A rectangular sheet of $20 \mathrm{~cm}$ by $30 \mathrm{~cm}$ was cut along the dashed lines $A C$ and $B D$ into four pieces: two equal triangles and two equal five-sided polygons, as shown in Figure I. The segments $A C$ and $B D$ have the same length and intersect at the center of the rectangle forming right angles.
![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-11.jpg?height=344&width=792&top_left_y=1284&top_left_x=744)

a) What is the length of the segment $A B$?

b) What is the area of one triangular piece? And of one five-sided piece?

c) With the four pieces, we can assemble a square with a rectangular hole, as shown in Figure

II. What is the area of the hole?
Solution

a) Let's represent the original sheet by the rectangle $P Q R S$ in the figure below. Let $M$ be the point where the segments $A C$ and $B D$ meet. Since the center of the rectangle is the center of symmetry of the figure, we conclude that $A M = M C = \frac{1}{2} A C$. On the other hand, we know that $A C = B D$, so $A M = B M = C M = D M$. Since the angles with vertex at $M$ are all right angles, the triangles $A M B, B M C, C M D$, and $D A M$ are congruent and, in particular, $A B = B C = C D = D A$ and the angles of these triangles at $A, B, C$, and $D$ are equal, so $A B C D$ is a square. Since $B P C Q$ is a rectangle, $B C = P Q = 20 \mathrm{~cm}$, so $A B = 20 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-33.jpg?height=389&width=525&top_left_y=2284&top_left_x=868)

b) The area of each of the triangles $A M B, B M C, C M D$, and $D A M$ is equal to $\frac{1}{4}$ of the area of the square $A B C D$, which is $20 \mathrm{~cm} \times 20 \mathrm{~cm} = 400 \mathrm{~cm}^{2}$; thus, the area of one of these triangles is $\frac{400}{4} = 100 \mathrm{~cm}^{2}$. Since the two five-sided pieces are equal, they have the same area. The original sheet has an area of $20 \times 30 = 600 \mathrm{~cm}^{2}$, and if we subtract from this area the areas of the two triangular pieces $A B M$ and $D M C$, the remaining area will be the area of the two five-sided pieces. Therefore, the area of each five-sided piece, in square centimeters, is equal to $\frac{600 - 2 \times 100}{2} = \frac{600 - 200}{2} = \frac{400}{2} = 200$.

Another solution for obtaining the area of the triangle: The base $A B$ of the triangle $A B M$ measures $20 \mathrm{~cm}$; the height relative to this base is half the height of the sheet, that is, $\frac{20 \mathrm{~cm}}{2} = 10 \mathrm{~cm}$. Therefore, the area of each of the two triangles is $\frac{20 \mathrm{~cm} \times 10 \mathrm{~cm}}{2} = 100 \mathrm{~cm}^{2}$.

Another solution for obtaining the area of the five-sided polygon: Each five-sided piece is formed by one of the four triangles above and a rectangle with a height of $20 \mathrm{~cm}$ and a width equal to $\frac{20 \mathrm{~cm} - 10 \mathrm{~cm}}{2} = \frac{10 \mathrm{~cm}}{2} = 5 \mathrm{~cm}$. Since the area of each triangle is $100 \mathrm{~cm}^{2}$ and the area of the rectangle is equal to $5 \times 20 \mathrm{~cm}^{2} = 100 \mathrm{~cm}^{2}$, we conclude that the area of each five-sided piece is equal to $100 \mathrm{~cm}^{2} + 100 \mathrm{~cm}^{2} = 200 \mathrm{~cm}^{2}$.

c) $1^{\text{st}}$ solution: The square formed by the four pieces and the hole has an area equal to 8 times the area of each triangular piece, as shown in the drawing below. Therefore, its area is equal to $8 \times 100 \mathrm{~cm}^{2} = 800 \mathrm{~cm}^{2}$. Since the sum of the areas of the four pieces is equal to the area of the original sheet, that is, $600 \mathrm{~cm}^{2}$, we conclude that the area of the hole is equal to $800 \mathrm{~cm}^{2} - 600 \mathrm{~cm}^{2} = 200 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-34.jpg?height=382&width=371&top_left_y=1017&top_left_x=748)

$2^{\text{nd}}$ solution: The hole is a rectangle whose height is equal to the height of the original sheet, that is, $20 \mathrm{~cm}$. Its length is the difference between the length of the original sheet and the segment $A B$, that is, $30 \mathrm{~cm} - 20 \mathrm{~cm} = 10 \mathrm{~cm}$. Therefore, the area of the hole is $20 \mathrm{~cm} \times 10 \mathrm{~cm} = 200 \mathrm{~cm}^{2}$.

$3^{\text{rd}}$ solution: Each right-angled triangle is isosceles with a hypotenuse of $20 \mathrm{~cm}$. If $a$ is the measure, in centimeters, of one of the legs, we have $20^{2} = a^{2} + a^{2} = 2 a^{2}$, so $a = \sqrt{20^{2} / 2} = \sqrt{200} = 10 \sqrt{2}$. Thus, the large square has a side equal to $10 \sqrt{2} \mathrm{~cm} + 10 \sqrt{2} \mathrm{~cm} = 20 \sqrt{2} \mathrm{~cm}$ and its area is $(20 \sqrt{2} \mathrm{~cm})^{2} = 800 \mathrm{~cm}^{2}$. Since the sum of the areas of the four pieces is equal to the area of the original sheet, that is, $600 \mathrm{~cm}^{2}$, we conclude that the area of the hole is equal to $800 \mathrm{~cm}^{2} - 600 \mathrm{~cm}^{2} = 200 \mathrm{~cm}^{2}$.