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

- $9:1$
- $3:1$
- $3:4$
- $1:3$

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$\textrm{according to the question}$

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

$\because$ $\textrm{area of square is $a^2$,where a is side of square.}$

$\textrm{assume that side of square 1 is $a_1$ & side of square 2 is $a_2$.}$

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

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

$\therefore$ $\textrm{side of square 1 is 3 and side of square 2 is 1}$

$\because$ $\textrm{perimeter of square is 4a,so there ratio is;}$

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

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

$\implies$ $3:1$

$\textrm{Option b is correct.}$

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

$\because$ $\textrm{area of square is $a^2$,where a is side of square.}$

$\textrm{assume that side of square 1 is $a_1$ & side of square 2 is $a_2$.}$

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

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

$\therefore$ $\textrm{side of square 1 is 3 and side of square 2 is 1}$

$\because$ $\textrm{perimeter of square is 4a,so there ratio is;}$

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

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

$\implies$ $3:1$

$\textrm{Option b is correct.}$