OABC is a tetrahedron of volume $\frac{2}{3}$cubic unit. Also $\mathrm{BC}=\sqrt{2},\angle \mathrm{AOB}=\frac{\mathrm{\pi }}{4}$and OA + OB = 4, then which is/are CORRECT ?

$\begin{array}{r}\because \mathrm{OA}+\mathrm{OB}=4\\ \Rightarrow \mathrm{OA}\cdot \mathrm{OB}\le 4\end{array}$

$\therefore \mathrm{ar}\left(\mathrm{\Delta AOB}\right)=\frac{1}{2}\left(\mathrm{OA}\right)\left(\mathrm{OB}\right)\times \frac{1}{\sqrt{2}}\le \sqrt{2}$

Let h = altitude drawn from the vertex C to the base AOB.

So, $\mathrm{h}\le \sqrt{2}$

$\therefore \mathrm{V}=\frac{1}{3}\times \mathrm{ar}\left(\mathrm{AOB}\right)\times \mathrm{h}\le \frac{1}{3}\times \sqrt{2}\times \sqrt{2}=\frac{2}{3}$

But $\mathrm{V}=\frac{2}{3}\Rightarrow$equality holds everywhere

So, OA = OB = 2 & $\mathrm{h}=\sqrt{2}$

$\text{ Also, }\mathrm{BC}\perp \text{ base }\mathrm{AOB}\Rightarrow \mathrm{\angle }\mathrm{OBC}={90}^{\circ }$

So, $\mathrm{OC}=\sqrt{6}$

And $\mathrm{AB}=\sqrt{4+4-8\times \frac{1}{\sqrt{2}}}=\sqrt{8-4\sqrt{2}}$

$\therefore \mathrm{AC}=\sqrt{8-4\sqrt{2}+2}=\sqrt{10-4\sqrt{2}}$

$>\sqrt{9-4\sqrt{2}}=(2\sqrt{2}-1)$