Daily Math Puzzle: 2026-04-22

20 units — Let's denote the squares as S1 (outermost), S2 (first inner), and S3 (second inner, which is the 'third' square in the sequence as per the question). 1. **Square S1 (Original):** Area = 100 square units. Since Area = side * side, the side length of S1 is `sqrt(100) = 10` units. 2. **Square S2 (First Inner Square):** This square is formed by connecting the midpoints of S1's sides. A fundamental geometric property (which can be proven using the Pythagorean theorem) is that when you connect the midpoints of a square's sides to form a new square, the new square has exactly half the area of the original square. So, Area of S2 = Area of S1 / 2 = 100 / 2 = 50 square units. (To confirm: if S1 has vertices at (0,0), (10,0), (10,10), (0,10), its midpoints are (5,0), (10,5), (5,10), (0,5). The side length of S2 would be `sqrt((5-0)^2 + (0-5)^2) = sqrt(25 + 25) = sqrt(50)` units. Area = `(sqrt(50))^2 = 50`.) 3. **Square S3 (Second Inner Square / 'Third' Square in the sequence):** This square is formed by connecting the midpoints of S2's sides. Following the same property, its area will be half that of S2. Area of S3 = Area of S2 / 2 = 50 / 2 = 25 square units. Now, we need to find the side length of S3: `side_S3 = sqrt(25) = 5` units. The question asks for the perimeter of this third square. Perimeter of S3 = `4 * side_S3 = 4 * 5 = 20` units.