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| 1. |
Find volume of double cones formed when a triangle of side 4m and 3m is revolved by hypotenuse . |
| Answer» AB = 3 m, AC = 4 mIn\xa0{tex}\\triangle{/tex}BAC, by pythagoras theoremBC2 = AB2 + AC2\xa0{tex}\\Rightarrow{/tex}BC2 = 32 + 42\xa0{tex}\\Rightarrow{/tex}BC2 = 25\xa0{tex}\\Rightarrow{/tex}BC =\xa0{tex}\\sqrt {25} {/tex}\xa0= 5 mIn\xa0{tex}\\triangle{/tex}AOB and\xa0{tex}\\triangle{/tex}CAB{tex}\\angle{/tex}ABO =\xa0{tex}\\angle{/tex}ABC [common]{tex}\\angle{/tex}AOB =\xa0{tex}\\angle{/tex}BAC [each 90o]\xa0Then,\xa0{tex}\\triangle{/tex}AOB -\xa0{tex}\\triangle{/tex}CAB [by AA similarity]{tex}\\therefore{/tex}\xa0{tex}\\frac { A O } { C A } = \\frac { O B } { A B } = \\frac { A B } { C B }{/tex}\xa0[c.p.s.t]{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { A O } { 4 } = \\frac { O B } { 3 } = \\frac { 3 } { 5 }{/tex}Then, AO =\xa0{tex}\\frac{{4 \\times 3}}{5}{/tex}\xa0and OB =\xa0{tex}\\frac{{3 \\times 3}}{5}{/tex}{tex}\\Rightarrow{/tex}\xa0AO =\xa0{tex}\\frac{12}{5}{/tex}\xa0m and OB =\xa0{tex}\\frac{9}{5}{/tex}\xa0m{tex}\\therefore{/tex}OC = 5 -\xa0{tex}\\frac{9}{5}{/tex}\xa0=\xa0{tex}\\frac{16}{5}{/tex}m{tex}\\therefore{/tex}\xa0Volume 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}\xa0=\xa0{tex}30 \\frac { 6 } { 35 } \\mathrm { m } ^ { 3 }{/tex}. | |