volume and surface area of double cone
Asked by sunil_0010 | 9th Mar, 2010, 11:06: PM
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
Answered by | 10th Mar, 2010, 07:37: AM
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