A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

Two formed cones (which are formed by revolving the right angle triangle) are given.
Hypotenuse AC = √(3² + 4²) = √25 = 5 cm
Area of right angled triangle ABC = 1/2 × AB × AC
⇒ 1/2 × AC × OB = 1/2 × 4 × 3
⇒ 1/2 × 5 × OB = 6
⇒ OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3πr²h₁ + 1/3πr²h₂
= 1/3πr²(h₁ + h₂) = 1/3πr²(OA + OC)
1/3 × 3.14 × (2.4)² (5)
= 30.14 cm³
Curved surface area of double cone = Curved surface area of cone 1 + curved surface area of cone 2
= πrl₁ + πrl₂ = πr[4 + 3]
= 3.14 × 2.4 × 7 = 52.75 cm².