A right angled triangle having hypotenuse 25 cm and sides are in the ratio 3:4 is made to revolve about its hypotenuse. The volume of the double cones of or med is

Hypotenuse is 25 cm and the sides are in the ratio 3:4, so the legs are 15 cm and 20 cm (since 3k, 4k, 5k with 5k = 25 gives k = 5). Area = ½ × 15 × 20 = 150 cm². The altitude to the hypotenuse is h = 2 × 150 / 25 = 12 cm. This altitude is the common radius of the double cone formed when the triangle revolves about its hypotenuse.

The hypotenuse is split into two parts by the foot of the altitude. Each part equals (leg)²/hypotenuse: 15²/25 = 9 cm and 20²/25 = 16 cm. These are the heights of the two cones.

• Volume of the two cones: (1/3)π(12)²(9) + (1/3)π(12)²(16) = 432π + 768π = 1200π cm³, which is option B.
• Options A (1000π), C (1500π) and D (1575π) do not match this sum. Only one cone alone gives 432π or 768π, and neither is listed.

The core concept is the solid of revolution: a right triangle rotated about its hypotenuse gives two cones sharing a base. This is a standard mensuration problem in the CSAT paper, and the same altitude-and-segment method appears in questions on solids made by rotating plane figures.

Trap: Using the legs as the cone heights instead of the segments of the hypotenuse, which gives a wrong answer.