# sum of the areas of two squares is 640 m2.if the difference of their perimeters is64m,find the sides of the two squares.

Let the length of the side of the smaller square be x metre.

Perimeter of the smaller square = 4x metre.

Now, perimeter of the bigger square is 64 metre more than the perimeter of the smaller square.

Therefore, perimeter of the bigger square = (4x + 64) metre

Length of the side of the bigger square = = (x + 16) metre

Now, sum of the area of the two squares is 640 m. • 20

Let the side of one of the square be- a and the other be- b.

ATQ, a^2+b^2=640 m^2...............(1)

4(a-b)=64 => a-b=16....................(2)

Substituting value of b from (2) in (1) we have,

(16+b)^2+b^2=640

2b^2+32b-384=0

b=8 or (-24)

As side of a figure cannot be in negative, so one of the square's side is 8 (b)

and the other is 24 (16+b).

THUMBS UP-IF CORRECT.

• 3

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

Ist case

area of square = (side)2

x2 + y2 =640 --------------------------   (1)

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

y+ 24y-8y -192=0

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

y=8

and x=24

• 72
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