In a convex quadrilateral, two adjacent sides, $a=28.29 \mathrm{~cm}$ and $b=11.71 \mathrm{~cm}$, are perpendicular to each other. The other two sides are equal and also form right angles. Calculate the area of the quadrilateral.
Our quadrilateral consists of two right-angled triangles, whose common hypotenuse is one of the diagonals of the quadrilateral. The legs of one right-angled triangle are $a$ and $b$, and the other right-angled triangle, according to the problem, is isosceles, with its legs denoted by $x$.

The area of the quadrilateral is therefore

$$
t=\frac{x^{2}}{2}+\frac{a b}{2}
$$

but according to Pythagoras' theorem, the square of the common hypotenuse is:

$$
2 x^{2}=a^{2}+b^{2}
$$

and thus the sought area is

$$
t=\frac{2 x^{2}+2 a b}{4}=\frac{a^{2}+b^{2}+2 a b}{4}=\left(\frac{a+b}{2}\right)^{2}=\left(\frac{40}{2}\right)^{2}=400 \mathrm{~cm}^{2}
$$

József Fischheller (Székesféhérvár, József A. g. I. o. t)

Remark: This problem is a school example of how much more advantageous it can be to solve a problem generally first, and only substitute the specific numerical values at the end, rather than calculating intermediate results with specific numerical values. (The above procedure is still the most advantageous even if $a+b$ is not a round number.) Unfortunately, the majority of our solvers calculated more clumsily, as can be seen from the following scoring.