2. The sides of a convex quadrilateral are equal to $6, 7, 8, 9$ in some order. It is known that a circle can be inscribed in this quadrilateral and a circle can be circumscribed around it. Find the area of the quadrilateral.
Solution. A circle can be inscribed in a convex quadrilateral $ABCD$ if and only if $AB + CD = BC + AD$. Therefore, without loss of generality, we can assume that $AB = 6, BC = 8, CD = 9, AD = 7$. By the cosine rule,

$$
BD^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cdot \cos A, \quad BD^2 = 8^2 + 9^2 - 2 \cdot 8 \cdot 9 \cdot \cos C
$$

$\cos C = -\cos A$, since $\angle A + \angle C = 180^\circ$. Therefore,

$$
85 - 84 \cos A = 145 + 144 \cos A, \quad \cos A = -\frac{5}{19}
$$

Then $\sin A = \sin C = \sqrt{1 - \frac{25}{361}} = \frac{4 \sqrt{21}}{19}$.

$$
S_{ABCD} = \frac{1}{2}(AB \cdot AD + BC \cdot CD) \sin A = 12 \sqrt{21}
$$