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