A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of p as found appropriate.)
Answers
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
Hope This Helps :)
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