Question

A right circular cone is divided into three parts by trisecting its height by two planes 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 ABC right circular cone of height of 3h and base radius be r.

The cone is trisected by two planes such that

AJ = JI = IH = h.

Now,

$\sf\triangle \: ABH \sim \triangle AFI \: \: \: \: (AA \: test \: of \: similarity)$

$\displaystyle \sf \implies \frac{AH}{ AI} = \frac{ BH}{ FI}$

$\displaystyle \sf \implies \: \frac{3 \cancel {h}}{2 \cancel{h}} = \frac{r}{r_{1}}$

$\displaystyle \sf \implies \: r_{1} \: = \frac{2r}{3} - - - - - (1)$

Now,

$\sf \triangle ABH \sim \triangle AJD \: (AA \: test \: of \: similarity) \:$

$\displaystyle \sf \implies \: \frac{AH }{ AJ} = \frac{BH}{JD}$

$\displaystyle \sf \implies \: \frac{3 \cancel{h}}{ \cancel{h}} = \frac{r}{r_{2}}$

$\displaystyle \sf \implies \: r_{2} \: = \frac{r}{3} - - - - - (2)$

Now , In cone ADE,

$\displaystyle \sf \implies \: V_{1} = \frac{1}{3} \pi r_{2}^{2} h$

$\sf \: V_{1} = \dfrac{1}{3} \pi \bigg( \dfrac{r}{3} \bigg)^{2} h - - - ( from \: (1))$

$\displaystyle \sf \: \displaystyle \sf \implies \: V_{1} = \frac{1}{27} \pi r^{2} h$

Now, In Frustum DEGF ,

$\displaystyle \sf \: V_{2} = \frac{1}{3} \pi \bigg(r_{1} {}^{2} + r_{2} {}^{2} + r_{1}r_{2} \bigg)h$

$\displaystyle \sf \: \: \: \: = \frac{1}{3} \pi \bigg( \frac{4 {r}^{2} }{9} + \frac{ {r}^{2} }{9} + \frac{2 {r}^{2} }{9} \bigg)h - - \: \: (from \: (1) \: and \: (2))$

$\displaystyle \sf \: V_{2} = \frac{7}{27} \pi {r}^{2} h$

In Frustum FGCB ,

$\displaystyle \sf \: V_{ 3} = \frac{1}{3} \pi \bigg( {r}^{1} + r_{1} {}^{2} + r_{1}r \bigg)h$

$\displaystyle \sf V_{ 3} = \frac{19}{27} \pi \: {r}^{2} h$

$\displaystyle \sf\therefore \: V_{ 1} :V_{ 2} : V_{ 3} = \frac{1}{27} \pi {r}^{2} h \: : \frac{7}{27} \pi {r}^{2} h \: : \: \frac{19}{27} \pi {r}^{2} h \:$

$\green{ \boxed{\displaystyle \sf\therefore \: V_{ 1} :V_{ 2} : V_{ 3} = \: 1 : 7 :19}}$