Question

The height of a cone and the radius of its base are respectively 9 and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of frustum of cone is 44 $\text{c}{\text{m}}^3$, then what is the radius of upper circular of frustum (Use $\pi ={\displaystyle \frac{22}{7}}$) ?
$$\begin{gathered} \text{A}\text{. }\sqrt[3]{12}\text{ cm} \\ \text{B}\text{. }\sqrt[3]{13}\text{ cm} \\ \text{C}\text{. }\sqrt[3]{6}\text{ cm} \\ \text{D}\text{. }\sqrt[3]{20}\text{ cm} \end{gathered}$$

Answer 1

Hint- Here, we will proceed by developing a relation between the height and the radius of the smaller cone obtained after the original cone is cut by a plane. This is achieved with the help of the concept of similar triangles.

Complete step by step answer:
Given, Height of the cone, H = 9 cm
Radius of the base of the cone, R = 3 cm
Now, this cone is cut by a plane parallel to its base so as to divide this original cone into a smaller cone and a frustum. Let the radius of the smaller (upper) circular end of the frustum be r cm and the height of the smaller cone obtained be h cm.
The radius of the larger (lower) circular end of frustum is equal to the radius of the original cone i.e., R = 3 cm
Height of frustum = (H – h) cm = (9-h) cm
In triangles ABC and ADE,
$\mathrm{\angle }\text{A}=\mathrm{\angle }\text{A}$ [Common angle in both the triangles]
$\mathrm{\angle }\text{ABC}=\mathrm{\angle }\text{ADE}$ [Corresponding angles (because BC is parallel to DE) are equal in measure]
$\mathrm{\angle }\text{ACB}=\mathrm{\angle }\text{AED}$ [Corresponding angles (because BC is parallel to DE) are equal in measure]
By AAA (Angle-Angle-Angle) similarity rule, we can say that the triangles ABC and ADE are similar to each other i.e., $△\text{ABC}\sim △\text{ADE}$.
For any two similar triangles, the ratio of their corresponding dimensions are always equal.
Using the above concept for the two similar triangles i.e., ABC and ADE, we have
$$\begin{gathered} \Rightarrow {\displaystyle \frac{\text{AF}}{\text{AG}}}={\displaystyle \frac{\text{BF}}{\text{DG}}} \\ \Rightarrow {\displaystyle \frac{h}{H}}={\displaystyle \frac{r}{R}} \\ \Rightarrow {\displaystyle \frac{h}{9}}={\displaystyle \frac{r}{3}} \\ \Rightarrow h={\displaystyle \frac{9r}{3}} \\ \Rightarrow h=3r \end{gathered}$$
Since, volume of any cone is given by
Volume of the cone = ${\displaystyle \frac{1}{3}}\pi {\left(\text{Base radius}\right)}^2\times \left(\text{Height}\right)$
As, we know that volume of the frustum will be given by
Volume of frustum = Volume of original cone – Volume of the smaller cone
$\Rightarrow$ Volume of frustum = ${\displaystyle \frac{1}{3}}\pi {\left(R\right)}^2\left(H\right)-{\displaystyle \frac{1}{3}}\pi {\left(r\right)}^2\left(h\right)$
By putting R = 3 cm, H = 9 cm, h = 3r and $\pi ={\displaystyle \frac{22}{7}}$ in the above equation, we get
 $\Rightarrow$ Volume of frustum = ${\displaystyle \frac{1}{3}}\pi {\left(3\right)}^2\left(9\right)-{\displaystyle \frac{1}{3}}\pi {\left(r\right)}^2\left(3r\right)=\left({\displaystyle \frac{22}{7}}\right)\left(3\right)\left(9\right)-\left({\displaystyle \frac{22}{7}}\right){\left(r\right)}^2\left(r\right)=\left({\displaystyle \frac{22}{7}}\right)\left(27-r^3\right)$
It is given that the volume of frustum is 44 $\text{c}{\text{m}}^3$
$$\begin{gathered} \Rightarrow 44=\left({\displaystyle \frac{22}{7}}\right)\left(27-r^3\right) \\ \Rightarrow 27-r^3={\displaystyle \frac{44\times 7}{22}}=2\times 7=14 \\ \Rightarrow r^3=27-14=13 \\ \Rightarrow r=\sqrt[3]{13}\text{ cm} \end{gathered}$$
Therefore, the radius of the upper circular of frustum is $\sqrt[3]{13}$cm.

Hence, option B is correct.

Note- In this particular problem, triangles ABC and ADE are similar triangles so, we can write the ratios of the corresponding sides will be equal i.e.,${\displaystyle \frac{\text{AB}}{\text{AD}}}={\displaystyle \frac{\text{BC}}{\text{DE}}}={\displaystyle \frac{\text{AC}}{\text{AE}}}$. In case of similar triangles, the ratio of the altitudes will be equal to the ratio of the corresponding sides i.e., ${\displaystyle \frac{\text{BC}}{\text{DE}}}={\displaystyle \frac{\text{AF}}{\text{AG}}}\Rightarrow {\displaystyle \frac{\text{2BF}}{\text{2DG}}}={\displaystyle \frac{\text{AF}}{\text{AG}}}\Rightarrow {\displaystyle \frac{\text{BF}}{\text{DG}}}={\displaystyle \frac{\text{AF}}{\text{AG}}}$.