Let's modify the convex quadrilateral $ABCD$ from the following data: $\angle DAB = 105^{\circ}, AB = 5 \text{~cm}, BC = 22 \text{~cm}$, $CD = 4 \text{~cm}, \angle CDA = 165^{\circ}$. Calculate the area of the quadrilateral.
1. Notice that the sum of the exterior angles at vertices $A$ and $D$ of the quadrilateral $ABCD$ is $75^{\circ} + 15^{\circ} = 90^{\circ}$, and thus the extensions of sides $AB$ and $CD$ form a right triangle. Let's denote the vertex of the right angle in this triangle as $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a3ba084b3d66dbbf72aag-1.jpg?height=602&width=1008&top_left_y=295&top_left_x=556)

The construction can be carried out as follows based on this. On the segment $AB$, measure the angle $BAD$ at point $A$, and draw a perpendicular to the segment at point $A$ (on the same side) with a length equal to $DC$. Translate the segment $AC'$ parallel to the side $AD$ until it intersects the circle centered at $B$ with radius $BC$, where we find the vertex $C$ of the quadrilateral. Finally, point $D$ is determined by the perpendicular from $C$ to $AB$. $C$ is formed because the parallel line intersects the circle at a point inside the circle.

2. The area of the quadrilateral $ABCD$ can be calculated if we complete it to form a right trapezoid by drawing perpendiculars from $C$ and $B$ to the line containing side $AD$. Let the feet of these perpendiculars be denoted by $N$ and $M$, respectively. The area of quadrilateral $ABCD$ is obtained by subtracting the sum of the areas of triangles $CND$ and $BMA$ from the area of trapezoid $CNMB$.

The altitude of triangle $CND$ to side $CD$ can be easily calculated. Let the foot of this altitude be $N'$ and the midpoint of side $CD$ be $F$. Since the triangle is right-angled, $FN = FC$, $\angle FDN = \angle DNF = 15^{\circ}$, $\angle FN'F = 30^{\circ}$, and thus $\angle FNN' = 60^{\circ}$, which is half of an equilateral triangle:

$$
\begin{gathered}
FN = 2, \quad NN' = 1, \quad N'F = \sqrt{3} \\
CND_{\text{area}} = \frac{4 \cdot 1}{2} = 2 \text{ cm}^2
\end{gathered}
$$

Since triangle $BMA$ is a $5:4$ enlargement of triangle $DNC$, its area is

$$
\begin{gathered}
\text{area } \left(\frac{5}{4}\right)^2 \text{ times larger, i.e.,} \\
BMA_{\text{area}} = \left(\frac{5}{4}\right)^2 \cdot 2 = \frac{25}{8} = 3.125 \text{ cm}^2
\end{gathered}
$$

To calculate the area of trapezoid $CNMB$, we first determine the necessary dimensions. To find the lengths of $CN$, $BM$, and $NM$, we use the Pythagorean theorem:

$$
CN = \sqrt{NN'^2 + N'C^2} = \sqrt{1^2 + (2 - \sqrt{3})^2} = 2 \sqrt{2 - \sqrt{3}}
$$

By the similarity mentioned, given $DN = 2 \sqrt{2 + \sqrt{3}}$,

$$
\begin{gathered}
BM = \frac{5}{2} \sqrt{2 + \sqrt{3}} \text{ and } \\
NM = \sqrt{BC^2 - (BM - CN)^2} = \sqrt{22^2 - (2.5 \sqrt{2 + \sqrt{3}} - 2 \sqrt{2 - \sqrt{3}})^2}.
\end{gathered}
$$

Thus, the area of the trapezoid is

$$
NCBM_{\text{area}} = \frac{5}{4} (\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}}) NM = 63.56 \text{ cm}^2
$$

Considering the data as accurate and using approximate values from tables where necessary, the area of quadrilateral $ABCD$ is $58.43 \text{ cm}^2$.

Note. The area of the quadrilateral can also be calculated as the difference between the areas of the right triangles $EBC$ and $EAD$, after calculating the segment $DE = x$ from the equation

$$
(x + 4)^2 + (x \tan 15^{\circ} + 5)^2 = 22^2
$$