Question

The height of a right circular cone is trisected by two places drawn parallel to the base. Show that the volumes of the three portions starting from the top are in the ratio $1:7:19$.

Answer 1

Let $VAB$ be a right circular cone of height $3h$ and base radius $r$.

This cone is cut by planes parallel to its base at points $O^{\prime }$ and $L$ such that $VL=L{O}^{\prime }=h$
Since triangles $VOA$ And $V{O}^{\prime }{A}^{\prime }$ are similar

$\therefore \frac{V{O}^{\prime }}{VO}=\frac{O^{\prime }{A}^{\prime }}{OA}\Rightarrow \frac{r}{r_1}=\frac{3h}{2h}\Rightarrow r_1=\frac{2r}{3}$

Also $△VOA\sim △VLC$

$\therefore \frac{VO}{VL}=\frac{OA}{LC}\Rightarrow \frac{3h}{h}=\frac{r}{r_2}\Rightarrow r_2=\frac{r}{3}$

Let $V_1$ be the volume of the cone VCD. Then

$V_1=\frac{1}{3}\pi r_2^{2}h=\frac{1}{3}\pi {\left(\frac{r}{3}\right)}^{2}h=\frac{1}{27}\pi r^{2}h$

Let $V_2$ be the volume of the frustum $A^{\prime }{B}^{\prime }{D}^{\prime }C$. Then

$V_2=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h=\frac{1}{3}\pi (\frac{4r^2}{9}+\frac{4r^2}{9}+\frac{{2r}^2}{9})h$

$\Rightarrow V_2=\frac{7}{27}\pi r^{2}h$

Let $V_3$ be the volume of the frustum $A{B}^{\prime }{A}^{\prime }$. then,

$V_3=\frac{1}{3}\pi (r_1^2+r^2+r_1{r}^{})h=\frac{1}{3}(r^2+\frac{4r^2}{9}+\frac{{2r}^2}{3})h\Rightarrow V_3=\frac{19\pi }{27}{r}^{2}h$

Required ratio $V_1:V_2:V_3=\frac{1}{27}\pi r^{2}h:\frac{7}{27}\pi r^{2}h:\frac{19\pi }{27}{r}^{2}h$

$\therefore$ $V_1:V_2:V_3=1:7:19$ $\left[Henceproved\right]$