Question

Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.

• A. 3:1
• B. 2:1
• C. 4:1
• D. 5:1

Hint:

When dealing with ratios of volumes or areas of similar geometric shapes, express the ratio of the formula variables and then substitute the given ratios. For example, for cones, $V \propto r^2h$, so $\frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^2 \left(\frac{h_1}{h_2}\right)$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks to find the ratio of the volumes of two cones, given the ratios of their heights and the radii of their bases.

Step 2: Key Formula or Approach:
The volume of a cone ($V$) is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Where $r$ is the radius of the base and $h$ is the height.

Step 3: Detailed Explanation:
Let the heights of the two cones be $h_1$ and $h_2$, and their radii be $r_1$ and $r_2$.
Given:
- Ratio of heights: $\frac{h_1}{h_2} = \frac{1}{3}$.
- Ratio of radii: $\frac{r_1}{r_2} = \frac{3}{1}$.
The volume of the first cone is $V_1 = \frac{1}{3} \pi r_1^2 h_1$.
The volume of the second cone is $V_2 = \frac{1}{3} \pi r_2^2 h_2$.
The ratio of their volumes is:
\[ \frac{V_1}{V_2} = \frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2} \]
Cancel out the common factors ($\frac{1}{3}\pi$):
\[ \frac{V_1}{V_2} = \frac{r_1^2 h_1}{r_2^2 h_2} \]
Rearrange the terms:
\[ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \]
Substitute the given ratios:
\[ \frac{V_1}{V_2} = \left(\frac{3}{1}\right)^2 \times \left(\frac{1}{3}\right) \]
\[ \frac{V_1}{V_2} = \left(\frac{9}{1}\right) \times \left(\frac{1}{3}\right) \]
\[ \frac{V_1}{V_2} = \frac{9}{3} = \frac{3}{1} \]
So, the ratio of their volumes is 3:1.

Step 4: Final Answer:
The ratio of their volumes is 3:1.