## Task 2 - 300822

A rectangle whose side lengths are in the ratio $1: 2$ is to be divided into eight congruent isosceles right triangles.

a) Draw and describe such a division! Explain why the eight triangles formed according to your description are isosceles right triangles and congruent to each other!

b) Determine the length of a leg of these triangles in terms of the smaller of the two side lengths of the rectangle!
a) The figure shows a possible decomposition of the required type for a rectangle $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0802.jpg?height=237&width=332&top_left_y=1600&top_left_x=862)

Description: If $E, F$ are the midpoints of $A B$ and $C D$ respectively, the decomposition is created by drawing the segments $E F, A F, D E, B F$, and $C E$.

Justification: Since $A B C D$ is a rectangle and $E, F$ bisect the segments $A B$ and $C D$ respectively, it follows from the condition $A D: A B = 1: 2$.

The segments $A E$ and $D F$ are of equal length and parallel to each other, and the segments $A E$ and $A D$ are of equal length and perpendicular to each other; thus, $A E F D$ is a square.

Similarly, it follows that $E B C F$ is a square. Each of these squares is divided into four triangles by the drawn diagonals.

Since in each square the diagonals are of equal length and perpendicular to each other and bisect each other, all eight resulting triangles are isosceles right triangles with equal leg lengths, and thus also congruent to each other.

b) In dependence on $a = A D$, the square $A E F D$ has an area of $a^{2}$. Each of its four mutually (congruent, thus) area-equal sub-triangles therefore has an area of $\frac{1}{4} a^{2}$.

If $x$ is the sought leg length of these triangles, then the area is also equal to $\frac{1}{2} x^{2}$. Therefore, we have

$$
\frac{1}{2} x^{2} = \frac{1}{4} a^{2} \quad ; \quad x = \sqrt{\frac{1}{2} a^{2}} = \frac{1}{2} a \sqrt{2}
$$