Question

A right triangle whose sides are 3 cm and 4 cm (other than hypotaneus) is made to revalve about to hypotaneus . Find the volume and surface area of the double cone so formed [ choose the value of pie as found appropriate ]

Answer 1

Answer:

Step-by-step explanation:

In the right angled triangle ABC (right angled at B) 

AB= 4cm and BC= 3cm 

AC= square root 3 square + 4 square  = Square root 9+16 = Square root 25 = 5cm 

Area of triangle ABC = ½ ABxBC = ½ ACxBO 

ABxBC = ACxBO 

4x3 = 5xBO 

BO= 12/5 = 2.4cm 

The volume of the required double cone = volume of the cone ABD + volume of the cone BCD 

= 1/3 pi r square h1 + 1/3 pi r square h square 

[where h1 and h2 are the heights of the cone ABD and BCD respectively] 

Surface area of double cone = surface area of cone ABD + surface area of cone BCD 

Pi r l1 + pi r l2 

Pi r (l1 + l2) 

22/7 x 2.4 x (4+3) 

22/7 x 2.4 x 7 

= 52.8 cm square 

So the volume of double cone is 30.17 cubic cm and surface area of double cone is 52.8 sq cm