total possible triangles with integral values How many triangles with altitudes 6,8,X(unknown value) can be formed such that the value of x is an integer?
 A: Let's assume corners
$$A=(0,0)\qquad B=(\beta,0)\qquad C=(\gamma,8)$$
These coordinates already ensure $h_c=8$. To get $h_b=6$ you have to choose
$$\gamma=\pm\frac43\sqrt{\beta^2 - 36}$$
So now your whole triangle is described in terms of a single parameter, $\beta>6$, and one choice of sign. You can compute the third height, $h_a$, in terms of this parameter:
$$h_a=\frac{8\beta}{\sqrt{(\beta-\gamma)^2+8^2}}$$

For the positive choice, $\gamma>0$, this height will assume all values $h_a>8\sqrt6\approx19.6$ (which is the value at $\beta=6$) so you can use it to obtain any integral height $h_a\ge20$. For the negative choice, the plot shows a minimum. To obtain details on that, you have to set the derivative to zero. You will find that the zero is for
$$\beta=3\sqrt{\frac{110-6\sqrt{137}}7}\approx7.15$$
and results in a height of
$$h_a=\sqrt[4]{\frac{3216236544-272719872\sqrt{137}}{343}}\approx16.3$$
so you can use that choice to obtain values $17\le h_a<20$ (and in fact any greater value as well, since that choice diverges to $\infty$ as well, just a lot more slowly than for $\gamma>0$). Integral values less than $17$ can not be obtained at all.
