The figure is a paper cutout formed by a square with sides measuring $10 \mathrm{~cm}$ and two isosceles triangles (isosceles triangles are triangles with two sides of the same length).

![](https://cdn.mathpix.com/cropped/2024_05_01_3fc4f3a6ef7302a792fcg-20.jpg?height=436&width=854&top_left_y=533&top_left_x=477)

a) By folding one of the triangles over the square, as shown in the figure, what is the area of the square that remains visible?

![](https://cdn.mathpix.com/cropped/2024_05_01_3fc4f3a6ef7302a792fcg-20.jpg?height=439&width=592&top_left_y=1269&top_left_x=611)

b) By folding both triangles over the square, as shown in the figure, what is the area of the square that remains visible?

![](https://cdn.mathpix.com/cropped/2024_05_01_3fc4f3a6ef7302a792fcg-20.jpg?height=280&width=293&top_left_y=2004&top_left_x=784)

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Solution

a) The visible area is the difference between the area of the square and the area of the triangle, that is, $10^{2}-\frac{10 \cdot 10}{2}=50 \mathrm{~cm}^{2}$.

b) Let's divide the figure into triangles of the same area, and for this, it is enough to draw a segment joining the midpoints of the vertical sides.

![](https://cdn.mathpix.com/cropped/2024_05_01_3fc4f3a6ef7302a792fcg-21.jpg?height=296&width=289&top_left_y=719&top_left_x=987)

The figure has been divided into 8 triangles of the same area, each with an area of $\frac{100}{8}=12.5 \mathrm{~cm}^{2}$. Therefore, the visible area is $2 \cdot 12.5=25 \mathrm{~cm}^{2}$.