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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