Q. 254.6( 8 Votes )

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.

Answer :

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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