Question

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 1

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  X = Y + 16 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.