC \times OB = \frac{1}{2} \times AB \times BC \) i.e., \( 25 \times OB = 15 \times 20 \) ⇒ \( OB = 12 \) cm When triangle ABC is rotated fully around the hypotenuse AC, the solid formed is a bicone (double cone). Each cone has radius OB = 12 cm ∴ Total volume of the bicone = \( \frac{1}{3} \pi \times OB^2 \times OA + \frac{1}{3} \pi \times OB^2 \times OC \) = \( \frac{1}{3} \pi \times OB^2 \times (OA + OC) \) = \( \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 25 \) cubic cm = \( \frac{26400}{7} \) cubic cm = \( 3771 \frac{3}{7} \) cubic cm