10.4. Let point $P$ be the intersection of the diagonals of a convex quadrilateral $K L M N$. The areas of triangles $K L M, L M N$, and $N K P$ are $8 \mathrm{~m}^{2}, 9 \mathrm{~m}^{2}$, and $10 \mathrm{~m}^{2}$, respectively. Find the area of quadrilateral $K L M N$.
# Solution:

Let $S_{P L M}=x$, and $\angle N P M=\angle K P L=\alpha$, then $S_{K L P}=8-x, S_{N P M}=9-x, \quad \angle K P N=\angle L P M=180^{\circ}-$

$$
\left\{\begin{array}{l}
x=\frac{1}{2} \sin \left(180^{\circ}-\alpha\right) P L \cdot P M \\
9-x=\frac{1}{2} \sin \alpha \cdot P M \cdot P N \\
10=\frac{1}{2} \sin \left(180^{\circ}-\alpha\right) P N \cdot P K \\
8-x=\frac{1}{2} \sin \alpha \cdot P K \cdot P L
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0a78038af02d099f4bb0g-2.jpg?height=342&width=417&top_left_y=600&top_left_x=1388)

$\alpha$.

By multiplying the corresponding parts of equations (1) and (3), and (2) and (4), considering that $\sin \alpha=\sin \left(180^{\circ}-\alpha\right)$, we get: $10 x=\frac{1}{2} \sin ^{2}\left(180^{\circ}-\alpha\right) P L \cdot P M \cdot P N \cdot P K=$ $\frac{1}{2} \sin ^{2} \alpha \cdot P L \cdot P M \cdot P N \cdot P K=(9-x)(8-x)$.

This relationship can be derived through similarity.

From which $x^{2}-27 x-72=0$. Solving the quadratic equation: $x_{1}=$ 24 (which is impossible, as $8-x>0$) and $x_{2}=3$.

$$
S_{K L M N}=x+(9-x)+(8-x)+10=3+6+5+10=24
$$

Answer: $S_{K L M N}=24 \mathrm{m}^{2}$.

| Criteria | points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| The area of one of the internal triangles is found, but the area of the entire quadrilateral is not calculated. | 5 |
| The equation is justified but not solved. | 3 |
| Incorrect solution. | 0 |