$\square ABCD$ is a square. If its perimeter is $40$, find $AC + BD$. (in $\sqrt{2}$)

In rhombus $ABCD$,$\overline{AC} \cap \overline{BD} = \{O\}$. Then the area of $\Delta OAB = \ldots \ldots \ldots \ldots \times$ area of $\square ABCD$.

In rectangle $ABCD$,$AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$,then $AC = \ldots$

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB^2 + AC^2 = 338$ and $AD = 5$,find $BC$.

In parallelogram $ABCD$,$AB^{2} + BC^{2} = 260$ and $AC = 18$. Then $BD = \ldots$

In $\Delta ABC$,$m \angle A + m \angle C = m \angle B$ and $\overline{BM}$ is an altitude to $\overline{AC}$. If $AM = 16$ and $CM = 9$,find $BM$,$AB$,and $BC$.