1. The problem is to find the side length and area of the inner square formed by connecting the midpoints of the sides of a larger square with side length $10$ cm. 2. The larger square has side length $10$ cm, so its area is $10^2 = 100$ cm$^2$. 3. The inner square's vertices are the midpoints of the larger square's sides. Each midpoint is $\frac{10}{2} = 5$ cm from the adjacent vertices. 4. The side length of the inner square is the distance between two adjacent midpoints of the larger square's sides. This forms a right triangle with legs of length $5$ cm each. 5. Using the Pythagorean theorem, the side length $s$ of the inner square is: $$ s = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} $$ 6. The area of the inner square is: $$ \text{Area} = s^2 = (5\sqrt{2})^2 = 25 \times 2 = 50 \text{ cm}^2 $$ 7. Therefore, the inner square has side length $5\sqrt{2}$ cm and area $50$ cm$^2$.