Question

If the right circular cone is separated into solids of volumes $v_1,v_2,v_3$ by two plane parallel to the base and trisect the altitude $v_1:v_2:v_3$ is

• A. $1:2:3$
• B. $1:4:6$
• C. $1:6:9$
• D. $1:7:19$

Answer 1

The correct option is D $1:7:19$

Let cone ABC be trisected by the planes PQR and XYZ such that $AQ=QY=YD=h$

and $QR=r$

In $△AQR$ and $△AYZ$,

$\angle A=\angle A$ [common]

$\angle AQR=\angle AYZ$ [each $90°$]

$\therefore △AQR\sim △AYZ$ [AA similarity]

$\Rightarrow \frac{AQ}{AY}=\frac{QR}{YZ}$

$\Rightarrow \frac{h}{2h}=\frac{r}{yz}\Rightarrow RYZ=2r$

Similarly, $△AQR\sim △ADC$

$\Rightarrow \frac{AQ}{AD}=\frac{QR}{DC}\Rightarrow \frac{h}{3h}=\frac{r}{YZ}\Rightarrow DC=3r$

$V_1=$Volume of cone $APR=\frac{1}{3}\pi r^{2}h\to 1$

$V_2=$Volume of frustum $DXZR=\frac{1}{3}\pi h[{\left(2r\right)}^2+2r\times r+r^2]=\frac{1}{3}\pi h\times 7r^2$$=7\times \frac{1}{3}\pi r^{2}h\to 2$

$V_3=$Volume of fristum $XBCZ=\frac{1}{3}\pi h[{\left(3r\right)}^2+3r\times 2r+{\left(2r\right)}^2]=\frac{1}{3}\pi h\times 19r^2$

$=19\times \frac{1}{3}\pi r^{2}h\to 3$

$V_1:V_2:V_3=\frac{1}{3}\pi r^{2}h:7\times \frac{1}{3}\pi r^{2}h:19\times \frac{1}{3}\pi r^{2}h$

$V_1:V_2:V_3=1:7:19$