$A$ $8 \, m$ long bamboo tree standing erect on the ground breaks at the height of $3 \, m$ from the ground. The broken part of the tree remains attached to the trunk. Find the distance in $m$ between the top of the tree on the ground and the base of the tree.

In a triangle $PQR$,$N$ is a point on $PR$ such that $QN \perp PR$. If $PN \cdot NR = QN^2$,prove that $\angle PQR = 90^{\circ}$.

The perimeter of an equilateral triangle is $18$. Then,the length of its altitude is $\ldots \ldots \ldots$

If $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$,then $BC^2 : YZ^2 = \ldots \ldots \ldots$ (Wait,the question asks for $ABC : XYZ = BC^2 : \ldots$ which refers to the ratio of areas). Given $\Delta ABC \sim \Delta XZY$,the ratio of their areas is equal to the ratio of the squares of their corresponding sides. Thus,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{BC^2}{ZY^2}$.

In $\Delta XYZ,$ the bisector of $\angle X$ intersects $\overline{YZ}$ at $T$. If $XY = 10, XZ = 14$ and $YT = 5,$ find $YZ.$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.