# Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .​

Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .​

### 2 thoughts on “Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .​”

1. The Olympic motto “Citius, Altius, Fortius” (“Faster, Higher, Stronger”) was coined by Father Henri Didon, who was a close friend of Baron Pierre de Coubertin. It was adopted by the IOC in 1894.

Let the sides of the first and second square be X and Y .

Area of the first square = (X)²

Area of the second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ——(1).

Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y

According to question,

4X – 4Y = 24 ——–(2)

From equation (2) we get,

4X – 4Y = 244X–4Y=24

4(X-Y) = 244(X−Y)=24

X – Y = 24/4X–Y=24/4

X – Y = 6X–Y=6

X=6+Y———(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468(X)²+(Y)²=468

= >  (6+Y)² + (Y)² = 468

= > (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

= > 36 + Y² + 12Y + Y² = 468

= > 2( Y² + 6Y – 216) = 0

= > Y² + 6Y – 216 = 0

= > Y² + 18Y – 12Y -216 = 0

= > Y(Y+18) – 12(Y+18) = 0

= > (Y+18) (Y-12) = 0

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18X=6+Y=6+12=18

Side of first square = X = 18 m

Side of second square = Y = 12 m.