How many metallic cones of radius 3 cm and height 13.5 cm are required to construct a metallic sphere of radius 4.5 cm?

Answer

To find out how many metallic cones of radius 3 cm and height 13.5 cm are required to construct a metallic sphere of radius 4.5 cm, we can follow these steps:

### Step 1: Calculate the Volume of the Sphere
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

For our sphere, the radius \( r = 4.5 \) cm. Plugging in the value:
\[ V = \frac{4}{3} \pi (4.5)^3 \]

### Step 2: Calculate \( (4.5)^3 \)
Calculating \( (4.5)^3 \):
\[ (4.5)^3 = 4.5 \times 4.5 \times 4.5 = 20.25 \times 4.5 = 91.125 \]
So,
\[ V = \frac{4}{3} \pi (91.125) \]

### Step 3: Calculate the Volume of One Cone
The formula for the volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

For our cone, the radius \( r = 3 \) cm and height \( h = 13.5 \) cm. Plugging in the values:
\[ V = \frac{1}{3} \pi (3)^2 (13.5) \]

### Step 4: Calculate \( (3)^2 \)
Calculating \( (3)^2 \):
\[ (3)^2 = 9 \]
So,
\[ V = \frac{1}{3} \pi (9) (13.5) \]

### Step 5: Calculate the Volume of the Cone
Now, calculate the volume:
\[ V = \frac{1}{3} \pi (9 \times 13.5) = \frac{1}{3} \pi (121.5) \]

### Step 6: Simplify the Volume of the Cone
\[ V = \frac{121.5}{3} \pi = 40.5 \pi \]

### Step 7: Find the Number of Cones Required
Let \( x \) be the number of cones required. We can set up the equation:
\[ x = \frac{\text{Volume of Sphere}}{\text{Volume of One Cone}} \]
Substituting the volumes we calculated:
\[ x = \frac{\frac{4}{3} \pi (91.125)}{40.5 \pi} \]

### Step 8: Cancel \( \pi \) and Simplify
Cancel \( \pi \) from the numerator and denominator:
\[ x = \frac{\frac{4}{3} (91.125)}{40.5} \]

### Step 9: Calculate \( \frac{91.125}{40.5} \)
Calculating \( \frac{91.125}{40.5} \):
\[ \frac{91.125}{40.5} = 2.25 \]
So,
\[ x = \frac{4}{3} \times 2.25 \]

### Step 10: Final Calculation
Calculating \( \frac{4 \times 2.25}{3} \):
\[ x = \frac{9}{3} = 3 \]

Thus, the number of metallic cones required is **3**.