a rt. triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is madeto revolve about its hypotenuse. find the volume and surface area of the double cone so formed.
volume and surface area of double cone
Let the triangle be ABC with B = 90
Draw a perpendicular from B onto AC, the hypotenuse, call it BN.
sin A = BC/AC = 3/5
But in ABN
sinA = BN/AB = 3/5
Hence BN = 3AB/5 = 3x4/5 = 12/5
Identify that this BN will be the radius of base for both the cones in the double cone.
Volume of double cone = Sum of individual cones,
= πR2H/3 + πR2H'/3
= (πR2/3)(H+H') = (πR2/3)(AC) ... since hypotenuse will be the addition of heights of the cones.
Put R = BN = 12/5 and AC = 5.
We find volume of double cone,
= (22x12x12x5)/(7x5x5x3) = 1056/35 = 30.17 cm2
CSA of double cone = πRL1 + πRL2 = πR AB + πR BC = πR(AB+BC) = 22x12x7/(7X5) =264/5 = 52.8 cm2