A right-angled 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 a value of π as found appropriate.)

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse, AC = √3² + 4² = √9 + 16 = √25 = 5 cm
Area of ΔABC = 1/2 × AB × AC ⇒ 1/2 × AC × DB = 1/2 × 4 × 3 ⇒ 1/2 × 5 × DB = 6
So, DB = 12/5 = 2.4 cm
The volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3 πr²h₁ + 1/3 πr²h₂
= 1/3 πr²[h₁ + h₂]
= 1/3 × πr²[DA + DC]
= 1/3 × 3.14 × 2.4² × 5
= 30.14 cm³
The surface area of double cone = Surface area of cone 1 + Surface area of cone 2
= πrl₁ + πrl₂
= πr[4 + 3]
= 3.14 × 2.4 × 7
= 52.75 cm²