The square $A B C D$ is divided into 6 isosceles right triangles as indicated in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-13.jpg?height=608&width=661&top_left_y=1430&top_left_x=777)

If the area of the shaded triangle is 2, calculate the area of the square.

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Solution

From the information given in the problem, the triangle $P Q R$ in the figure has an area of 2.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-37.jpg?height=505&width=523&top_left_y=424&top_left_x=648)

However, since it is a right triangle, the area of triangle $P Q R$ can be calculated as $P R \times Q R / 2$. Since $P R = Q R$, we conclude that $P R = Q R = 2$. Now, we know that $P R S$ is also an isosceles right triangle, and therefore $R S = P R = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-37.jpg?height=505&width=526&top_left_y=1223&top_left_x=646)

Since the right triangle $Q S D$ is isosceles, we can conclude that $S D = Q S = 2 + 2 = 4$. On the other hand, since triangles $P Q R$ and $A P S$ are isosceles right triangles, then $\measuredangle P A S = \measuredangle P Q R = 45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-37.jpg?height=500&width=526&top_left_y=1981&top_left_x=646)

This implies that triangle $A Q S$ is isosceles, and therefore $A S = S Q = 4$. Up to this point, we have that the side of the square measures $A D = A S + S D = 8$. The area of the square is therefore $8^{2} = 64$.