Each triangle has an area of

\frac{\sqrt{3}}{4}

. Clearly if I can make a square out of such shapes then its area will be a whole number multiple of

\frac{\sqrt{3}}{4}

Now, the triangle has three lengths:

1, 0.5, \frac{\sqrt{3}}{2}

This means that the side of the square will be of length

N + \frac{M \sqrt{3}}{2} + \frac{P}{2}

for some whole numbers

N, M, P

Squaring this side length gives us a number of the form

A + B \sqrt{3}

where A and B are rational numbers with A not zero (B might be zero, but that does not concern us).

This contradicts the statement that the area of the square is a rational multiple of

\sqrt{3}

and, therefore, no such square is possible.