A wire of length 30 m is cut into two pieces. One of the pieces is folded into a square and the other into a circle. If 'x' m is the length of one of the pieces, the combined area of the square and the circle is given by:

$\frac{x^2}{4 \pi}+\frac{1}{16}(30-x)^2$

The correct answer is Option (2) → $\frac{x^2}{4 \pi}+\frac{1}{16}(30-x)^2$

$\text{Let } x \text{ be the length used for the circle, then } (30 - x) \text{ for the square}$

$2\pi r = x$

$r = \frac{x}{2\pi}$

$\text{Area of circle} = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}$

$\text{Side of square} = \frac{30 - x}{4}$

$\text{Area of square} = \left(\frac{30 - x}{4}\right)^2 = \frac{(30 - x)^2}{16}$

$\text{Total area} = \frac{x^2}{4\pi} + \frac{(30 - x)^2}{16}$

The combined area is $\frac{x^2}{4\pi} + \frac{(30 - x)^2}{16}$.