Question

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

Answer 1

Hey mate !!!.


HERE IS THE ANSWER


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

Ist 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

and x=24

$hope \: it \: helps \: you$
■■■Thanx

Answer 2

Step-by-step explanation:

Let the sides of the two squares be x and y [x > y]

Perimeter = 4 x side

                = 4x - 4y = 64

                = x - y = 16

                = x = 16 + y

x^2 + y^2 = 640

[16 + y]^2 + y^2 = 640

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

2y^2 + 32y + 256 - 640 = 0

2y^2 + 32y - 384 = 0

y^2 + 16y - 192 = 0/2 = 0

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

y [y+24] -8 [y+24] = 0

y = -24 [rejected]         y = 8

   = y = 8m

  = x = y + 16 = 8 + 16 = 24m