2. $[\mathbf{2 0}]$ The four sides of quadrilateral $A B C D$ are equal in length. Determine the perimeter of $A B C D$ given that it has area 120 and $A C=10$.
Answer: $\square$ Let $M$ be the midpoint of $A C$. Then triangles $A M B, B M C, C M D$, and $D M A$ are all right triangles having legs 5 and $h$ for some $h$. The area of $A B C D$ is 120 , but also $4 \cdot\left(\frac{1}{2} \cdot 5 \cdot h\right)=10 h$, so $h=12$. Then $A B=B C=C D=D A=\sqrt{12^{2}+5^{2}}=13$, and the perimeter of $A B C D$ is 52 .