10.3. In the quadrilateral $A B C D$, $A B=2, B C=4, C D=5$. Find its area, given that it is both inscribed and circumscribed.
Answer: $2 \sqrt{30}$.

Solution. Since the quadrilateral $A B C D$ is circumscribed, then $A B+C D=$ $B C+A D$, from which $A D=3$. By the cosine theorem from triangle $A B D$ we get $B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos A$, and from triangle $B C D$

$$
B D^{2}=B C^{2}+C D^{2}-2 B C \cdot C D \cos C
$$

Since the quadrilateral $A B C D$ is inscribed, then $\angle A+\angle C=180^{\circ}$, and $\cos A=-\cos \mathrm{C}$.

Therefore,

$$
2^{2}+3^{2}+2 \cdot 2 \cdot 3 \cos C=4^{2}+5^{2}-2 \cdot 4 \cdot 5 \cos C
$$

From which $\cos \mathrm{C}=\frac{7}{13}$, and $\sin \mathrm{C}=\sqrt{1-\cos ^{2} \mathrm{C}}=\frac{2 \sqrt{30}}{13}$. Since $\sin A=\sin C$, then

$$
S_{A B C D}=S_{A B D}+S_{B C D}=\frac{1}{2} A B \cdot A D \cdot \sin A+\frac{1}{2} B C \cdot C D \cdot \sin C=2 \sqrt{30}
$$