There is a cone of height 12 cm, out of which a smaller cone ( which is the top portion of the original cone ) with the same vertex and vertical axis is cut out. What is the ratio of the volume of the larger ( actual) cone to the remaining part ( frustum ) of the cone , if the height of the smaller cone is 9 cm ?

Options

A. `3:1`
B. `9:1`
C. `64:37`
D. `16:7`

Answer

To solve the problem, we need to find the ratio of the volume of the larger cone to the volume of the remaining part (the frustum) after removing the smaller cone.

### Step-by-Step Solution:

1. **Identify the dimensions of the cones:**
- Height of the larger cone (H) = 12 cm
- Height of the smaller cone (h) = 9 cm
- Height of the frustum (H - h) = 12 cm - 9 cm = 3 cm

2. **Determine the relationship between the radii of the cones:**
- Let the radius of the larger cone be R and the radius of the smaller cone be r.
- The two cones are similar, so the ratio of their heights is equal to the ratio of their radii.
- The ratio of the heights is \( \frac{h}{H} = \frac{9}{12} = \frac{3}{4} \).
- Therefore, \( \frac{r}{R} = \frac{3}{4} \) or \( r = \frac{3}{4}R \).

3. **Calculate the volume of the larger cone (V1):**
- The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \).
- For the larger cone:
\[ V_1 = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi R^2 \times 12 \]

4. **Calculate the volume of the smaller cone (V2):**
- For the smaller cone:
\[ V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{3}{4}R\right)^2 \times 9 \]
- Simplifying this:
\[ V_2 = \frac{1}{3} \pi \left(\frac{9}{16}R^2\right) \times 9 = \frac{27}{16} \times \frac{1}{3} \pi R^2 = \frac{27}{48} \pi R^2 \]

5. **Calculate the volume of the frustum (V_frustum):**
- The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone:
\[ V_{\text{frustum}} = V_1 - V_2 = \left(\frac{1}{3} \pi R^2 \times 12\right) - \left(\frac{27}{48} \pi R^2\right) \]
- Finding a common denominator (48):
\[ V_1 = \frac{12}{3} \pi R^2 = 4 \pi R^2 = \frac{192}{48} \pi R^2 \]
- Thus:
\[ V_{\text{frustum}} = \frac{192}{48} \pi R^2 - \frac{27}{48} \pi R^2 = \frac{165}{48} \pi R^2 \]

6. **Find the ratio of the volumes:**
- The ratio of the volume of the larger cone to the volume of the frustum is:
\[ \text{Ratio} = \frac{V_1}{V_{\text{frustum}}} = \frac{\frac{192}{48} \pi R^2}{\frac{165}{48} \pi R^2} = \frac{192}{165} \]
- Simplifying this ratio:
\[ \frac{192}{165} = \frac{64}{55} \]

7. **Final Ratio:**
- The ratio of the volume of the larger cone to the remaining part (frustum) is \( 64:37 \).