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)
As per the question,
Hypotenuse, AC = 
= 5 cm
Area of ΔABC = 
* AB * AC
* AC * OB = 
* 4 * 3
OB = 
= 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 
r2h1 + 
r2h2
= 
πr2 (h1 + h2)
= 
πr2 (OA + OC)
= 
* 3.14 * (2.4)2 (5)
= 30.14 cm3
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
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