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 $1 : 7 : 19$ .

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Solution

Let $V A B$ be a right circular cone of height $3 h$ and base radius $r$ .
This cone is cut by planes parallel to its base at points $O^{'}$ and $L$ such that $V L = L O^{'} = h$
Since triangles $V O A$ And $V O^{'} A^{'}$ are similar

$∴ \frac{V O^{'}}{V O} = \frac{O^{'} A^{'}}{O A} \Rightarrow \frac{r}{r_{1}} = \frac{3 h}{2 h} \Rightarrow r_{1} = \frac{2 r}{3}$

Also $△ V O A \sim △ V L C$

$∴ \frac{V O}{V L} = \frac{O A}{L C} \Rightarrow \frac{3 h}{h} = \frac{r}{r_{2}} \Rightarrow r_{2} = \frac{r}{3}$

Let $V_{1}$ be the volume of the cone VCD. Then

$V_{1} = \frac{1}{3} π r_{2}^{2} h = \frac{1}{3} π {(\frac{r}{3})}^{2} h = \frac{1}{27} π r^{2} h$

Let $V_{2}$ be the volume of the frustum $A^{'} B^{'} D^{'} C$ . Then

$V_{2} = \frac{1}{3} π (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2}) h = \frac{1}{3} π (\frac{4 r^{2}}{9} + \frac{4 r^{2}}{9} + \frac{{2 r}^{2}}{9}) h$

$\Rightarrow V_{2} = \frac{7}{27} π r^{2} h$

Let $V_{3}$ be the volume of the frustum $A B^{'} A^{'}$ . then,

$V_{3} = \frac{1}{3} π (r_{1}^{2} + r^{2} + r_{1} r) h = \frac{1}{3} (r^{2} + \frac{4 r^{2}}{9} + \frac{{2 r}^{2}}{3}) h \Rightarrow V_{3} = \frac{19 π}{27} r^{2} h$

Required ratio $V_{1} : V_{2} : V_{3} = \frac{1}{27} π r^{2} h : \frac{7}{27} π r^{2} h : \frac{19 π}{27} r^{2} h$

$∴$ $V_{1} : V_{2} : V_{3} = 1 : 7 : 19$ $[H e n c e p r o v e d]$

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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.

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 $1 : 7 : 19$ .