Question

If the ratio of areas of two squares is 81: 121, then the ratio of their perimeters is?

Answer 1

Answer:

Answer:

9:11

Hope is it help full

Step-by-step explanation:

Answer 2

The ratio of perimeters of the given two squares with the ratio in the area as 81 : 121 is 9 : 11.  

Given :

The ratio of areas of 2 squares = 81 : 121

To Find :

Ratio of perimeters of the 2 squares

Solution :

Let the length of a side of the first square S1 be "a" and the length of a side of the second square i.e. S2 be "b".

Area of square S1 with side "a"   $= a * a$

                                                      $= a^{2}$

Area of square S2 with side "b"  $= b * b$

                                                      $= b^{2}$

Then the ratio of the areas of the 2 squares as compared with the given ratio is :

                    $81 : 121 = Area S1 : Area S2$

                    $81 : 121 = a^{2} : b^{2}$

                     $81/121 = a^{2}/b^{2}$

                     $9^{2}/11^{2} = a^{2} /b^{2}$

                     $(9/11)^{2} = (a/b)^{2}$

                       $9/11 = a/b$

            Thus, $a : b = 9 : 11$

The perimeter of square S1 with side "a"   $= 4 * a$

                                                                     $= 4a$

The perimeter of square S2 with side "b"  $= 4 * b$

                                                                      $= 4b$

Thus, the ratio of perimeters of the given two squares is :

                  $P1 : P2 = 4a : 4b$

                               $= 4a / 4b$

                                $= a/b$

                                $= a : b$

                                $= 9 : 11$

Hence, the ratio of perimeters of the given two squares according to the question is 9 : 11.                                

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