10.8. Find the maximum possible area of the quadrilateral whose side lengths are equal to $1 \mathrm{~cm}$, $2 \sqrt{2} \mathrm{~cm}$, $3 \mathrm{~cm}$, and $4 \mathrm{~cm}$.
Solution: It is evident that the quadrilateral of maximum area is a convex quadrilateral.

Let $ABCD$ be a convex quadrilateral. If we change the configuration of the quadrilateral by only swapping two adjacent sides, the area of the quadrilateral does not change. Therefore, we can consider that the sides are arranged in increasing order of their lengths.

$A_{ABCD} = A_{ABC} + A_{ADC} = \frac{1}{2} \cdot 2\sqrt{2} \cdot 3 \cdot \sin(\angle B) + \frac{1}{2} \cdot 1 \cdot 4 \cdot \sin(\angle D) \leq 3\sqrt{2} + 2$.

If we deform the quadrilateral so that the side of length 1

![](https://cdn.mathpix.com/cropped/2024_06_06_cc93ce82125d03b91556g-2.jpg?height=299&width=423&top_left_y=1821&top_left_x=1426)
is perpendicular to the side of length 4, then the side of length $2\sqrt{2}$ will also be perpendicular to the side of length 3, because $1^2 + 4^2 = 17 = (2\sqrt{2})^2 + 3^2$. Then the area of the quadrilateral will be $\frac{1}{2} \cdot 1 \cdot 4 + \frac{1}{2} \cdot 2\sqrt{2} \cdot 3 = 2 + 3\sqrt{2}$. Therefore, the maximum area is $2 + 3\sqrt{2} \, \text{cm}^2$.

Answer: $2 + 3\sqrt{2} \, \text{cm}^2$.