Question 1

We will analyze the dimensions of the given pieces and then arrange them to form a new square with double the area.

Step 1 — Understand the pieces

Let $s$ be the side length of Square 1. The area of Square 1 is $s^2$. Square 1 is divided into four identical right-angled isosceles triangles (1, 2, 3, 4) by its diagonals. Each of these triangles has two equal sides (legs) of length $s/\sqrt{2}$. The third side (hypotenuse) of each triangle is a side of Square 1, so its length is $s$. The area of each triangle (1, 2, 3, or 4) is calculated as half the product of its legs. $\text{Area of one triangle} = \frac{1}{2} \times \left(\frac{s}{\sqrt{2}}\right) \times \left(\frac{s}{\sqrt{2}}\right)$ $= \frac{1}{2} \times \frac{s^2}{2}$ $= \frac{s^2}{4}$ The problem states that Square 2 is identical to Square 1. Pieces 5, 6, 7, and 8 are also identical to the triangles cut from Square 1. So, each of pieces 5, 6, 7, and 8 is a right-angled isosceles triangle with legs $s/\sqrt{2}$ and hypotenuse $s$. We need to place pieces 5, 6, 7, and 8 around Square 1. The total area of the new square will be the area of Square 1 plus the areas of pieces 5, 6, 7, and 8. $\text{Total Area} = \text{Area(Square 1)} + \text{Area(5)} + \text{Area(6)} + \text{Area(7)} + \text{Area(8)}$ $= s^2 + 4 \times \left(\frac{s^2}{4}\right)$ $= s^2 + s^2$ $= 2s^2$ The new square will have an area of $2s^2$, which is double the area of Square 1. Let $S_{\text{new}}$ be the side length of this new square. $S_{\text{new}}^2 = 2s^2$ $S_{\text{new}} = \sqrt{2s^2}$

$\boxed{S_{\text{new}} = s\sqrt{2}}$ The new square must have a side length of $s\sqrt{2}$.

Step 2 — Arrange the pieces

We need to arrange Square 1 and the four triangles (5, 6, 7, 8) to form a square with side length $s\sqrt{2}$. We can achieve this by placing the four triangles (5, 6, 7, 8) in the corners of the new, larger square. The right-angle vertex of each triangle should be positioned at a corner of the larger square. The legs of each triangle, which are $s/\sqrt{2}$ long, will lie along the sides of the larger square. The total side length of the larger square will be the sum of the lengths of two legs. $\text{Side length of large square} = \frac{s}{\sqrt{2}} + \frac{s}{\sqrt{2}}$ $= \frac{2s}{\sqrt{2}}$ $= s\sqrt{2}$ This matches the required side length for a square with double the area. The hypotenuse of each triangle (5, 6, 7, or 8), which is $s$ long, will face inwards. These four hypotenuses will form the sides of the central shape. Since all four hypotenuses are of length $s$ and they meet at right angles (due to the arrangement of the outer triangles), they form a square of side length $s$. This central square is exactly Square 1.