Question -
Answer -
Let side of first square = x m
and of second squares = y m
According to the given conditions,
4x – 4y = 64
=> x – y = 16 ….(i)
and x² + y² = 640 ….(ii)
From (i), x = 16 + y
In (ii)
(16 + y)² + y² = 640
=> 256 + 32y + y² + y² = 640
=> 2y² + 32y + 256 – 640 = 0
=> y² + 16y – 192 = 0 (Dividing by 2)
=> y² + 24y – 8y – 192 = 0
=> y (y + 24) – 8 (y + 24) = 0
=> (y + 24)(y – 8) = 0
Either y + 24 = 0, then y = -24, which is not possible as it is negative
or y – 8 = 0, then y = 8
x = 16 + y = 16 + 8 = 24
Side of first square = 24 m
and side of second square = 8m