Question #6b9d0

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Question #6b9d0

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Answer 1 · 2016-02-17T18:30:14

In fact the ratio is $\frac{7}{19}$ $7/19$

Explanation:

Refer to the figure below

I created this figure using MS Excel

Since the 2 planes divide the cone in 3 parts we can obtain 3 ratios. But I presume that $\frac{V_{1}}{V_{2}}$ $V_1/V_2$ is what is intended.

Just not to create confusion, be noticed that $b, b_{1} and b_{2}$ $b, b_1 and b_2$ are the areas (in square units) of the bases of their respective cones.

The volume of a cone is given as
$V = \frac{S_{b a s e} \cdot h e i g h t}{3}$ $V=(S_(base)*height)/3$

Finding $V_{2}$ $V_2$

$V_{2} = V - (V_{0} + V_{1}) = \frac{b h}{3} - \frac{b_{2} \cdot \frac{2 h}{3}}{3} = \frac{b h}{3} - \frac{2}{9} \cdot (b_{2} \cdot h)$ $V_2=V-(V_0+V_1)=(bh)/3-(b_2*(2h)/3)/3=(bh)/3-2/9*(b_2*h)$
Notice that
$\to \frac{b_{2}}{b} = \frac{π \cdot r_{2}^{2}}{π \cdot r^{2}} = {(\frac{r_{2}}{r})}^{2}$ $->(b_2)/b=(pi*r_2^2)/(pi*r^2)=(r_2/r)^2$ and since $r_2/r=((2cancel(h))/3)/cancel(h)=2/3$
$=>(b_2)/b=(2/3)^2$ => $b_2=4/9*b$
So
$V_2=(bh)/3-2/9*(b_2*4/9b)=(27bh-8bh)/81$ => $V_2=(19*bh)/81$

Finding $V_1$

$V_1=V-V_0-V_2=(bh)/3-(b_1*h/3)/3-(19*bh)/81$
But as we saw above
$b_1/b=((cancel(h)/3)/cancel(h))^2=(1/3)^2$ => $b_1=b/9$
So
$V_1=(bh)/3-b/9*h/9-(19*bh)/81=(27*bh-bh-19*bh)/81$ => $V_1=(7*bh)/81$

So the asked ratio is

$V_1/V_2=((7*cancel(bh))/cancel(81))/((19*cancel(bh))/cancel(81))=7/19$

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Question #6b9d0

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