Question

The sum of the areas of two squares is is 468 m square and the difference between their perimeter is 24. find their lengths

Answer 1

Answer:

Please mark as brainlist

Answer 2

Answer:

SQUARE 1 = 18 m , SQUARE 2 = 12 m

Step-by-step explanation:

let side of square 1 be x & square 2 be y where x > y

4x - 4y = 24 , $x^2 + y^2$ = 468 -----> (2)

⇒4(x-y) = 24

⇒x-y = 6

⇒y = x-6 -----> (1)

on sub. (1) in (2)

⇒ $x^2 + x^2 + 36$ -12x = 468

⇒ $2x^2$ -12x -432 = 0

⇒ $x^2$ - 6x -216 = 0

⇒ $x^2$ -18x + 12x - 216 = 0

⇒ x(x - 18) + 12(x -18) = 0

⇒(x -18)(x +12) = 0

⇒ x = 18

⇒ y = 18 - 6 = 12