It's a poorly worded question alright. Here's my take on it:

Consider the values of z along the side OA:

z = a + 0i, 0 < a < 1,

=> z^2 = a^2, 0 < a < 1. These points obviously lie on part of the real axis. The part from 0 to 1.

Consider the values of z along the side OB:

z = a + ai, 0 < a < 1,

=> z^2 = 2ai, 0 < a < 1. These points obviously lie on part of the imaginary axis. The part from 0 to 2i.

Consider the values of z along the side AB:

z = 1 + ai, 0 < a < 1,

=> z^2 = (1 + ai)^2 = 1 - a^2 + 2ai, 0 < a < 1. So:

x = 1 - a^2 .... (1)

y = 2a => y/2 = a .... (2)

Treat (1) and (2) as parametric equations and eliminate a by substituting (2) into (1):

x = 1 - (y/2)^2, which is the equation of a sideways parabola. The domain is 0 < x < 1. The range is 0 < y < 2.

So as z TRACES the perimeter of the triangle, the locus of z^2 does indeed consist of part of the real axis, part of a parabola and part of the imaginary axis. The sketch of this locus should be straightforward.