Request a call back

volume and surface area of double cone
Asked by sunil_0010 | 09 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

Regards,

Team,

TopperLearning.

Answered by | 10 Mar, 2010, 07:37: AM
CBSE 10 - Maths
Asked by naveenkumarbana1 | 08 Jan, 2020, 07:38: AM
CBSE 10 - Maths
Asked by | 02 Jan, 2013, 09:56: AM
CBSE 10 - Maths
CBSE 10 - Maths
CBSE 10 - Maths
CBSE 10 - Maths
CBSE 10 - Maths