2. Let $P$ be the intersection of the diagonals of a convex quadrilateral $A B C D$. Calculate its area if the areas of triangles $A B C, B C D$ and $D A P$ are $8 \mathrm{~cm}^{2}, 9 \mathrm{~cm}^{2}$, $10 \mathrm{~cm}^{2}$, respectively.
2. Let $S_{XYZ}$ denote the area of triangle $XYZ$ in $\mathrm{cm}^{2}$, and let $S=S_{ABP}$. According to the problem, we have $S_{ADP}=10$, $S+S_{BCP}=8$, and $S_{BCP}+S_{CDP}=9$. From the second equation, it follows that $S_{BCP}=8-S$. Substituting this into the third equation, we get $S_{CDP}=1+S$ (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_04_17_e205f11fab915c7d36c0g-2.jpg?height=342&width=758&top_left_y=1885&top_left_x=658)

Fig. 1

Triangles $ABP$ and $ADP$ have the same height from vertex $A$. Therefore, the ratio of their areas is $S: S_{ADP} = |BP|:|DP|$. Similarly, for triangles $BCP$ and $CDP$, we get $S_{BCP}: S_{CDP} = |BP|:|DP|$. Hence, it follows that $S: S_{ADP} = S_{BCP}: S_{CDP}$, which, given the derived relationships, means

$$
\frac{S}{10} = \frac{8-S}{1+S}
$$

After rearranging, we get a quadratic equation for $S$:

$$
S^2 + 11S - 80 = (S + 16)(S - 5) = 0,
$$

which has two roots, -16 and 5. Since the area $S$ of triangle $ABP$ is a non-negative number, only $S=5$ is valid. From the above relationships, we can easily calculate $S_{BCP}=3$ and $S_{CDP}=6$. The area of the quadrilateral $ABCD$ in $\mathrm{cm}^{2}$ is therefore

$$
S + S_{BCP} + S_{CDP} + S_{ADP} = 5 + 3 + 6 + 10 = 24.
$$

Conclusion. The area of quadrilateral $ABCD$ is $24 \mathrm{~cm}^{2}$.

Alternative Solution. Let the lengths of segments $PA, PB, PC, PD$ (in $\mathrm{cm}$) be denoted by $a, b, c, d$ respectively. According to the known formula $S = \frac{1}{2}uv \sin \omega$ for the area of a triangle with sides of lengths $u$ and $v$ forming an angle of size $\omega$, we express the conditions of the problem with the equations

$$
8 = \frac{1}{2}ab \sin \varphi + \frac{1}{2}bc \sin (\pi - \varphi), \quad 9 = \frac{1}{2}bc \sin (\pi - \varphi) + \frac{1}{2}cd \sin \varphi, \quad 10 = \frac{1}{2}da \sin (\pi - \varphi),
$$

where $\varphi = |\angle APB|$. For simplicity, let $k = \frac{1}{2} \sin \varphi = \frac{1}{2} \sin (\pi - \varphi)$. We then get the system of equations

$$
8 = kab + kbc, \quad 9 = kbc + kcd, \quad 10 = kda,
$$

from which we will find the area $S$ of quadrilateral $ABCD$, which in $\mathrm{cm}^{2}$ is given by the formula

$$
S = kab + kbc + kcd + kda.
$$

First, we eliminate $d$ by a suitable combination of the second and third equations (1):

$$
9a - 10c = kab c.
$$

By combining this equation with the first equation (1), whose right-hand side we rewrite as the product $k b(a + c)$, we eliminate $b$:

$$
(9a - 10c)(a + c) - 8ac = 0.
$$

Using the identity

$$
(9a - 10c)(a + c) - 8ac = 9a^2 - 9ac - 10c^2 = (3a - 5c)(3a + 2c),
$$

we see that for (positive) numbers $a, c$, we have $3a = 5c$ or $9a = 15c$, and thus equation (3) can be rewritten as $5c = kab c$, from which $kab = 5$. From equations (1), it follows that $kbc = 3$ and $kcd = 6$, which, when substituted into (2) along with the value $kda = 10$, leads to the result $S = 24$.

For a complete solution, award 6 points, of which 1 point for expressing the area using one of the three triangles $ABP, BCP$, or $CDP$, 2 points for justifying the proportionality between the areas of the four triangles, 1 point for setting up the corresponding equation, and 1 point for solving it. If the solver correctly calculates the area only for a specially chosen suitable quadrilateral $ABCD$ (e.g., with perpendicular diagonals), award no more than 2 points.