## Task 4 - 240814

From three congruent isosceles right triangles with given leg length $a$, a trapezoid can be formed.

a) Draw such a trapezoid!

b) Such a trapezoid can be divided into four congruent trapezoids. Draw such a division!

c) Determine the lengths of the parallel sides, the height, and the area of one of these smaller trapezoids in terms of $a$!
Possible solutions to (a) and (b) are shown in the figure:
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0748.jpg?height=292&width=988&top_left_y=1276&top_left_x=540)

(c) Each of the three isosceles right triangles with leg length \(a\) has an area of \(\frac{1}{2} a^{2}\). The trapezoid composed of these three triangles consequently has an area of \(\frac{3}{2} a^{2}\).

Each of the four (congruent, and thus equal in area) sub-trapezoids therefore has an area of \(\frac{3}{8} a^{2}\).

If \(ABCD\) is the composite trapezoid as shown in the right figure, and \(E\) is the midpoint of \(CD\), then \(DA\) and \(EC\) are each a parallel side in one of the sub-trapezoids, and \(DE\) is the height of one sub-trapezoid.

Therefore, the lengths of the parallel sides \(DA = a\) and \(EC = \frac{a}{2}\), and the length of the height is \(DE = \frac{a}{2}\).

The desired area \(F\) can also be found from these lengths:

\[
F = \frac{1}{2} \left(a + \frac{a}{2}\right) \cdot \frac{a}{2} = \frac{3}{8} a^{2}
\]

Solutions from the first round of 1984 adopted from [5]

### 5.26.2 Second Round 1984, Class 8