# Question #6b9d0

Feb 17, 2016

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

#### Explanation:

Refer to the figure below

Since the 2 planes divide the cone in 3 parts we can obtain 3 ratios. But I presume that ${V}_{1} / {V}_{2}$ is what is intended.

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

The volume of a cone is given as
$V = \frac{{S}_{b a s e} \cdot h e i g h t}{3}$

Finding ${V}_{2}$

${V}_{2} = V - \left({V}_{0} + {V}_{1}\right) = \frac{b h}{3} - \frac{{b}_{2} \cdot \frac{2 h}{3}}{3} = \frac{b h}{3} - \frac{2}{9} \cdot \left({b}_{2} \cdot h\right)$
Notice that
$\to \frac{{b}_{2}}{b} = \frac{\pi \cdot {r}_{2}^{2}}{\pi \cdot {r}^{2}} = {\left({r}_{2} / r\right)}^{2}$ and since ${r}_{2} / r = \frac{\frac{2 \cancel{h}}{3}}{\cancel{h}} = \frac{2}{3}$
$\implies \frac{{b}_{2}}{b} = {\left(\frac{2}{3}\right)}^{2}$ => ${b}_{2} = \frac{4}{9} \cdot b$
So
${V}_{2} = \frac{b h}{3} - \frac{2}{9} \cdot \left({b}_{2} \cdot \frac{4}{9} b\right) = \frac{27 b h - 8 b h}{81}$ => ${V}_{2} = \frac{19 \cdot b h}{81}$

Finding ${V}_{1}$

${V}_{1} = V - {V}_{0} - {V}_{2} = \frac{b h}{3} - \frac{{b}_{1} \cdot \frac{h}{3}}{3} - \frac{19 \cdot b h}{81}$
But as we saw above
${b}_{1} / b = {\left(\frac{\frac{\cancel{h}}{3}}{\cancel{h}}\right)}^{2} = {\left(\frac{1}{3}\right)}^{2}$ => ${b}_{1} = \frac{b}{9}$
So
${V}_{1} = \frac{b h}{3} - \frac{b}{9} \cdot \frac{h}{9} - \frac{19 \cdot b h}{81} = \frac{27 \cdot b h - b h - 19 \cdot b h}{81}$ => ${V}_{1} = \frac{7 \cdot b h}{81}$

${V}_{1} / {V}_{2} = \frac{\frac{7 \cdot \cancel{b h}}{\cancel{81}}}{\frac{19 \cdot \cancel{b h}}{\cancel{81}}} = \frac{7}{19}$