Find vol.of double cone formed when …

Let ABC be the right triangle right angled at A
whose sides AB and AC measure 3 cm
and 4 cm respectively,
{tex}\therefore \text { hypotenuse } \mathrm { BC } = \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = 5 \mathrm { cm }{/tex}

Let AO (A'O) is the radius of the common base of the double cone formed by revolving the triangle around BC.
Now, the area of {tex}\triangle A B C = \frac { 1 } { 2 } \times 4 \times 3 = \frac { 1 } { 2 } \times B C \times A O{/tex}
{tex}\Rightarrow 6 = A 0 \times \frac { 5 } { 2 } \Rightarrow A O = \frac { 12 } { 5 } = 2.4 \mathrm { cm }{/tex}
Now, the volume of the double cone
{tex}= \frac { 1 } { 3 } \times \pi \times \mathrm { AO } ^ { 2 } \times \mathrm { BO } + \frac { 1 } { 3 } \times \pi \times \mathrm { A } \mathrm { O } ^ { 2 } \times \mathrm { CO }{/tex}
{tex}= \frac { 1 } { 3 } \pi \times A O ^ { 2 } \times ( B O + C O ){/tex}
{tex}= \frac { 1 } { 3 } \times \pi \times \mathrm { AO } ^ { 2 } \times \mathrm { BC } [ \because \mathrm { BO } + \mathrm { CO } = \mathrm { BC } ]{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 2.4 \times 2.4 \times 5 = 30.17 \mathrm { cm } ^ { 3 }{/tex}
The surface area of the double cone = {tex}\pi \times A O \times A B + \pi \times A O \times A C{/tex}
{tex}= \pi \times A O [ A B + A C ]{/tex}
{tex}= \frac { 22 } { 7 } \times 2.4 \times ( 3 + 4 ) = 22 \times 2.4 = 52.8 \mathrm { cm } ^ { 2 }{/tex}