Finding the distance between two triangles, one inside the other I have two right triangles One is a $6$-$8$-$10$ and inside is a $3$-$4$-$5$ and the space between the two triangles is a uniform amount.
I made a really awkward and weird pic of the diagram and I need to solve for $X$

How would I go about solving this? I couldn't figure out any good approaches.

Draw lines connecting the "respective vertices". Add up the areas to solve for $x$.

Area of a right triangle with the non-hypotenuse sides being $a$ and $b$ is $\frac{1}{2}(a \times b)$ while the area of a trapezium with parallel sides being $a$ and $b$ and the height being $h$ is $\frac{1}{2}h(a+b)$
$$\frac{1}{2}x(5+10) + \frac{1}{2}x(3+6) + \frac{1}{2}x(4+8) + \frac{1}{2}(3 \times 4) = \frac{1}{2} (6 \times 8)$$
$$36x + 12 = 48 \Rightarrow x = 1$$
EDIT:
Let us try to look at a slightly general case. Take a triangle and scale it to another similar triangle with the scale factor being $t$.

Let the sides of the inner triangle be $a$, $b$ and $c$.
The perimeter and area of the inner triangle is $P$ and $A$ respectively.
The sides of the outer triangle are $ta$, $tb$ and $tc$ while the perimeter and area are $tP$ and $t^2A$.
As before join the "respective" vertices and summing the areas give us,
$$A + \frac{1}{2}x(ta + a) + \frac{1}{2}x(tb + b) + \frac{1}{2}x(tc + c) = t^2A$$
$$A + \frac{1}{2}x(t+1)P = t^2A \Rightarrow \frac{1}{2}x(t+1)P = (t^2-1)A$$
$$x = \frac{2A}{P}(t-1)$$
In the problem asked, $t=2$ with $P = 2A = 12$ and hence we get $x=1$.
Also, $t=1$ gives $x=0$ as expected.
This also gives a nice proof that the radius of the incircle of a triangle is $$r_{in} = \frac{2A}{P}$$
This is got by plugging in $t=0$ and realizing that $\left| x \right|$ is nothing but the radius of the incircle.
Hence, $$x = (t-1)r_{in}$$
