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Question

A solid cone of base radius $$10\ cm$$ is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.


Solution


$$\triangle OAB\sim \triangle OCD$$ (By AA similarity)

$$\therefore \quad \dfrac { OA }{ OB } =\dfrac { AB }{ CD } \Rightarrow \dfrac { h/2 }{ h } =\dfrac { r }{ 10 } \Rightarrow r=\dfrac { 10 }{ 2 } =5cm$$

Given that cone cut into two parts through the mid point of its height $$H = \dfrac{h}{2}$$

Volume of frustum of cone $$=\dfrac { \pi H }{ 3 } \left[ { R }^{ 2 }+{ r }^{ 2 }+Rr \right] $$
                                             
                                              $$=\dfrac { \pi h }{ 3\times 2 } \left[ { 10 }^{ 2 }+{ 5 }^{ 2 }+10\times 5 \right] =\dfrac { 175\pi h }{ 6 } $$

Volume of cone $$OAB=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }\dfrac { h }{ 2 } =\dfrac { 1 }{ 3 } \times \pi \times { 5 }^{ 2 }\times \dfrac { h }{ 2 } =\dfrac { 25\pi h }{ 6 } $$

Required ratio $$=\dfrac { \dfrac { 25\pi h }{ 6 }  }{ \dfrac { 175\pi h }{ 6 }  } =\dfrac { 1 }{ 7 } $$

956518_975517_ans_57023581857f4d82bacee896ac72d89f.png

Mathematics

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