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

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AB = \sqrt{10}$ and $AM = 2.5$,then $MC = \ldots$

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 $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$,$PQ = 7$ and $QR = 25$. Find the perimeter of $\Delta PQR$.

In the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that:
$AB^{2} = AD^{2} - BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$

$A$ $5\, m$ long ladder is placed leaning against a vertical wall such that it reaches the wall at a point $4 \,m$ high. If the foot of the ladder is moved $1.6 \,m$ towards the wall,find the distance by which the top of the ladder would slide upwards on the wall (in $m$).