Question

right circular cone is divided by a plane parallel to its base into small cones at the top and frustum at the bottom . Both parts have the volume 1:3 . Find the ratio of the height of small cone and frustum​

Answer 1

Answer:

The ratio is $\dfrac{h}{h_1} = \dfrac{3r^2}{R^2 + r^2 + rR}$

Step-by-step explanation:

* Refer the figure, we can understand that,

• The volume of cone = $\dfrac{1}{3} \pi r^2 h_1$

• Volume of frustum = $\dfrac{1}{3} \pi h (R^2 + r^2 + r\times R)$

It is given that the ratio of volume of cone to frustum is 1:3.

So,

• $\dfrac{\dfrac{1}{3} \pi r^2 h_1}{\dfrac{1}{3} \pi h (R^2 + r^2 + r \times R)} = \dfrac{1}{3}$

By simplifying it, we get.

• $h(R^2 + r^2 + r\times R) = 3r^2 h_1$

                      $\dfrac{h}{h_1} = \dfrac{3r^2}{R^2 + r^2 + rR}$