Hi Kathy!

On problem 21, a good strategy is working with pairs of squares of different lengths at the same time. Also, the diagram in the proof of the Pythagorean Theorem might give an idea of how to cut out the pieces.

For problem 29, you know that equilateral triangles have all angles $60^\circ$, so the smaller triangles formed when the square is inscribed in the bigger triangle are both $30-60-90$ triangles. The sides of a $30-60-90$ triangle are in ratio $1:\sqrt{3}:2$. As we are assuming the side of the square to have length $x$, the sides of the triangle have length $\frac{x\sqrt{3}}{3}$, $x$, and $\frac{2x\sqrt{3}}{3}$. The equilateral triangle has side length $6$, so we can set up the equation $6 = \frac{x\sqrt{3}}{3}+x+\frac{x\sqrt{3}}{3}$, from which we have that $$x = \frac{18}{2\sqrt{3}+3} = \frac{18}{2\sqrt{3}+3}\cdot\frac{2\sqrt{3}-3}{2\sqrt{3}-3} = 12\sqrt{3} -18.$$

We are also looking at the other post you made about the placement test.