A Wire in the form of a circle, encloses an area 3118.5 $$cm^2$$. It is now bent to form a rectangle whose length and breadth are very nearly in the ratio 7 : 4. The length of the rectangle, in cm, is:(Take $$\pi = \frac{22}{7}$$)

Area enclosed by wire = 3118.5 $$cm^2$$

As we know, 

Area of circle = $$\pi\ r^2$$

$$\therefore\ \frac{22}{7}\times\ r^2=3118.5$$

$$\therefore\ \ r^2=992.25$$

$$\therefore\ \ r =31.5\ cm$$

Now, 

Let the length and breadth of the rectangle be 7x and 4x respectively.

$$\therefore\ \ $$The circumference of the circle = Perimeter of the rectangle formed by bending

$$2\pi\ r=2\left(l+b\right)$$

$$\therefore\ 2\times\ \frac{22}{7}\times\ \ 31.5=2\left(7x+4x\right)$$

$$\therefore\ x=9$$

Therefore, 

Length of the rectangle = 7x = $$7\times\ 9=63\ cm$$

Hence, Option D is correct.