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Question

The height of a cone is $$10 cm$$. The cone is divided into two parts by drawing a plane through the midpoint of the axis of the cone, parallel to the base. Compare the volume of the two parts.


A
1:7
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B
2:9
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C
3:11
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D
3:5
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Solution

The correct option is A $$1 : 7$$
Let the height of the cone be $$H$$ and the radius be $$R$$.
This cone is divided into two parts through the midpoint of its axis. therefore $$AQ= \cfrac 12 AP$$
Since $$QD \parallel PC$$, therefore $$\triangle AQD \sim \triangle APC$$ (by the condition of similarity)
$$\cfrac { QD }{ PC } =\cfrac { AQ }{ AP } =\cfrac { AQ }{ 2AQ }$$ 
$$\cfrac { QD }{ R } =\cfrac { 1 }{ 2 } $$
$$QD=\cfrac { R }{ 2 } $$
Volume of the cone ABC $$= \cfrac 13 \pi{ r }^{ 2 }h$$
Volume of the frustum $$=$$ Volume of the cone ABC $$-$$ Volume of the cone $$AED$$
$$=\cfrac 13 \pi { r }^{ 2 }h- \cfrac 13\pi \left({ \cfrac r2 }\right)^2 \cfrac h2$$
$$= \cfrac 13 \pi { r }^{ 2 }h \left(1-\cfrac 18\right)$$
$$= \cfrac 13\pi { r }^{ 2 }h \times \cfrac 78$$
Volume of the cone AED $$=\cfrac 13\pi { r }^{ 2 }h \times \cfrac 18$$
$$\cfrac { \text {Volume of part taken out} }{ \text {Volume of remaining part of the cone} } =\cfrac { 1/3\pi { r }^{ r }h \times 1/8 }{ 1/3\pi { r }^{ r }h \times 7/8 } =\cfrac { 1 }{ 7 } $$

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Mathematics

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