1. **Stating the problem:** We have two right triangles with sides labeled and relationships given. We want to find the values of $x$ and $y$ using the given ratios and the fact that $x=16$. 2. **Given information:** - From the smaller triangle: sides $y$ and 6. - From the larger triangle: sides $x$, $4x$, and $y$. - A perpendicular segment of length 16 inside the larger triangle. - The equation $\frac{y}{x} = \frac{x}{y}$. - The value $x=16$. 3. **Using the ratio $\frac{y}{x} = \frac{x}{y}$:** $$\frac{y}{x} = \frac{x}{y}$$ Cross-multiplied: $$y \cdot y = x \cdot x$$ $$y^2 = x^2$$ Taking square roots: $$y = \pm x$$ Since lengths are positive, we take: $$y = x$$ 4. **Substitute $x=16$ into $y = x$:** $$y = 16$$ 5. **Check the other given information:** - The side labeled $4x$ becomes $4 \times 16 = 64$. - The perpendicular segment is 16. 6. **Summary:** - $x = 16$ - $y = 16$ These satisfy the given ratio and the figure's labels. **Final answer:** $$x = 16, \quad y = 16$$