In $\Delta XYZ$,$m\angle X = 90^{\circ}$,$XY = 8$ and $YZ = 17$. Then,the area of $\Delta XYZ$ is.............

In $\Delta PQR, m\angle Q = 90^{\circ}$ and $\overline{QM}$ is a median. If $PQ = 20$ and $QR = 21$,find $QM$.

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 $\Delta PQR,$ the midpoints of the sides are $X, Y, Z.$ In $\Delta XYZ,$ the midpoints of the sides are $A, B, C.$ If the area of $\Delta PQR$ is $240,$ find the area of $\Delta XYZ$ and $\Delta ABC.$

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $M$ is the midpoint of $\overline{XZ}$. If $XY = 3$ and $YZ = 4$,then $YM = \ldots$

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $PR$. If $QD = 15$ and $PR = 34$,find $PQ$.