Question

From the top of a cone of base radius $$24$$cm and height $$45$$cm, a cone of slant height $$17$$cm is cut off. What is the volume of the remaining frustum of the cone?

Solution

$${ L }^{ 2 }={ R }^{ 2 }+{ H }^{ 2 }\Rightarrow L=\sqrt { { 24 }^{ 2 }+{ 45 }^{ 2 } } =\sqrt { 2601 } =51cm$$we know that, $$\triangle OAB\sim \triangle OCD$$, therefore$$\dfrac { r }{ 24 } =\dfrac { h }{ 45 } =\dfrac { 17 }{ 51 }$$$$\Rightarrow \quad \dfrac { r }{ 24 } =\dfrac { 17 }{ 51 } \Rightarrow r=\dfrac { 24 }{ 3 } =8cm$$$$\Rightarrow \quad \dfrac { h }{ 45 } =\dfrac { 17 }{ 51 } \Rightarrow h=\dfrac { 45 }{ 3 } =15cm$$$$\therefore$$ Height of frustum $$= 45 - 15 = 30cm$$$$\therefore$$ Volume of frustum = $$\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) h=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 30\left( { 24 }^{ 2 }+{ 8 }^{ 2 }+24\times 8 \right)$$                                    = $$\dfrac { 220 }{ 7 } \times 832=26148.57{ cm }^{ 3 }$$Mathematics

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