1. **State the problem:** We have two right triangles inside a larger triangle. The left triangle has one leg of length 4 and a hypotenuse of length 10. The right triangle has one leg of length $x$ and a hypotenuse of length 8. We need to solve for $x$. 2. **Use the Pythagorean theorem:** For a right triangle with legs $a$ and $b$ and hypotenuse $c$, the relationship is $$a^2 + b^2 = c^2$$ 3. **Find the other leg of the left triangle:** Let the other leg be $h$ (the height). Using the Pythagorean theorem: $$4^2 + h^2 = 10^2$$ $$16 + h^2 = 100$$ $$h^2 = 100 - 16 = 84$$ $$h = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$$ 4. **Use the height $h$ for the right triangle:** The right triangle shares the same height $h$, so: $$x^2 + h^2 = 8^2$$ $$x^2 + (2\sqrt{21})^2 = 64$$ $$x^2 + 4 \times 21 = 64$$ $$x^2 + 84 = 64$$ 5. **Solve for $x^2$:** $$x^2 = 64 - 84 = -20$$ 6. **Interpret the result:** Since $x^2$ is negative, there is no real solution for $x$ with the given dimensions. This suggests the triangle configuration is not possible with these lengths. **Final answer:** No real solution for $x$ exists with the given triangle dimensions.