The given figure shows a right triangle ABD with sides 15 cm, 20 cm and 25 cm. The triangle is revolved about its hypotenuse, find the volume of the double cone so formed. (Take $\pi = 3\dfrac{1}{7}$)

From figure,

Let DO = x cm and OB = (25 - x) cm

In right angle triangle AOD,

⇒ AD² = AO² + OD²

⇒ 20² = AO² + x²

⇒ AO² = 20² - x² ………..(1)

In right angle triangle AOB,

⇒ AB² = AO² + OB²

⇒ 15² = AO² + (25 - x)²

⇒ AO² = 15² - (25 - x)² ………..(2)

From equation (1) and (2), we get :

⇒ 20² - x² = 15² - (25 - x)²

⇒ 400 - x² = 225 - (625 + x² - 50x)

⇒ 400 - x² = 225 - 625 - x² + 50x

⇒ 50x - x² + x² = 400 + 625 - 225

⇒ 50x = 800

⇒ x = $\dfrac{800}{50}$ = 16 cm.

Substituting value of x in equation (1), we get :

⇒ AO² = 20² - x²

⇒ AO² = 400 - 16²

⇒ AO² = 400 - 256

⇒ AO² = 144

⇒ AO = $\sqrt{144}$ = 12 cm.

Since triangle ADB is rotated around hypotenuse BD.

∴ AO = OC = 12 cm.

In cone BAC,

Radius (r) = AO = OC = 12 cm

Height (h) = BO = (25 - x) = (25 - 16) = 9 cm.

$\text{Volume } = \dfrac{1}{3}πr^2h \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 12^2 \times 9 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 144 \times 9 \\[1em] = 1357.71 \text{ cm}^3$

In cone ACD,

Radius (r) = AO = OC = 12 cm

Height (H) = DO = x = 16 cm.

$\text{Volume } = \dfrac{1}{3}πr^2H \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 12^2 \times 16 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 144 \times 16 \\[1em] = 2413.71 \text{ cm}^3$

Total volume = 2413.71 + 1357.71 = 3771.42 cm³.

Hence, volume of double cone formed = 3771.42 cm³.