4. In a convex quadrilateral $A B C D$, the segments connecting the midpoints of opposite sides have lengths 2 and 3 and form an angle of $45^{\circ}$. Calculate the area of quadrilateral $A B C D$.
4. Let the midpoints of sides $AB$, $BC$, $CD$, $DA$ be $M$, $N$, $P$, $Q$, respectively, and let the segments $MP$ (length 2) and $NQ$ (length 3) intersect at point $O$. Then $MN$, $NP$, $PQ$, $QM$ are the midlines of $\triangle ABC$, $\triangle BCD$, $\triangle CDA$, and $\triangle DAB$, respectively, so we have $P(\triangle MBN) = \frac{1}{4} P(\triangle ABC)$, $P(\triangle NCP) = \frac{1}{4} P(\triangle BCD)$, $P(\triangle PDQ) = \frac{1}{4} P(\triangle CDA)$, and $P(\triangle QAM) = \frac{1}{4} P(\triangle DAB)$. By summing these four equalities, we get

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-47.jpg?height=324&width=586&top_left_y=1501&top_left_x=916)

Ok 2019 3B 4

\[
\begin{aligned}
& P(\triangle MBN) + P(\triangle NCP) + P(\triangle PDQ) + P(\triangle QAM) \\
& \quad = \frac{1}{4} (P(\triangle ABC) + P(\triangle CDA) + P(\triangle BCD) + P(\triangle DAB)) \\
& \quad = \frac{1}{4} \cdot 2 P(ABCD) = \frac{1}{2} P(ABCD)
\end{aligned}
\]

From this, it follows that $P(MNPQ) = \frac{1}{2} P(ABCD)$. Also, due to the observed midlines, we have $MN \parallel AC \parallel QP$ and $NP \parallel BD \parallel MQ$, so $MNPQ$ is a parallelogram. Let $h$ be the height from point $P$ to the diagonal $NQ$. Then, due to the $45^\circ$ angle, we have $h = \frac{PO}{\sqrt{2}} = \frac{\frac{MP}{2}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ (the middle equality is obtained based on the fact that the diagonals of a parallelogram bisect each other). From this, we calculate $P(MNPQ) = 2 P(\triangle NPQ) = 2 \cdot \frac{QN \cdot h}{2} = 3 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, and finally $P(ABCD) = 2 P(MNPQ) = 3 \sqrt{2}$.