Let $ABCD$ be a regular tetrahedron with side length $1$. Let $EF GH$ be another regular tetrahedron such that the volume of $EF GH$ is $\tfrac{1}{8}\text{-th}$ the volume of $ABCD$. The height of $EF GH$ (the minimum distance from any of the vertices to its opposing face) can be written as $\sqrt{\tfrac{a}{b}}$, where $a$ and $b$ are positive coprime integers. What is $a + b$?

1. **Volume of a Regular Tetrahedron:**
   The volume \( V \) of a regular tetrahedron with side length \( s \) is given by:
   \[
   V = \frac{s^3}{6\sqrt{2}}
   \]
   For the tetrahedron \(ABCD\) with side length \( s = 1 \):
   \[
   V_{ABCD} = \frac{1^3}{6\sqrt{2}} = \frac{1}{6\sqrt{2}}
   \]

2. **Volume of Tetrahedron \(EFGH\):**
   Given that the volume of \(EFGH\) is \(\frac{1}{8}\) of the volume of \(ABCD\):
   \[
   V_{EFGH} = \frac{1}{8} \times \frac{1}{6\sqrt{2}} = \frac{1}{48\sqrt{2}}
   \]

3. **Height of a Regular Tetrahedron:**
   The height \( h \) of a regular tetrahedron with side length \( s \) can be derived from the volume formula. The volume can also be expressed as:
   \[
   V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
   \]
   The base of a regular tetrahedron is an equilateral triangle with area:
   \[
   \text{Base Area} = \frac{\sqrt{3}}{4} s^2
   \]
   For \( s = 1 \):
   \[
   \text{Base Area} = \frac{\sqrt{3}}{4}
   \]
   Therefore, the volume can be written as:
   \[
   \frac{1}{3} \times \frac{\sqrt{3}}{4} \times h = \frac{1}{6\sqrt{2}}
   \]
   Solving for \( h \):
   \[
   h = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}
   \]

4. **Height of Tetrahedron \(EFGH\):**
   Since the volume of \(EFGH\) is \(\frac{1}{8}\) of the volume of \(ABCD\), the side length of \(EFGH\) must be scaled by a factor of \(\sqrt[3]{\frac{1}{8}} = \frac{1}{2}\). Therefore, the height of \(EFGH\) is half the height of \(ABCD\):
   \[
   h_{EFGH} = \frac{1}{2} \times \frac{\sqrt{6}}{3} = \frac{\sqrt{6}}{6}
   \]

5. **Expressing the Height in the Given Form:**
   The height of \(EFGH\) can be written as:
   \[
   h_{EFGH} = \sqrt{\frac{6}{36}} = \sqrt{\frac{1}{6}}
   \]
   Here, \( a = 1 \) and \( b = 6 \), so:
   \[
   a + b = 1 + 6 = 7
   \]

The final answer is \( \boxed{7} \)