Exploring Some Geometric Themes — Question 3

(a) Let the side of the square of Sierpinski’s Carpet be 1 square unit. As one-ninth of the square is removed, and one-eighth remains in each step. Step 0: Area of whole square = (1) 2 = 1 Step 1: Area of remaining region = \(\frac{8}{9}(1)^2=\frac{8}{9}\) Step 2: Area of remaining region = \(\frac{8}{9} \times \frac{8}{9}=\frac{64}{81}\) Step 3: Area of remaining region = \(\frac{8}{9} \times \frac{64}{81}=\frac{8^3}{9^3}=\frac{512}{729}\) Area of remaining region after nth step = \(\left(\frac{8}{9}\right)^n\) sq. units (b) Let the area of the triangle of Sierpinski’s Gasket be 1 square unit. One-fourth of the triangle is removed in each step, and three-fourth remains. Step 0: Area of the whole triangle = 1 sq. unit. Step 1: Area of remaining region = \(\frac {3}{4}\) sq. units. Step 2: Area of remaining region = \(\frac{3}{4} \times \frac{3}{4}\) square units = \(\frac {9}{16}\) sq. units Area of remaining region after nth step = \(\left(\frac{3}{4}\right)^n\) sq. units Figure It Out (Page 73)