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If the ratio of the areas of two squares is $9:1$, the ratios of their perimeters is:

  1. $9:1$
  2. $3:1$
  3. $3:4$
  4. $1:3$
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According to the question:

$\frac{\text{area of 1st square}}{\text{area of 2nd square}}=\frac{9}{1}$

$\because$ Area of  square is $a^2$,where $a$ is side of square.

Assume that side of square1 is $a_1$ & side of square 2 is $a_2$.

$\implies$ $\frac{(a_1)^2}{(a_2)^2}=\frac{9}{1}$

$\implies$ $\frac{a_1}{a_2}=\frac{3}{1}$

$\therefore$ Side of square1 is $3$ and side of square2 is $1$

$\because$ perimeter of square is $4*a$,so there ratio is;

$\implies$ $\frac{4*a_1}{4*a_2}$

$\implies$ $\frac{4*3}{4*1}$

$\implies$ $3:1$

Option (b) is correct.
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$\frac{a_1}{a_2}=\frac{3}{1}$

Does the above equation mean side length of square 1 is 3 and side length of square 2 is 1?
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