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.

Answer 2

Two cones have heights in the ratio 1:3 and base radii in the ratio 3:1. Rather than manipulating the ratio symbolically, we can assign convenient actual numbers that satisfy both ratios and compute real volumes to see which option the numbers support.

1. 3:1: Let the first cone have radius \( 3 \) and height \( 1 \), and the second cone have radius \( 1 \) and height \( 3 \), matching both given ratios. Volume of cone 1: \( \frac{1}{3}\pi (3)^2 (1) = 3\pi \). Volume of cone 2: \( \frac{1}{3}\pi (1)^2 (3) = \pi \). The ratio \( V_1 : V_2 = 3\pi : \pi = 3:1 \), matching this option exactly.
2. 2:1: For this ratio to hold, \( V_1 \) would need to be only twice \( V_2 \), but the actual computed volumes above give \( 3\pi \) and \( \pi \), a ratio of 3, not 2, so this option does not fit the given dimensions.
3. 4:1: This would require \( V_1 = 4\pi \) for the same \( V_2 = \pi \), but the radius-squared and height combination for cone 1 only produces \( 3\pi \), not \( 4\pi \), so this overstates the ratio.
4. 5:1: Similarly, this would need \( V_1 = 5\pi \), which is well above the \( 3\pi \) that the given height and radius ratios actually produce.

Assigning concrete values consistent with both given ratios and computing the volumes directly confirms that the first cone's volume is exactly three times the second's.

Therefore, the correct answer is 3:1.