Given that in a right angled triangle, whose sides are 3 cm and 4 cm(other than hypotenuse).
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse .
Hypotenuse AC = Ö( 32 + 42) = 5 cm
Area of ΔABC = (1/2) x AB xAC
⇒ (1/2) x AC x OB = (1/2)x 4 x3
⇒ (1/2) x 5 x OB = 6
⇒ OB = 12 / 5 = 2.4 cm.
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(OA +OC)
= (1/3) x 3.14x(2.4)2 x(5)
= 30.14 cm3
surface area of double cone = surface area of cone ABD + surface area of cone BCD
= πrl1 + πrl2 [where l1 and l2 are the slant heights of the cone ABD and BCD respectively]
thus the volume of the double cone is 30.14 cubic cm and surface area of double cone is 52.8 sq cm