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.)

Answers (1)

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

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