Using $1 \mathrm{~cm}$ side squares, stairs are constructed as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-22.jpg?height=468&width=1288&top_left_y=497&top_left_x=287)

a) Calculate the total area and the perimeter of the fifth stair constructed.

b) We need a stair with an area of $78 \mathrm{~cm}^{2}$. Which stair should we choose?

c) We need a stair with a perimeter of $100 \mathrm{~cm}$. Which stair should we choose?
Solution

a) We start with an observation: The first stair is made up of only one square. The second stair is obtained from the first by adding a new level containing two squares. Thus, it has $1+2=3$ squares. The third stair is obtained from the second by adding a new level containing three squares, so it has $1+2+3=6$ squares. This same reasoning applies to the other stairs.

Thus, to calculate the area of the fifth stair, we observe that there are 5 squares on the first level, 4 squares on the second level, 3 squares on the third level, 2 squares on the second level, and 1 square on the first level. In total, the stair is made up of $1+2+3+4+5=15$ squares. Each square has an area of $1 \mathrm{~cm}^{2}$. Therefore, the total area of the stair is $15 \mathrm{~cm}^{2}$.

To calculate the perimeter, we can count the number of vertical segments and the number of horizontal segments that make up the contour of the fifth stair.

Since there are five steps, there are 5 vertical segments and 5 horizontal segments making up the steps. Note that for each horizontal segment on a step, there is a horizontal segment at the base of the stair. Similarly, for each vertical segment on a step, there is a vertical segment on the right side of the stair. In total, we have $5+5$ vertical segments and $5+5$ horizontal segments. Therefore, the total perimeter of the stair is $(5+5+5+5)=20 \mathrm{~cm}$.

Observation: We can generalize the argument to calculate the area and perimeter of the $n$-th stair. In this case, at the base of the stair, there will be $n$ squares, on the second level there will be $n-1$, and so on, until there is 1 square at the top. Thus, the $n$-th stair is made up of $1+2+\cdots+n$ squares.

To calculate this sum, we write the numbers as follows, repeating the numbers of the sum in reverse:

$$
\begin{array}{ccccccccccccc}
1 & + & 2 & + & 3 & + & \cdots & + & n-2 & + & n-1 & + & n \\
n & + & n-1 & + & n-2 & + & \cdots & + & 3 & + & 2 & + & 1
\end{array}
$$

Observe that the sum of two numbers in any column is always $n+1$. Thus, if we sum all the numbers above, since there are $n$ rows, we will get $n(n+1)$ as the answer. Since each number appears twice in the sum above, the result $n(n+1)$ corresponds to twice the sum we want. That is,

$$
1+2+\cdots+n=\frac{n(n+1)}{2}
$$

Thus, the stair will be made up of $\frac{n(n+1)}{2}$ squares. We conclude that

$$
\text { The area of the } n \text {-th stair is } \frac{n(n+1)}{2} \mathbf{c m}^{2}
$$

On the other hand, in the $n$-th stair, we can count $n+n$ vertical lines and $n+n$ horizontal lines. Thus, we conclude that

$$
\text { the perimeter of the } n \text {-th stair is } 4 n \mathrm{~cm} \text {. }
$$

b) By (.5), to get a stair of $78 \mathrm{~cm}^{2}$, we need $n(n+1)=2 \times 78=$ $12 \times 13$. Therefore, the 12th stair is the one that has $78 \mathrm{~cm}^{2}$ of area.

c) By (.6), to get a stair of $100 \mathrm{~cm}$, we need $4 n=100$. Therefore, the 25th stair is the stair that has $100 \mathrm{~cm}$ of perimeter.