Question

sum of the areas of two square is 468m square. if the difference of their perimeter is 24m then find the sides of two square?

Answer 1

Let the sides of the two squares be x and y.

Given that the sum of areas of two squares = 468m^2.

x^2 + y^2 = 468   ------ (1)

We know that perimeter of a square = 4 * Side.

Given that if their difference of their perimeter is 24m.

4x - 4y = 24

x - y = 6   

x = y + 6  ------ (2)


substitute (2) in (1), we get

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

y^2 + 36 + 12y + y^2 = 468

2y^2 + 12y + 36 = 468

2y^2 + 12y + 36 - 468 = 0

2y^2 + 12y - 432 = 0

y^2 + 6y - 216 = 0

y^2 + 18y - 12y - 216 = 0

y(y + 18) -12(y + 18) = 0

(y + 18)(y - 12) = 0

y = -18,12.

The side of the square cannot be negative.

Therefore the value of y = 12.


Substitute y = 12 in (2), we get

x = y + 6

x = 12 + 6

x = 18.


Therefore the sides of the two squares = 12 and 18.

Verification:

12^2 + 18^2 = 468

144 + 324 = 468

468 = 468.


4x - 4y = 24

4(18) - 4(12) = 24

72 - 48 = 24

24 = 24.


Hope this helps!