If one side of a square be a, then area of the largest circle in it, will be :

Question

Accepted Answer

To determine the area of the largest circle that can fit inside a square with side length $a$, we need to consider the geometry of the situation.

1. **Explanation for Option 2: $\frac{1}{4}\pi a^2$**

- The largest circle that can fit inside a square will have its diameter equal to the side of the square. Therefore, the diameter of the circle is $a$.
   - The radius $r$ of the circle is half of the diameter, so $r = \frac{a}{2}$.
   - The area $A$ of a circle is given by the formula $A = \pi r^2$.
   - Substituting the radius into the formula, we get:
     $$
     A = \pi \left(\frac{a}{2}\right)^2 = \pi \cdot \frac{a^2}{4} = \frac{1}{4}\pi a^2
     $$
   - Therefore, the area of the largest circle that can fit inside the square is $\frac{1}{4}\pi a^2$, which corresponds to Option 2.

2. **Explanation for why other options are incorrect:**

- **Option 1: $\frac{1}{2}\pi a^2$**
     - This option suggests that the area of the circle is half of $\pi a^2$. However, as calculated, the correct area is $\frac{1}{4}\pi a^2$. The factor of $\frac{1}{2}$ would imply a larger circle than can fit within the square, which is not possible given the constraints of the square's dimensions.
   
   - **Option 3: $\frac{1}{6}\pi a^2$**
     - This option suggests a smaller area than the actual largest circle that can fit inside the square. The calculation shows that the area is $\frac{1}{4}\pi a^2$, so $\frac{1}{6}\pi a^2$ underestimates the area of the circle.
   
   - **Option 4: $4\pi a^2$**
     - This option suggests an area much larger than the square itself, which is impossible. The area of the square is $a^2$, and a circle with an area of $4\pi a^2$ would not fit inside the square. This option is clearly incorrect based on the geometric constraints.

In conclusion, the correct answer is Option 2: $\frac{1}{4}\pi a^2$, as it accurately represents the area of the largest circle that can fit inside a square with side length $a$.