1. **Problem statement:** We have a right triangle with points G, H, and F, where GH is parallel to DE. Given that GH is 7 more than EH, FH = 6, and DE = 15, we need to find the length of EH. 2. **Understanding the problem:** Since GH is parallel to DE and DE = 15, by similarity of triangles, the segments relate proportionally. 3. **Assign variables:** Let EH = $x$. Then GH = $x + 7$. 4. **Use the Pythagorean theorem:** Since FH is perpendicular to GH, triangle FGH is right-angled at F. We know FH = 6. 5. **Express GF:** Since GH = $x + 7$ and EH = $x$, and GH is parallel to DE, the length GF corresponds to DE - GH = $15 - (x + 7) = 8 - x$. 6. **Apply Pythagoras in triangle FGH:** $$GF^2 + FH^2 = GH^2$$ $$ (8 - x)^2 + 6^2 = (x + 7)^2$$ 7. **Expand and simplify:** $$ (8 - x)^2 + 36 = (x + 7)^2$$ $$ (64 - 16x + x^2) + 36 = x^2 + 14x + 49$$ $$ 100 - 16x + x^2 = x^2 + 14x + 49$$ 8. **Cancel $x^2$ on both sides:** $$ 100 - 16x = 14x + 49$$ 9. **Bring all terms to one side:** $$ 100 - 16x - 14x = 49$$ $$ 100 - 30x = 49$$ 10. **Isolate $x$:** $$ 100 - 49 = 30x$$ $$ 51 = 30x$$ $$ x = \frac{51}{30} = \frac{17}{10} = 1.7$$ 11. **Answer:** The length of EH is $\boxed{\frac{17}{10}}$ or 1.7 units.