1. Let's denote the number of sides of the two polygons as $x$ and $y$, with $x \le y$. 2. Given that the sum of their sides is 9, we have the equation: $$x + y = 9$$ 3. The number of diagonals $d$ in a polygon with $n$ sides is given by the formula: $$d = \frac{n(n-3)}{2}$$ 4. The sum of the diagonals of the two polygons is 7, so: $$\frac{x(x-3)}{2} + \frac{y(y-3)}{2} = 7$$ 5. Multiply both sides by 2 to eliminate denominators: $$x(x-3) + y(y-3) = 14$$ 6. Expand both terms: $$x^2 - 3x + y^2 - 3y = 14$$ 7. Using $y = 9 - x$ from step 2, substitute $y$: $$x^2 - 3x + (9 - x)^2 - 3(9 - x) = 14$$ 8. Expand $(9 - x)^2$ and simplify: $$x^2 - 3x + (81 - 18x + x^2) - 27 + 3x = 14$$ 9. Combine like terms: $$x^2 - 3x + 81 - 18x + x^2 - 27 + 3x = 14$$ $$2x^2 - 18x + 54 = 14$$ 10. Simplify the constant terms: $$2x^2 - 18x + 40 = 0$$ 11. Divide entire equation by 2: $$x^2 - 9x + 20 = 0$$ 12. Factor the quadratic: $$(x - 5)(x - 4) = 0$$ 13. So, $x = 5$ or $x = 4$. 14. Since $x \le y$, and $x + y = 9$, the smaller polygon has $4$ sides. **Final answer:** The smaller polygon has **4** sides.