Q. 83
Question
The figure shown is called a Sierpinski triangle. It may be constructed recursively as follows: We start with a large black equilateral triangle. Every time we see a black equilateral triangle, we inscribe a white equilateral triangle with each vertex at the midpoint of a side of the black triangle. If the side of the largest triangle is \(1\) unit and this process is repeated recursively, what is the area of the white-shaded region?
Step-by-Step Solution
VerifiedTotal area of white-shaded region is \(\frac{165\sqrt3}{1024} \text{ unit}^2\).
The length of the largest equilateral triangle is \(1\) unit.
The length of the largest white equilateral triangle is \(\frac{1}{2}\) unit.
Area of equilateral triangle \(=\frac{\sqrt3}{4}\left (\text{side} \right )^2\)
Area \(A_1\) \(=\frac{\sqrt3}{4}\left (\text{side} \right )^2\)
\(=\frac{\sqrt3}{4}\left (\frac{1}{2}\right )^2\)
\(=\frac{\sqrt3}{16}\text{ unit} ^2\)
There are \(3\) equilateral triangles of side length \(\frac{1}{2}\times\frac{1}{2} \) unit.
Area of \(3\) equilateral triangles \(A_2\) \(=\frac{\sqrt3}{4}\left (\frac{1}{4}\right )^2\)
\(=\frac{3\sqrt3}{64}\text{ unit} ^2\)
There are \(9\) equilateral triangles of side length \(\frac{1}{2}\times \frac{1}{4} \) unit.
Area of \(9\) equilateral triangles \(A_3\) \(=\frac{\sqrt3}{4}\left (\frac{1}{8}\right )^2\)
\(=\frac{9\sqrt3}{256}\text{ unit} ^2\)
There are \(27\) equilateral triangles of side length \(\frac{1}{2}*\frac{1}{8} \) unit.
Area of \(27\) equilateral triangles \(A_4\) \(=\frac{\sqrt3}{4}\left (\frac{1}{16}\right )^2\)
\(=\frac{27\sqrt3}{1024}\text{ unit} ^2\)
Hence, the total area of white shaded region \(=A_1+A_2+A_3+A_4\)
\(=A_1+A_2+A_3+A_4\)
\(=\frac{\sqrt3}{16}+\frac{3\sqrt3}{64}+\frac{9\sqrt3}{256}+\frac{27\sqrt3}{1024} \text{ unit} ^2\)
\(=\frac{165\sqrt3}{1024} \text{ unit} ^2 \)