The area of a square and circle is same and the perimeter of square and equilateral triangle is same, then the ratio between the area of circle and the area of equilateral triangle is:

Options

A. `pi:3`
B. `9:4sqrt3`
C. `4:9sqrt3`
D. none of these

Answer

To solve the problem, we need to find the ratio between the area of a circle and the area of an equilateral triangle given that the area of a square and a circle are the same, and the perimeter of a square and an equilateral triangle are also the same.

### Step-by-Step Solution:

1. **Let the side of the square be \( a \)**.
- The area of the square \( A_s \) is given by:
\[ A_s = a^2 \]

2. **Since the area of the square is equal to the area of the circle, we can express the area of the circle**:
- Let the radius of the circle be \( r \). The area of the circle \( A_c \) is given by:
\[ A_c = \pi r^2 \]
- Setting the areas equal gives:
\[ a^2 = \pi r^2 \]
- From this, we can express \( r \) in terms of \( a \):
\[ r = \sqrt{\frac{a^2}{\pi}} = \frac{a}{\sqrt{\pi}} \]

3. **Next, we find the perimeter of the square**:
- The perimeter \( P_s \) of the square is:
\[ P_s = 4a \]

4. **Let the side of the equilateral triangle be \( b \)**. The perimeter \( P_t \) of the equilateral triangle is:
- The perimeter of the triangle is:
\[ P_t = 3b \]
- Since the perimeters of the square and the triangle are equal, we have:
\[ 4a = 3b \]
- From this, we can express \( b \) in terms of \( a \):
\[ b = \frac{4a}{3} \]

5. **Now, we calculate the area of the equilateral triangle**:
- The area \( A_t \) of an equilateral triangle is given by:
\[ A_t = \frac{\sqrt{3}}{4} b^2 \]
- Substituting \( b = \frac{4a}{3} \):
\[ A_t = \frac{\sqrt{3}}{4} \left(\frac{4a}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{16a^2}{9} = \frac{4\sqrt{3}}{9} a^2 \]

6. **Now we have both areas**:
- Area of the circle \( A_c = \pi r^2 = a^2 \)
- Area of the equilateral triangle \( A_t = \frac{4\sqrt{3}}{9} a^2 \)

7. **Finding the ratio of the area of the circle to the area of the equilateral triangle**:
\[ \text{Ratio} = \frac{A_c}{A_t} = \frac{a^2}{\frac{4\sqrt{3}}{9} a^2} = \frac{9}{4\sqrt{3}} \]

8. **Simplifying the ratio**:
- The ratio can be expressed as:
\[ \text{Ratio} = \frac{9}{4\sqrt{3}} \implies \text{Ratio} = 9 : 4\sqrt{3} \]

### Final Answer:
The ratio between the area of the circle and the area of the equilateral triangle is \( 9 : 4\sqrt{3} \).