Question

Sum of the areas of two squares is $640m^2.$ If the difference of their perimeters is 64m, find the sides of two squares.

Answer 1

Let side of first square be x and that of another be y

I^st case

area of square = (side)²

x² + y² =640 -------------------------- (i)

2nd case

4x -4y =64

=> x-y=16

x=16+y

put value of x in 1

(16+y)²+y²=640

256+y²+32y+y² =640

2y² +32y =384

y²+16y=192

y² + 24y-8y -192=0

y(y+24) -8(y+24)=0

y=8 , y=-24 (since length can not be negative, we will neglect y= -24 )

and x= $\sqrt{640-8^2}$ = $\sqrt{576}$ = 24m

hence the sides of the two square is 8m and 24m