# The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Let the sides of the squares be ‘(X)’cms and ‘(Y) cms.

X2 + Y2 = 640 –––––(i)

Area = (Side)2

Perimeter = 4 (Side)

4X – 4Y = 64

On simplifying further,

X – Y = 16 ––––––(ii)

Squaring the above mentioned equation, i.e., equation (ii)

X2 + Y2 – 2XY = 256

Using the identity of a2 + b2 – 2ab = (a – b)2

Putting the value of equation (i) in equation (ii)

640 – 2XY = 256

2XY = 384

XY = 192

Putting the value of in equation (i)

(Y+16)2 + Y2 = 640

Y2 + 256 + 32Y + Y2 = 640

2Y2 + 32Y – 384 = 0

Y2 + 16Y – 192 = 0

On applying Sreedhracharya formula

X = – 12 or 24.

the side is X = 24 cms (Only positive values), other square’s side is Y=X–16 i.e., 8 cms.

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