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  • 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 pi as found appropriate.)

2. 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 \pi as found appropriate.)

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The volume of the double cone will be   =    Volume of cone 1  +  Volume of cone 2.

                                                 =\ \frac{1}{3}\ \pi r^2h_1\ +\ \frac{1}{3}\ \pi r^2h_2

                                                =\ \frac{1}{3}\ \pi \times 2.4^2\times 5        (Note that sum of heights of both the cone is 5 cm  -  hypotenuse).

                                                 =\ 30.14\ cm^3

Now the surface area of a double cone is :

                                                    =\ \pi rl_1\ +\ \pi rl_2

                                                   =\ \pi \times 2.4 \left ( 4\ +\ 3 \right )

                                                  =\ 52.8\ cm^2

Posted by

Devendra Khairwa

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