The areas of two similar triangles are $25$ and $16$. Then,the ratio of their perimeters is $\ldots \ldots \ldots \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. Prove that $\frac{AB^2}{BC^2} = \frac{AM}{CM}$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude. If $PD = 9 DR$,then $PQ = \ldots \ldots \ldots \times QR$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 25$ and $AM = 16$,then $BM = \dots$

In $\Delta ABC$,$m \angle A = 90^{\circ}$,$AB = 8$ and $AC = 15$. Then,$BC = \ldots$.

In $\Delta ABC$,$m\angle A + m\angle C = m\angle B$. If $AB = 7$ and $BC = 24$,then $AC = \ldots$