10. $[8] A B C D$ is a convex quadrilateral such that $A B=2, B C=3, C D=7$, and $A D=6$. It also has an incircle. Given that $\angle A B C$ is right, determine the radius of this incircle.
Answer: $\frac{1+\sqrt{13}}{3}$. Note that $A C^{2}=A B^{2}+B C^{2}=13=C D^{2}-D A^{2}$. It follows that $\angle D A C$ is right, and so
$$
[A B C D]=[A B C]+[D A C]=2 \cdot 3 / 2+6 \cdot \sqrt{13} / 2=3+3 \sqrt{13}
$$

On the other hand, if $I$ denotes the incenter and $r$ denotes the inradius,
$$
[A B C D]=[A I B]+[B I C]+[C I D]+[D I A]=A B \cdot r / 2+B C \cdot r / 2+C D \cdot r / 2+D A \cdot r / 2=9 r
$$

Therefore, $r=(3+3 \sqrt{13}) / 9=\frac{1+\sqrt{13}}{3}$.