Solution. Let $MK$ and $PH$ be segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$, with $MK = PH$, $AC = 18$, and $BD = 7$.

We have: $MP \| AC$, $MP = \frac{1}{2} AC$ (as the midline of $\triangle ABC$); $HK \| AC$, $HK = \frac{1}{2} AC$ (as the midline of $\triangle ADC$). $\Rightarrow MP \| HK$, $MP = HK \Rightarrow MPKH$ is a parallelogram. 

![](https://cdn.mathpix.com/cropped/2024_05_06_8c1ab4103bc5f9777fe4g-01.jpg?height=285&width=529&top_left_y=2256&top_left_x=1363)
Since $MK = PH$, the quadrilateral $MPKH$ is a rectangle, with sides parallel to the diagonals $AC$ and $BD$ of the given quadrilateral $ABCD$, therefore $AC \perp BD$. This means that $S_{ABCD} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \cdot 18 \cdot 7 = 63$ (sq.units).

Answer: 63.

| Criterion Description | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 15 |
| The solution contains a computational error, possibly leading to an incorrect answer, but the solution has a correct sequence of all steps. | 10 |
| In the solution, one or two formulas are correctly written, which could be the beginning of a possible solution. | 5 |
| The solution does not meet any of the criteria described above. | 0 |
| Maximum score | 15 |
2. (15 points) Find the area of a convex quadrilateral with equal diagonals, if the lengths of the segments connecting the midpoints of its opposite sides are 13 and 7.