The areas of two similar triangles are $100 {\mathrm{cm}}^2$ and $49 {\mathrm{cm}}^2$ respectively. If the altitude of the bigger triangle is $5 \mathrm{cm}$, find corresponding altitude of the other.

The areas of two similar triangles are $81 {\mathrm{cm}}^2$ and $49 {\mathrm{cm}}^2$ respectively. If the altitude of the first triangle is $6.3 \mathrm{cm}$, find corresponding altitude of the other.

In a right triangle $ABC$, right angled at $C, P$ and $Q$ are the points of the sides $CA$ and $CB$ respectively, which divide these sides in the ratio $2:1$. Prove that $9A{Q}^2=9A{C}^2+4B{C}^2$.

In a right triangle $ABC$, right angled at $C, P$ and $Q$ are the points of the sides $CA$ and $CB$ respectively, which divide these sides in the ratio $2:1$. Prove that $9B{P}^2=9B{C}^2+4A{C}^2$

In a right triangle $ABC$, right angled at $C, P$ and $Q$ are the points of the sides $CA$ and $CB$ respectively, which divide these sides in the ratio $2:1$. Prove that $9\left(A{Q}^2+B{P}^2\right)=13A{B}^2$.

$P$ and $Q$ are the mid points on the sides $CA$ and $CB$ respectively of $△ABC$ right angled at $C$. Prove that $4\left(A{Q}^2+B{P}^2\right)=5A{B}^2$.