In $\Delta ABC$,$\overline{AD}$ is a median and $AB^2 + AC^2 = 148$. If $AD = 7$,then $BC = \ldots$

In the figure,two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6 \, cm$,$PB = 3 \, cm$,$PC = 2.5 \, cm$,$PD = 5 \, cm$,$\angle APB = 50^{\circ}$ and $\angle CDP = 30^{\circ}$. Then,$\angle PBA$ is equal to: (in $^{\circ}$)

As shown in the adjoining diagram of a staircase,the width of each step is $20 \, cm$ and the height is $15 \, cm$. Find the distance between points $A$ and $B$ in meters.

The lengths of the diagonals of a rhombus are $16 \, cm$ and $12 \, cm$. Then,the length of the side of the rhombus is (in $cm$)

In the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that $AC^2 + AB^2 = 2AD^2 + \frac{1}{2} BC^2$.

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude. If $AB = \sqrt{5}$ and $BD = 2$,then the length of hypotenuse $\overline{BC}$ is.........