Question

A right circular cone is divided by a plane parallel to its base into small cone of volume v1 at the top and a frustum of volume v2 as second part at bottom. If v1:v2=1:3, then find the ratio of the height of the altitude of small cone and that of frustum.

Answer 1

Answer:

Let V be the volume of Original cone, and $V_{1} {\text{and}} V_{2}$ be the volume of small cone and frustum respectively.

$\frac{V_{1}}{V_{2}}=\frac{1}{3}\\\\\frac{\frac{1}{3}\pi\times r^2h}{\frac{\pi}{3}H(R^2+r^2+Rr)}=\frac{1}{3}$

                                 -----------------------------------(1)

As, when you cut the cone, by a plane, the two larger cone and smaller cone will be similar by angle angle similarity, having same semi vertical angle, and one angle =90°.

So, when triangles are similar , their sides are proportional.

$\frac{r}{R}=\frac{h}{P}$

h+H=P

Dividing numerator and denominator by r²

$=\frac{h}{H(1+\frac{R^2}{r^2}+\frac{R}{r})}=\frac{1}{3}\\\\(x-1)(x^2+x+1)=3\\\\ x^3-1=3\\\\x^3=4\\\\ x=\pm2\\\\{\text{here}}\frac{R}{r}=\frac{H}{h}+1,{\text{and}} \frac{R}{r}=x, so, \frac{H}{h}=2+1=3$

Here, taken positive value of, x, which is 2, as ratio of height can't be negative.

And, used the identity

x³-1=(x-1)(x²+x+1)

Answer 2

Answer:

Step-by-step explanation: