Area of a circle is `61m^(2)`. Find the area of the greatest square constructed in this circle.

Answer

To solve the problem of finding the area of the greatest square that can be inscribed in a circle with an area of \(61 \, m^2\), we can follow these steps:

### Step-by-Step Solution

1. **Find the radius of the circle**:
The area of a circle is given by the formula:
\[ A = \pi r^2 \]
Given that the area \(A = 61 \, m^2\), we can set up the equation:
\[ \pi r^2 = 61 \]
To find \(r^2\), we rearrange the equation:
\[ r^2 = \frac{61}{\pi} \]

2. **Calculate \(r\)**:
Using the approximate value of \(\pi \approx 3.14\), we can calculate \(r^2\):
\[ r^2 = \frac{61}{3.14} \approx 19.48 \]
Now, take the square root to find \(r\):
\[ r \approx \sqrt{19.48} \approx 4.42 \, m \]

3. **Find the diameter of the circle**:
The diameter \(d\) of the circle is twice the radius:
\[ d = 2r \approx 2 \times 4.42 \approx 8.84 \, m \]

4. **Relate the diameter to the square**:
The greatest square that can be inscribed in the circle has its diagonal equal to the diameter of the circle. Thus, the diagonal \(d\) of the square is:
\[ d = s\sqrt{2} \]
where \(s\) is the side length of the square. Setting the two expressions for the diagonal equal gives:
\[ s\sqrt{2} = 8.84 \]

5. **Solve for the side length \(s\)**:
Rearranging the equation to find \(s\):
\[ s = \frac{8.84}{\sqrt{2}} \approx \frac{8.84}{1.414} \approx 6.25 \, m \]

6. **Calculate the area of the square**:
The area \(A_s\) of the square is given by:
\[ A_s = s^2 \approx (6.25)^2 \approx 39.06 \, m^2 \]

### Final Answer
The area of the greatest square that can be constructed in the circle is approximately:
\[ \boxed{39.06 \, m^2} \]