A quadrilateral $A B C D$ has the coordinates of three of its vertices as $A(0 ; 0), B(6 ;-1), C(7 ; 7)$. The lengths of the sides meeting at the fourth vertex are 5 units, and at this vertex, the angle of the quadrilateral is greater than $180^{\circ}$. What is the area? And by how much does the area increase if we only change the fact that the mentioned angle is less than $180^{\circ}$?
Let the projection of point $C$ onto the $x$-axis be denoted by $E$. Instead of calculating the area of quadrilateral $ABCD$, we will calculate the area of quadrilateral $AEC D$, which has the same area. First, however, let's prove that these two areas are indeed equal.

Draw a line through point $B$ parallel to the diagonal $AC$. This line will also be diagonal in direction, passing through the point opposite to $B$, which is $E$. The triangles $ABC$ and $AEC$ share the base $AC$, and the heights corresponding to this base are equal, as they fall between parallel lines. Therefore, the areas of the two triangles are equal, $\frac{49}{2}$ square units. This proves the equality of the areas of the two quadrilaterals.

We calculate the length of segment $AC$ using the Pythagorean theorem:

$$
\overline{AC}=\sqrt{7^{2}+7^{2}}=\sqrt{98}
$$

Since $AC < 10$, the circular arcs with radius 5 units drawn from points $A$ and $C$ indeed intersect at points $D_1$ and $D_2$, which are symmetric with respect to the line $AC$.

A concave quadrilateral is formed if the intersection point lies on the upper side of the $x$-axis relative to the line $AC$. In this case, $T_{ABCD_1}=T_{AEC}-T_{AD_1C}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_45e677b15ef981b5a95cg-1.jpg?height=794&width=718&top_left_y=779&top_left_x=690)

We still need to calculate the area of triangle $AD_1C$, where $\overline{AC}=\sqrt{98}$, $\overline{AD_1}=\overline{D_1C}=5$. The height $m$ corresponding to side $AC$ is calculated using the Pythagorean theorem.

$$
\begin{aligned}
& m=\sqrt{5^{2}-\left(\frac{\sqrt{98}}{2}\right)^{2}}=\sqrt{\frac{1}{2}} \\
& t=\frac{AC \cdot m}{2}=\frac{\sqrt{98} \cdot \sqrt{\frac{1}{2}}}{2}=\frac{7}{2}
\end{aligned}
$$

Thus, the area of the quadrilateral is

$$
\frac{49}{2}-\frac{7}{2}=21
$$

square units.

Due to symmetry, the area of triangle $AD_2C$ is equal to the area of triangle $AD_1C$, and in the convex case, the area of quadrilateral $AD_1CB$ will be exactly the area of triangle $AD_1C$ larger than the area of triangle $AEC$.

$AD_2CB$ area $=\frac{49}{2}+\frac{7}{2}=28$ square units.