Adriana likes to paint geometric figures. She painted the front of a wall of her house as follows:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-07.jpg?height=363&width=1360&top_left_y=504&top_left_x=448)

a) The wall is rectangular, has a height of $1 \mathrm{~m}$ and a width of $4 \mathrm{~m}$. The four inner squares have a side length of $1 \mathrm{~m}$. What is the area of the region she painted in gray?

b) Adriana enjoyed the game and decided to paint, in a similar way, the front of another rectangular wall, which is $1 \mathrm{~m}$ high and $32 \mathrm{~m}$ wide. She divides this rectangular wall into 32 squares with a side length of $1 \mathrm{~m}$ and then draws a diagonal. She then paints the regions in an alternating pattern, as in the previous wall. Below is the leftmost part of the wall:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-07.jpg?height=425&width=1222&top_left_y=1324&top_left_x=517)

What is the area of the wall that is painted gray? (The total area painted gray, not just the part of the wall shown above).

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Solution

a) The diagonal of the rectangle bisects the segment $J H$ at its midpoint, the point $I$, as shown in the following figure. In addition, points $B$ and $P$ are the midpoints of the sides $A C$ and $O Q$ of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-28.jpg?height=371&width=1100&top_left_y=797&top_left_x=381)

Also, observe that $F$ is the midpoint of $E G$ and $M$ is the midpoint of $L N$. Therefore, the lengths of segments $G F, F E, L M$, and $M N$ are all equal to $\frac{1}{4} \mathrm{~m}$. There are several ways to calculate the areas of the four gray regions (which are two triangles and two trapezoids). We will show a slightly different method from the usual ones.

By sliding the trapezoid $F D J I$ up onto the triangle $A G F$, and sliding the triangle $M N Q$ up onto the trapezoid $H K M I$, we obtain the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-28.jpg?height=271&width=1025&top_left_y=1766&top_left_x=413)

where we observe two rectangles with a base of $1 \mathrm{~m}$ and a height of $\frac{3}{4} \mathrm{~m}$. Therefore, the total area of the gray regions is, in square meters, equal to $2 \times \frac{3}{4}=1.5$.

b) Let's apply the same idea as in the previous question. First, we divide the squares into pairs. The first and second, then the third and fourth, then the fifth and sixth, and so on. Next, we take each gray figure in an even-ordered square and slide it to the left onto the gray figure of its left neighboring square. For example, the gray figure in the second square is slid to the left and placed on top of the gray figure in the first square, forming the figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-29.jpg?height=405&width=682&top_left_y=237&top_left_x=790)

The gray region now forms a rectangle with a base of $1 \mathrm{~m}$. Let's determine the height of this rectangle. Since the wall is $32 \mathrm{~m}$ wide and $1 \mathrm{~m}$ high, and each square inside the wall has a side of $1 \mathrm{~m}$, we have the figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-29.jpg?height=448&width=1453&top_left_y=901&top_left_x=404)

By similarity of triangles, $\frac{x}{1}=\frac{1}{32} \mathrm{~m}$.

Therefore, $x=\frac{1}{32}$. Thus, the height of the rectangle formed by the gray regions in the first square is, in meters, equal to $1-\frac{1}{32}=\frac{31}{32}$. Since there are 32 squares, we will have $\frac{32}{2}=16$ gray rectangles, all identical (one in the first square, another in the third square, another in the fifth, etc.). Therefore, the total area of the wall painted gray will be, in square meters, equal to $\frac{31}{32} \times 16=\frac{31}{2}$.