1. **State the problem:** We have two right triangles sharing a common horizontal side. The upper triangle has a hypotenuse of length $7\sqrt{3}$ and angles $45^\circ$ and $90^\circ$. The lower triangle has angles $60^\circ$ and $90^\circ$, and side $x$ opposite the $60^\circ$ angle. We need to find $x$. 2. **Analyze the upper triangle:** It is a right triangle with a $45^\circ$ angle, so it is a 45-45-90 triangle. The sides of a 45-45-90 triangle are in the ratio $1:1:\sqrt{2}$. 3. **Find the legs of the upper triangle:** The hypotenuse is $7\sqrt{3}$. Let each leg be $a$. Then: $$a\sqrt{2} = 7\sqrt{3}$$ 4. **Solve for $a$:** $$a = \frac{7\sqrt{3}}{\sqrt{2}} = 7 \frac{\sqrt{3}}{\sqrt{2}} = 7 \sqrt{\frac{3}{2}}$$ 5. **Simplify $a$:** $$a = 7 \sqrt{\frac{3}{2}} = 7 \frac{\sqrt{6}}{2} = \frac{7\sqrt{6}}{2}$$ 6. **Interpret $a$:** This leg $a$ is the common horizontal side shared by both triangles. 7. **Analyze the lower triangle:** It is a right triangle with angles $60^\circ$ and $30^\circ$ (since $90^\circ - 60^\circ = 30^\circ$). The sides of a 30-60-90 triangle are in the ratio $1: \sqrt{3}: 2$, where the side opposite $30^\circ$ is the shortest. 8. **Identify sides in the lower triangle:** The side $x$ is opposite the $60^\circ$ angle, so it corresponds to the side of length $\sqrt{3}k$ if the shortest side is $k$. 9. **The horizontal side $a$ is adjacent to $60^\circ$ and opposite $30^\circ$, so it corresponds to $k$ in the ratio. Therefore:** $$k = a = \frac{7\sqrt{6}}{2}$$ 10. **Find $x$ using the ratio:** $$x = \sqrt{3} k = \sqrt{3} \times \frac{7\sqrt{6}}{2} = \frac{7}{2} \sqrt{3 \times 6} = \frac{7}{2} \sqrt{18}$$ 11. **Simplify $x$:** $$\sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2}$$ 12. **Final expression for $x$:** $$x = \frac{7}{2} \times 3 \sqrt{2} = \frac{21}{2} \sqrt{2}$$ **Answer:** $$\boxed{x = \frac{21}{2} \sqrt{2}}$$