The equations of the sides of a quadrilateral are:

$$
y=-x+7, \quad y=\frac{x}{2}+1, \quad y=-\frac{3}{2} x+2 \quad \text { and } \quad y=\frac{7}{4} x+\frac{3}{2} .
$$

Determine the area of the quadrilateral.
I. solution: If we successively plot the given lines, we see that the vertices of our quadrilateral are:

the intersection of the 4th and 1st line $A(2,5)$,

the intersection of the 1st and 2nd line $B(4,3)$,

the intersection of the 2nd and 3rd line $C\left(\frac{1}{2}, \frac{5}{4}\right)$

the intersection of the 3rd and 4th line $D\left(\frac{2}{13}, \frac{23}{13}\right)$

The double area of $\triangle ABC$ is

$$
2 t_{1}=\left|2\left(3-\frac{5}{4}\right)+4\left(\frac{5}{4}-5\right)+\frac{1}{2}(5-3)\right|=\left|\frac{7}{2}-15+1\right|=\left|-\frac{21}{2}\right|=\frac{21}{2}
$$

and thus

$$
t_{1}=\frac{21}{4}
$$

The double area of $\triangle ACD$ is

$$
\begin{gathered}
2 t_{2}=\left|2\left(\frac{5}{4}-\frac{23}{13}\right)+\frac{1}{2}\left(\frac{23}{13}-5\right)+\frac{2}{13}\left(5-\frac{5}{4}\right)\right|=\left|2 \cdot \frac{65-92}{52}+\frac{1}{2} \cdot \frac{23-65}{13}+\frac{2}{13} \cdot \frac{15}{4}\right|= \\
=\left|-\frac{27}{26}-\frac{42}{26}+\frac{15}{16}\right|=\left|-\frac{54}{26}\right|=\frac{27}{13}
\end{gathered}
$$

and thus

$$
t_{2}=\frac{27}{26}
$$

Therefore, the area of the quadrilateral is

$$
t=t_{1}+t_{2}=\frac{21}{4}+\frac{27}{26}=\frac{327}{52}\left(=6+\frac{15}{52} \approx 6.288\right)
$$

Zoltán Szilágyi (Győr, Révai g. IV. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_61dbba006f3a4a4afb19g-1.jpg?height=650&width=455&top_left_y=1361&top_left_x=824)

II. solution: If we denote the intersection points of the given sides of the quadrilateral with the $y$-axis as $B_{1}, B_{2}, B_{3}$, and $B_{4}$, respectively, then the area of the quadrilateral can also be obtained by subtracting the area of $\triangle B_{1} B_{4} A$ and $\triangle B_{3} B_{2} C$ from the area of $\triangle B_{1} B_{2} B$ and adding the area of $\triangle B_{3} B_{4} D$, which is the common part of the latter two triangles (which was subtracted twice). However, the area of these 4 triangles can be calculated very simply, because one side of each triangle lies on the $y$-axis, and the height corresponding to this side is the abscissa of one of the vertices (there is no need for the ordinates of the vertices), and the ordinates of the points $B_{1}, B_{2}, B_{3}$, and $B_{4}$ are the segments cut from the $y$-axis by the sides of the given quadrilateral. Therefore, the sought area is:

$$
\frac{1}{2}(7-1) 4-\frac{1}{2}\left(7-\frac{3}{2}\right) 2-\frac{1}{2}(2-1) \frac{1}{2}+\frac{1}{2}\left(2-\frac{3}{2}\right) \frac{2}{13}=12-\frac{11}{2}-\frac{1}{4}+\frac{1}{26}=\frac{327}{52}
$$

László Németh (Gyula, Erkel F. g. IV. o. t.)

Remark: No fewer than 29 solvers made a numerical calculation error.