Answer :


Let ABC be the triangle, with
AB = 3 cm
BC = 4 cm
and is revolved around hypotenuse.
By Pythagoras theorem,
Hypotenuse, AC =

= 5 cm

Area of ΔABC = × AB × BC

× AC × OB = × 3 × 4

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

l1 = AB = 3cm
l2  = BC = 4 cm

= πr [3 + 4]

= 3.14 × 2.4 × 7

= 52.75 cm2

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