A cone is cut half way through its axis and parallel to the base, the volumes of two portions are in the ratio

Foundation

Mathematics Foundation

Surface Area and Volume

Question

1 : 7

1 : 4

1 : 1

1 : 3

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volume of cone ABCF
$v_{1} = \frac{1}{3} π {(r_{2})}^{2} h$
   vol. of frustum
$v_{2} = \frac{1}{3} πh [r_{1}^{2} + r_{2}^{2} + r_{1} r_{2}]$
$\frac{r_{1}}{r_{2}} = \frac{2 h}{h}$
   r₁   = 2r₂
   So, $v_{2} = \frac{1}{3} πh [7 r_{2}^{2})$
$∴ v_{2} = \frac{1}{3} π {(r_{2})}^{2} h$
$\frac{V_{1}}{V_{2}} = \frac{\frac{1}{3} π {(r_{2})}^{2} h}{\frac{1}{3} π (7 r_{2}^{2})}$

$\begin{matrix} \frac{v_{1}}{v_{2}} = \frac{1}{7} \end{matrix}$

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