[tex]\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \cfrac{small}{large}\qquad \stackrel{sides}{\cfrac{s^2}{s^2}}=\stackrel{areas}{\cfrac{6}{150}}\implies \cfrac{\stackrel{\textit{hypotenuse of small}}{5^2}}{\stackrel{\textit{hypotenuse of large}}{s^2}}=\cfrac{6}{150}
\\\\\\
\cfrac{5^2\cdot 150}{6}=s^2\implies 625=s^2\implies \sqrt{625}=s\implies \boxed{25=s}\\\\
-------------------------------\\\\
\textit{thus the sides are at a }\cfrac{5}{25}\implies \cfrac{1}{5}\implies 1:5~ratio[/tex]
now, notice, the ratio is from small to large, so the larger right-triangle sides are 5 times larger than the small one then.
now, recall, a right-triangle with a hypotenuse of 5, will be a 3,4,5 triangle, you can check using the pythagorean theorem, thus, check the picture below.
