\(\mathrm{Kè} \mathrm{IH} \perp \mathrm{AC}, \mathrm{IK} \perp \mathrm{BD} \Rightarrow \mathrm{HA}=\mathrm{HC}=\frac{1}{2} \mathrm{AC}\) và \(\mathrm{KB}=\mathrm{KD}=\frac{1}{2} \mathrm{BD}\)

\(\triangle \mathrm{AIH}\) có \(\mathrm{AH}^{2}=\mathrm{R}^{2}-\mathrm{IH}^{2}=4 \mathrm{a}^{2}-\mathrm{IH}^{2} \Rightarrow \mathrm{AC}^{2}=16 \mathrm{a}^{2}-4 \mathrm{IH}^{2}\)

\(\triangle B I K\) có \(B K^{2}=R^{2}-I K^{2}=4 a^{2}-\mathrm{IK}^{2} \Rightarrow \mathrm{BD}^{2}=16 \mathrm{a}^{2}-4 \mathrm{IK}^{2}\)

\(IHMK\) là hình chữ nhật ( 3 góc vuông) \(\Rightarrow \mathrm{IH}^{2}+\mathrm{IK}^{2}=\mathrm{IM}^{2}=\mathrm{a}^{2}\)

\(\begin{array}{l}\mathrm{AC}^{2}+\mathrm{BD}^{2}=32 \mathrm{a}^{2}-4\left(\mathrm{IH}^{2}+\mathrm{IK}^{2}\right)=32 \mathrm{a}^{2}-4 \mathrm{a}^{2}=28 \mathrm{a}^{2} \\\mathrm{~S}_{\mathrm{ABCD}}=\frac{1}{2} \cdot \mathrm{AC} \cdot \mathrm{BD} \leq \frac{\mathrm{AC}^{2}+\mathrm{BD}^{2}}{4}=\frac{28 \mathrm{a}^{2}}{4}=7 \mathrm{a}^{2}\end{array}\)

\(\Rightarrow \text{Max}\left(\mathrm{S}_{\mathrm{ABCD}}\right)=7 \mathrm{a}^{2}\) khi \(\mathrm{AC}=\mathrm{BD}\) và hai dây cách tâm \(\mathrm{I}\) một khoảng \(\mathrm{IH}=\mathrm{IK}=\frac{\sqrt{2}}{2} \mathrm{a}\)

Vậy : \(\text{Max}\left(S_{A B C D}\right)=7 a^{2}\).