construct a right angle triangle whose base is 6cm and difference between hypotaneous and other side is 8cm

In the given question, let AB be the base and BC be the hypotenuse.
So,
AB = 6 cm
Let the hypotenuse be x. We are given that the difference between AC and BC is 8. So,
AC = x-8

Now, using the Pythagoras theorem in triangle ABC;
$$\begin{gathered} B{C}^2=A{B}^2+A{C}^2 \\ {x}^2=6^2+{\left(x-8\right)}^2 \\ {x}^2=36+x^2+64-16x \\ 16x=100 \\ x=\frac{100}{16} \\ x=6.25 \end{gathered}$$

So,
$$\begin{gathered} AC=x-8 \\ AC=6.25-8 \\ AC=-1.75 \end{gathered}$$

As the side of a triangle cannot be negative, so a right angled triangle whose base is 6 cm and difference between hypotenuse and other side as 8 cm cannot be constructed.