Question

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.

Answer 1

We have,

Radius of solid cone, R = CP = 10 cm,

Let the height of the solid cone be, AP = H,

the radius of the smaller cone, QD = r and

the height of the smaller cone be, AQ = h.

Also, ${\displaystyle \text{AQ}=\frac{\text{AP}}{2} \text{i.e} \text{h}=\frac{\text{H}}{2}\text{or}\text{H = 2h}.......\left(i\right)}$

Now, in Δ AQD and Δ APC

∠QAD = ∠PAC     (Common angle)

∠AQD = ∠APC = 90°

So, by AA criteria
ΔAQD _˜ ΔAPC

${\displaystyle \Rightarrow \frac{\text{AQ}}{\text{AP}}=\left(\text{QD}\right)}$

${\displaystyle \Rightarrow \frac{\text{h}}{\text{H}}=\frac{\text{r}}{\text{R}}}$   

${\displaystyle \Rightarrow \frac{\text{h}}{\text{2h}}=\frac{r}{\text{R}}}$    [Using (i)]

${\displaystyle \Rightarrow \frac{1}{2}=\frac{\text{r}}{\text{R}}}$

${\displaystyle \Rightarrow \text{R}=2\text{r}}$        ..............(ii)

As,

Volume of smaller cone ${\displaystyle =\frac{1}{3}\pi {\text{r}}^2\text{h}}$

And, 

Volume of solid cone ${\displaystyle =\frac{1}{3}\pi {\text{R}}^{2}H}$

${\displaystyle =\frac{1}{3}\pi {\left(2\text{r}\right)}^2\times \left(2\text{h}\right)}$          [Using (i) and (ii)]

${\displaystyle =\frac{8}{3}\pi r^{2}h}$

so, Volume frustum = Volume of solid cone - Volume of smaller cone 

${\displaystyle =\frac{8}{3}\pi {\text{r}}^2\text{h}-\frac{1}{3}\pi {\text{r}}^2\text{h}}$

${\displaystyle =\frac{7}{3}\pi {\text{r}}^2\text{h}}$

Now, the ratio of the volumes of the two parts ${\displaystyle =\frac{\text{Volume of the smaller cone}}{\text{Volume of the frustum}}}$

${\displaystyle =\frac{\frac{1}{3}\pi {\text{r}}^2\text{h}}{\frac{7}{3}\pi {\text{r}}^2\text{h}}}$

${\displaystyle =\frac{1}{7}}$

= 1 : 7

So, the ratio of the volume of the two parts of the cone is 1 : 7.