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# A right angled triangle with sides …

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A right angled triangle with sides 3 cm and 4 cm is revolved around hypotenuse. Find the volume of double cone thus generated.

Preeti Dabral 2 months, 2 weeks ago

AB = 3 cm, AC = 4 cm
In {tex}\triangle{/tex}BAC, by pythagoras theorem
BC2 = AB2 + AC2
{tex}\Rightarrow{/tex}BC2 = 32 + 42
{tex}\Rightarrow{/tex}BC2 = 25
{tex}\Rightarrow{/tex}BC = {tex}\sqrt {25} {/tex} = 5 cm
In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}CAB
{tex}\angle{/tex}ABO = {tex}\angle{/tex}ABC [common]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}BAC [each 90o
Then, {tex}\triangle{/tex}AOB - {tex}\triangle{/tex}CAB [by AA similarity]
{tex}\therefore{/tex} {tex}\frac { A O } { C A } = \frac { O B } { A B } = \frac { A B } { C B }{/tex} [c.p.s.t]
{tex}\Rightarrow{/tex} {tex}\frac { A O } { 4 } = \frac { O B } { 3 } = \frac { 3 } { 5 }{/tex}
Then, AO = {tex}\frac{{4 \times 3}}{5}{/tex} and OB = {tex}\frac{{3 \times 3}}{5}{/tex}
{tex}\Rightarrow{/tex} AO = {tex}\frac{12}{5}{/tex} cm and OB = {tex}\frac{9}{5}{/tex} cm
{tex}\therefore{/tex}OC = 5 - {tex}\frac{9}{5}{/tex} = {tex}\frac{16}{5}{/tex}cm
{tex}\therefore{/tex} Volume of double cone thus generated = volume of first cone + volume of second cone
{tex}= \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times B O + \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times O C{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 9 } { 5 } + \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 16 } { 5 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \left[ \frac { 9 } { 5 } + \frac { 16 } { 5 } \right]{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \times 5{/tex}
={tex}\frac{1056}{35}{/tex} = {tex}30 \frac { 6 } { 35 } \mathrm { cm } ^ { 3 }{/tex}.

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