. Cho hình chóp S.ABCD có SA vuông góc với đáy, $SA=a\sqrt{6}.$...

Ta có: $\text{CE // AB}\Rightarrow \text{CE}\bot A\text{D}$
Mặt khác $CE\bot \text{S}A\Rightarrow CE\bot \left( SE\text{D} \right)$
$\Rightarrow {{R}_{C.SE\text{D}}}=\sqrt{\dfrac{C{{E}^{2}}}{4}+{{\left( {{R}_{S\text{D}E}} \right)}^{2}}}$
Lại có $CE=AB=a,\sin \widehat{SE\text{A}}=\sin \widehat{SE\text{D}}$
$=\dfrac{SA}{SE}=\dfrac{a\sqrt{6}}{\sqrt{{{a}^{2}}+6{{\text{a}}^{2}}}}=\dfrac{a\sqrt{6}}{\sqrt{7}}$
$\Rightarrow {{R}_{SE\text{D}}}=\dfrac{S\text{D}}{2\sin \widehat{SE\text{D}}}=\dfrac{a\sqrt{10}}{2.\dfrac{a\sqrt{6}}{\sqrt{7}}}=\dfrac{a\sqrt{105}}{6}$
Vậy ${{R}_{S.C\text{D}E}}=a\sqrt{\dfrac{19}{6}}$.
Cách 2: Do $\left( SE\text{D} \right)\bot \left( CE\text{D} \right)\Rightarrow R=\sqrt{R_{1}^{2}+R_{2}^{2}-\dfrac{G{{T}^{2}}}{4}}$ trong đó ${{R}_{1}}={{R}_{SE\text{D}}}=\dfrac{a\sqrt{105}}{6}$,
${{R}_{2}}={{R}_{CE\text{D}}}=\dfrac{C\text{D}}{2}=\dfrac{a\sqrt{2}}{2}$ và $GT=E\text{D}=a\Rightarrow R=a\sqrt{\dfrac{19}{6}}$.