設圓之半徑為 $R = \dfrac{7}{2}$。設 $\overline{AC}$ 與 $\overline{BD}$ 交於 $P$ 點，且 $\overline{AC} \perp \overline{BD}$。 由圓內垂直相交弦性質知：對應弧長滿足 $\text{arc}(AB) + \text{arc}(CD) = 180^\circ$。 設 $\angle AOB = \alpha$，$\angle COD = \beta$，則 $\dfrac{\alpha}{2} + \dfrac{\beta}{2} = 90^\circ \implies \beta = 180^\circ - \alpha$。 弦長公式：$\overline{AB} = 2R \sin\dfrac{\alpha}{2} = 5$ $\overline{CD} = 2R \sin\dfrac{\beta}{2} = 2R \sin(90^\circ - \dfrac{\alpha}{2}) = 2R \cos\dfrac{\alpha}{2}$ 由於 $\sin^2\dfrac{\alpha}{2} + \cos^2\dfrac{\alpha}{2} = 1$，得： $$\left( \dfrac{5}{2R} \right)^2 + \left( \dfrac{\overline{CD}}{2R} \right)^2 = 1 \implies 5^2 + \overline{CD}^2 = (2R)^2$$ 代入 $2R = 7$： $$25 + \overline{CD}^2 = 49 \implies \overline{CD}^2 = 24 \implies \overline{CD} = \sqrt{24} = 2\sqrt{6}$$ 故填 $2\sqrt{6}$。