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

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Let the sides of the squares be x m and y m. Then,
Area of the first square = x² m
Area of the second square = y² m
Perimeter of the first square = 4x m
and Perimeter of the second square = 4y m
According to the given condition,
x² + y² = 640 ...(i)
and 4x – 4y = 64 or x – y = 16 ...(ii)
From (ii), y = x – 16 ...(iii)
Substituting the value of y in (i), we get
x² + (x – 16)² = 640
⇒ x² + x² + 256 – 32x = 640
⇒ 2x² – 32x + 256 = 640
⇒ 2x² – 32x – 384 = 0
⇒ x² – 16x – 192 = 0
⇒ x² – 24x + 8x – 192 = 0
⇒ x (x – 24) +8 (x – 24) = 0
⇒ (x + 8) (x – 24) = 0
x = 24 or x = –8
⇒ x = –8 rejected as the side of a square cannot be negative.
Hence, side of the first square,
x = 24 m
and side of the second square,
y = (24 – 16) m ...[Using (iii)]
= 8 m.