Question

sum of the areas of two squares is 468 m2. if difference of their perimeters is 24 m. find tha sides of the two squares ​

Answer 1

Let us say that the sides of the two squares are 'a' and 'b'

Sum of their areas = a^2 + b^2 = 468

Difference of their perimeters = 4a - 4b = 24

=> a - b = 6

=> a = b + 6

So, we get the equation

(b + 6)^2 + b^2 = 468

=> 2b^2 + 12b + 36 = 468

=> b^2 + 6b - 216 = 0

=> b = 12

=> a = 18

The sides of the two squares are 12 and 18

Answer 2

Here is ur answer hope this helps if so mark as brainliest.