The height of right circular cone is trisected by two plane parallel to its base Show that the volume - Maths - Surface Areas and Volumes

Question

The height of right circular cone is trisected by two plane
parallel to its base Show that the volume of the three propotion
from top are in ratio 1:7:19

Aakash Sharma · Accepted Answer

Let VAB be a right circular cone of height 3h and base radius r. Thus cone is cut by planes parallel to base at point of and L such that VL = L O´ = h.

Since ∆VOA and ∆VB´A´ are similar

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

Also ∆VOA ~ ∆VLC

$∴ \frac{VO}{VL} = \frac{OA}{LC} \Rightarrow \frac{3 h}{2 h} = \frac{r}{r_{2}} \Rightarrow r_{2} = \frac{r}{3}$

Let V₁ be the volume of 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 V₂ be the volume of the frustum A´B´DC. Then

$V_{2} = \frac{1}{3} π (r_{1}^{2} + r_{2}^{2} + 4 r_{2}) h \Rightarrow V_{2} = \frac{1}{3} π (\frac{4 r^{2}}{9} + \frac{r^{2}}{9} + \frac{2 r^{2}}{9}) h (∵ r_{1} = \frac{2 r}{3} {and r}_{2} = \frac{r}{3}) \Rightarrow V_{2} = \frac{7}{27} π r^{2} h$

Let V₃ be the volume of the frustum ABB´A´. Then

$V_{3} = \frac{1}{3} π (r^{2} + r_{1}^{2} + r_{1} r) h \Rightarrow V_{3} = \frac{1}{3} π (r^{2} + \frac{4 r^{2}}{9} + \frac{2 r^{2}}{3}) h \Rightarrow V_{3} = \frac{19}{27} π r^{2} h ∴ Required ratio = V_{1} : V_{2} : V_{3} = \frac{1}{27} π r^{2} h : \frac{7}{27} π r^{2} h : \frac{19}{27} π r^{2} h = 1 : 7 : 19$

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The height of right circular cone is trisected by two plane parallel to its base. Show that the volume of the three propotion from top are in ratio 1:7:19.