A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled

Answer

To solve the problem, we need to find the height of an inverted cone that is filled with chocolate sauce poured from a hemispherical bowl. We will use the formulas for the volumes of a hemisphere and a cone.

### Step-by-Step Solution:

1. **Find the radius of the hemispherical bowl:**
- The diameter of the bowl is given as 7.2 cm.
- The radius \( r \) is half of the diameter.
\[ r = \frac{7.2}{2} = 3.6 \text{ cm} \]

**Hint:** Remember that the radius is half of the diameter.

2. **Calculate the volume of the hemispherical bowl:**
- The formula for the volume \( V \) of a hemisphere is:
\[ V = \frac{2}{3} \pi r^3 \]
- Substituting the radius:
\[ V = \frac{2}{3} \pi (3.6)^3 \]

3. **Calculate \( (3.6)^3 \):**
- First, calculate \( 3.6 \times 3.6 = 12.96 \).
- Then, \( 12.96 \times 3.6 = 46.656 \).
- Therefore, \( (3.6)^3 = 46.656 \).

4. **Substitute back to find the volume of the hemisphere:**
\[ V = \frac{2}{3} \pi (46.656) = \frac{93.312}{3} \pi = 31.104 \pi \text{ cm}^3 \]

5. **Find the volume of the cone:**
- The formula for the volume \( V \) of a cone is:
\[ V = \frac{1}{3} \pi r^2 h \]
- The radius of the cone is given as 4.8 cm. Therefore:
\[ V = \frac{1}{3} \pi (4.8)^2 h \]

6. **Calculate \( (4.8)^2 \):**
- \( (4.8)^2 = 23.04 \).

7. **Substitute back to find the volume of the cone:**
\[ V = \frac{1}{3} \pi (23.04) h = \frac{23.04}{3} \pi h \]

8. **Set the volumes equal to each other:**
- Since the volume of the chocolate sauce in the hemisphere is equal to the volume of the cone:
\[ 31.104 \pi = \frac{23.04}{3} \pi h \]

9. **Cancel \( \pi \) from both sides:**
\[ 31.104 = \frac{23.04}{3} h \]

10. **Multiply both sides by 3 to isolate \( h \):**
\[ 93.312 = 23.04 h \]

11. **Solve for \( h \):**
\[ h = \frac{93.312}{23.04} \approx 4.05 \text{ cm} \]

### Final Answer:
The height of the cone is approximately **4.05 cm**.