Question

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is \[\dfrac{1}{{27}}\] of the volume of the given cone, at what height, above the base is the section cut?
A. 20
B. 24
C. 25
D. 34

Answer 1

Hint: In this question first of all draw the diagram with the given data which will give us a clear idea of what we have to find, then find the volume of the smaller cone and the volume of the bigger cone. Next find the height of the smaller cone by using the given conditions. The height at which the small cone was made from the base is given by the difference of total height of the cone and height of the smaller cone.

Complete step-by-step answer:
Given height of the cone \[H = 30{\text{ cm}}\]
Let the small cone be cut off at a height ‘\[h\]’ from the top.
Let the radius of the bigger cone be \[R\] and the radius of the smaller cone be \[r\].
The figure of the cone is shown as below:

We know that the volume of the cone with radius \[R\] and height \[H\] is given by \[V = \dfrac{1}{3}\pi {R^2}H\]
So, volume of the bigger cone is given by
\[
   \Rightarrow {V_1} = \dfrac{1}{3}\pi {R^2}H \\
   \Rightarrow {V_1} = \dfrac{1}{3}\pi {R^2}\left( {30} \right) \\
  \therefore {V_1} = 10\pi {R^2}{\text{ cu}}{\text{.cm}} \\
\]
And the volume of smaller cone is given by
\[
   \Rightarrow {V_2} = \dfrac{1}{3}\pi {r^2}h \\
  \therefore {V_2} = \dfrac{1}{3}\pi {r^2}h{\text{ cu}}{\text{.cm}} \\
\]
But given that volume of small cone is \[\dfrac{1}{{27}}\] times of volume of big cone i.e.,
\[
   \Rightarrow {V_2} = \dfrac{1}{{27}}{V_1} \\
   \Rightarrow \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{{27}}\left( {10\pi {R^2}} \right) \\
   \Rightarrow \dfrac{{{r^2}}}{{{R^2}}} = \dfrac{3}{{27}} \times \dfrac{{10}}{h} \\
  \therefore {\left( {\dfrac{r}{R}} \right)^2} \times \left( {\dfrac{h}{{30}}} \right) = \dfrac{1}{{27}}...........................................\left( 1 \right) \\
\]
From the figure, we have
\[
   \Rightarrow \angle ADC = \angle ABO \\
   \Rightarrow \angle CAD = \angle OAB \\
\]
Thus, by AA similarity criterion \[\Delta ACD \cong \Delta AOB\]
We know that the ratio of sides of similar triangles are equal.
So, we have \[\dfrac{r}{R} = \dfrac{h}{{30}}...................................\left( 2 \right)\]
From equation (1) and (2), we have
\[
   \Rightarrow {\left( {\dfrac{h}{{30}}} \right)^2} \times \left( {\dfrac{h}{{30}}} \right) = \dfrac{1}{{27}} \\
   \Rightarrow {\left( {\dfrac{h}{{30}}} \right)^3} = {\left( {\dfrac{1}{3}} \right)^3} \\
\]
Taking cube root on both sides, we have
\[
   \Rightarrow \dfrac{h}{{30}} = \dfrac{1}{3} \\
  \therefore h = \dfrac{{30}}{3} = 10{\text{ cm}} \\
\]
So, the height of the small cone above the base = 30 – 10 = 20 cm
Thus, at a height of 20 cm above the base a small cone is made.

Note: The volume of the cone with radius \[R\] and height \[H\] is given by \[V = \dfrac{1}{3}\pi {R^2}H\]. The AA criterion tells us that two triangles are similar if two corresponding angles are equal to each other.