The side of a square is equal to the diameter of a circle. If the side and radius change at the same rate, then the ratio of the change of their areas, is 

Let x be the length of a side of the square and r be the radius of the circle. 

Then,$x=2r$ [Given] and $\frac{dx}{dt}=\frac{dr}{dt}$

Let $A_1$ and $A_2$ be the areas of the square and circle respectively. Then, 

$A_1=x^2$ and $A_2=\pi r^2$

$\Rightarrow \frac{d{A}_1}{dt}=2x\frac{dx}{dt}$ and $\frac{d{A}_2}{dt}=2\pi r\frac{dr}{dt}$

$\Rightarrow \frac{{\displaystyle \frac{d{A}_1}{dt}}}{\frac{d{A}_2}{dt}}=\frac{2x\frac{dx}{dt}}{2\pi r\frac{dr}{dt}}=\frac{2}{\pi } \left[\because x=2r\text{ and }\frac{dx}{dt}=\frac{dr}{dt}\right]$