Question

The sum of the areas of two squares is 640 $m^2$. If the difference in the perimeters be 64 m, find the sides of the two squares.

Answer 1

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

Ist case

area of square = $(side{)}^2$

$x^2+y^2=640$ -------------------------- (i)

2nd case

$4x-4y=64$

=> $x-y=16$

$x=16+y$

put the value of x in 1

$(16+y{)}^2+y^2=640$

$256+y^2+32y+y^2=640$

$2y^2+32y=384$

$y^2+16y=192$

$y^2+24y-8y-192=0$

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

y = 8

and x = 24