Question

if the ratio of the areas of two squares is 9 : 1 . then the ratio of their perimeters is ​

Answer 1

Answer:

1/3

Step-by-step explanation:

area of a square with side a is a²

and perimeter of a square with side a is 4a

given a²/b²=9/1

therfore a/b=1/3

ratio of their perimeters is 4a/4b =a/b=1/3

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Answer 2

Answer:

Ratio = 3 : 1

Step-by-step explanation:

Let the side of one square be a and side of other square be a'.

According to the question;

$\frac{a {}^{2} }{a' {}^{2} } = \frac{9}{1} \\ \\ (\frac{a }{a' }){}^{2} = ( \frac{3}{1} ) {}^{2} \\ \\ \frac{a }{a'} = \frac{3}{1}$

Now, we know that;

Perimeter of Square = 4 * side,

$\frac{a }{a'} = \frac{3}{1} \\ \\ \frac{a }{a'} = \frac{3}{1} \\ \\ \frac{4 \times a}{4 \times a'} = \frac{4 \times 3}{4 \times 1} \\ \\ \frac{4a}{4a'} = \frac{12}{4} \\ \\ \frac{4a}{4a'} = \frac{3}{1} \\ \\ \boxed { \bf 4a : 4a' = 3:1}$

Hence, the ratio of the perimeters of both the squares is 3:1.