Let x = length of one piece, in meters.

So, (10 -x) = lentgh of the other piece, in meters.

say the x-piece be bent into a squre. Then the (10 -x) piece be bent into a circle.

For the square:

Perimeter = 4s = x

So, s = x/4

And so, area enclosed, A1 = s^2 = (x/4)^2 = (1/16)x^2

For the circle:

Perimeter = circumference = (2pi)r = 10 -x

So, r = (10 -x)/(2pi)

And so, area enclosed, A2 = (pi)r^2 = (pi)[(10 -x)/(2pi)]^2

A2 = (1/(4pi))[10 -x)^2]

Hence, total enclosed area,

A = A1 +A2

A = (1/16)x^2 +(1/(4pi))[(10 -x)^2] ------**

You now take over.