The perimeter of a right-angled triangle is 4 times larger than the shortest side.
Find the ratio of other sides.
I tried approaching it this way . Let's have a right angled triangle $ABC$ (right angled at $B$) and $BC=x$ .Then $P=5x$ and on adding the three sides $AB+AC=4x$.
But couldn't move after this as how to get ratios.
Let the sides be $a,b,c$ such that: $$\begin{cases}4a=a+b+c \\ a^2+b^2=c^2 \end{cases} \Rightarrow \begin{cases}9a^2=b^2+2bc+c^2 \\ 9a^2=9c^2-9b^2\end{cases} \Rightarrow 10b^2+2bc-8c^2=0 \Rightarrow \\ 5\left(\frac bc\right)^2+\frac bc-4=0 \Rightarrow \frac bc=\frac{-1+9}{10}=\frac45.$$