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The height of a cone is 10 cm . The cone is divide into 2 parts |
| Answer» Let the radius of original cone be r2Radius of\xa0cut of cone be r1According to the questionHeight of the original\xa0cone = 10 cm (given)The cone is cut off from the midpoint of the height,therefore, height the cone cut off = 5 cm{tex}\\Delta A O C \\sim \\Delta A\' O\' C{/tex}OA = Radius of original cone = r2O\'A\' = Radius of cutoff cone = r1Ratio\xa0of radius of two cones = Ratio of the height of cones{tex}\\therefore \\quad \\frac { A O } { A\' O ^ { \\prime } } = \\frac { r _ { 2 } } { r _ { 1 } } = \\frac { 10 } { 5 }{/tex}{tex}\\Rightarrow \\quad r _ { 2 } = 2 r _ { 1 }{/tex}Radius of original cone = 2 (radius of the cut off cone)Volume of the cut off cone\xa0{tex}= \\frac { 1 } { 3 } \\pi r _ { 1 } ^ { 2 } \\times 5{/tex}{tex}= \\frac { 5 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsVolume of original cone\xa0{tex}= \\frac { 1 } { 3 } \\pi \\left( 2 r _ { 1 } \\right) ^ { 2 } \\times 10{/tex}{tex}= \\frac { 40 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsVolume of frustum = Volume of an original cone - Volume of cut off cone{tex}= \\frac { 40 } { 3 } \\pi r _ { 1 } ^ { 2 } - \\frac { 5 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}{tex}= \\frac { 35 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsRequired ratio = Volume of frustum: Volume of cut off cone{tex}= \\frac { 35 \\pi r _ { 1 } ^ { 2 } } { 5 \\pi r _ { 1 } ^ { 2 } } = \\frac { 7 } { 1 }{/tex}Therefore, the required ratio\xa0= 7: 1. | |