The figure below shows a "staircase" formed by two squares, one with a side of $8 \mathrm{~cm}$ and one with a side of $6 \mathrm{~cm}$. The task is to cut the figure into three pieces and reassemble them to form a square without any gaps.

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-22.jpg?height=503&width=1019&top_left_y=985&top_left_x=427)

(a) What is the side length of the square that should be formed at the end?

(b) Using only a pencil, a $20 \mathrm{~cm}$ ruler with markings every $1 \mathrm{~cm}$, and a pair of scissors that can only cut along a straight line, show how to perform the desired task.
Solution

Cutting the staircase to form a square

(a) Since the square should not have any holes, the final area must be equal to the original area. If we call $L$ the side of the square, we have:

$$
\begin{aligned}
L^{2} & =8^{2}+6^{2} \\
L^{2} & =64+36 \\
L^{2} & =100 \\
L & =10
\end{aligned}
$$

(b) By the Pythagorean Theorem, 8, 6, and 10 are the sides of a right triangle, since $6^{2}+8^{2}=10^{2}$. Taking the longest side of the figure above, which has a length of $8+6=14$, mark the point $R$ that divides it into pieces of lengths 6 and 8. This can be done with a ruler. With a pencil, draw segments from this point to the opposite left and right ends, denoted by $P$ and $Q$, as shown in the following figure. Using the Pythagorean Theorem, we have $P R=Q R=10$. These segments separate the figure into pieces $A, B$, and $C$. Pieces $A$ and $C$ are congruent triangles, and using the sums of their internal angles, we can conclude that $\angle P R Q=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-23.jpg?height=574&width=1025&top_left_y=798&top_left_x=615)

With scissors, the figure is separated into pieces $A, B$, and $C$. Then, they are rearranged to form the square with side 10 as shown in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-23.jpg?height=691&width=694&top_left_y=1733&top_left_x=778)