The sum of volumes of right circular cone (N) and right circular cylinder(L) both of height 18cm each is approximately `981cm^3`. If N's radius is half of L's radius , What is the approximate value of the area of L's base (in `cm^2`)

Options

A. 25
B. 50
C. 200
D. 100

Answer

To solve the problem, we need to find the approximate value of the area of the base of the right circular cylinder (L) given the volumes of both the cone (N) and cylinder (L) and their dimensions.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a right circular cone (N) and a right circular cylinder (L) both with a height of 18 cm.
- The radius of the cone (N) is half of the radius of the cylinder (L).
- The total volume of both shapes is approximately 981 cm³.

2. **Define Variables**:
- Let the radius of the cylinder (L) be \( R \).
- Therefore, the radius of the cone (N) will be \( \frac{R}{2} \).

3. **Volume Formulas**:
- The volume of the cone (N) is given by:
\[ V_{cone} = \frac{1}{3} \pi \left(\frac{R}{2}\right)^2 \cdot h = \frac{1}{3} \pi \left(\frac{R^2}{4}\right) \cdot 18 = \frac{3 \pi R^2}{8} \]
- The volume of the cylinder (L) is given by:
\[ V_{cylinder} = \pi R^2 \cdot h = \pi R^2 \cdot 18 = 18 \pi R^2 \]

4. **Setting Up the Equation**:
- The sum of the volumes of the cone and cylinder is:
\[ V_{cone} + V_{cylinder} = \frac{3 \pi R^2}{8} + 18 \pi R^2 = 981 \]
- Combine the terms:
\[ \frac{3 \pi R^2}{8} + \frac{144 \pi R^2}{8} = 981 \]
\[ \frac{147 \pi R^2}{8} = 981 \]

5. **Solving for \( R^2 \)**:
- Multiply both sides by 8:
\[ 147 \pi R^2 = 7848 \]
- Divide by \( 147 \pi \):
\[ R^2 = \frac{7848}{147 \pi} \]

6. **Calculating the Area of the Base of Cylinder (L)**:
- The area of the base of the cylinder (L) is:
\[ A = \pi R^2 \]
- Substitute \( R^2 \):
\[ A = \pi \left(\frac{7848}{147 \pi}\right) = \frac{7848}{147} \]

7. **Final Calculation**:
- Calculate \( \frac{7848}{147} \):
\[ A \approx 53.4 \text{ cm}^2 \]

8. **Conclusion**:
- The approximate value of the area of L's base is \( \approx 53.4 \text{ cm}^2 \).