Geometry Help Please!!

Let's first find the length of one of the sides of the equilateral triangles. Since each side of the square has length 2, the altitude of one of the equilateral triangles is also 2. Therefore, the length of one of the sides of the equilateral triangle is:

$$s = \frac{2}{\sqrt{3}}$$

Now let's consider one of the smaller squares formed by joining two adjacent vertices of the equilateral triangles. Since the angles of an equilateral triangle are all 60 degrees, the angle between two adjacent sides of the square is 120 degrees. Therefore, the length of one of the sides of the smaller square is:

$$2s \cos 120^\circ = -s$$

The negative sign is because the cosine of 120 degrees is negative.

Now let's consider the diagonal of the large square. This diagonal is the sum of the diagonal of the smaller square and two sides of the equilateral triangles. Since the sides of the equilateral triangles have length $s$, their diagonals have length $2s$. Therefore, the length of the diagonal of the large square is:

$$2(-s) + 2s + 2s = 4s = \frac{8}{\sqrt{3}}$$

The length of a side of the large square is the square root of half the length of its diagonal, so:

$$\text{side length of large square} = \frac{1}{\sqrt{2}} \cdot \frac{8}{\sqrt{3}} = \boxed{\frac{4\sqrt{6}}{3}}$$

Let the square be ABCD and the other vertices of the equilateral triangles be E, F, G, and H. (So that the four equilateral triangles are ABE, BCF, CDG, and ADH.) Now consider the angles HAE, EBF, FCG, and GDH. (It would definitely help to draw the diagram yourself.) We know that these angles are 360 - 90 - 60 - 60 = 150 degrees. Now, let the side length of the outer square be x. By Law of Cosines, we have

\(x^2 = 2^2 + 2^2 - 2 \cdot 2 \cdot 2 \cdot \cos150^\circ = 8-8 \cdot (-\frac{\sqrt{3}}{2}) = 8+4\sqrt{3}\)

so we see that the side length of the outer square is \(\sqrt{8+4\sqrt{3}}\)