In the below picture, author provides an example of a continuous, closed, non-intersecting curve which lies in a bounded region $R$ but which has infinite length.
While calculating area bounded by curve, I think there is a mistake!
The third term of an infinite sum must be $(12)(\frac{1}{9})^2\frac{√3}{4}$ but author writes $(9)(\frac{1}{9})^2\frac{√3}{4}$.
Reason: since in a third step, if we trisect the, sides DE, EF, FB, BG, GH, etc. and constructing equilateral triangles as before. Then $12$ new equilateral triangles with side length $\frac{1}{9}$ are created (since there are 12 sides which we are trisecting DE,EF, FB, BG, GH, HJ, JC, CK KL, LM, MA & AD as shown in fig 4.22 of image)
so the area bounded by new figure (in third step of the process) will be
$=\text{previous area}+\text{Area of these 12 new equilateral triangles}$
$= Area(∆ABC) + Area(∆ DEF) + Area(∆GHJ) + Area(∆KLM)+ \text{Area of these 12 new equilateral triangles}$
$= \frac{√3}{4} + \frac{√3}{4}(\frac{1}{3})^2+\frac{√3}{4}(\frac{1}{3})^2+\frac{√3}{4}(\frac{1}{3})^2+\text{Area of these 12 new equilateral triangles}$
$=\frac{√3}{4} + 3\frac{√3}{4}(\frac{1}{3})^2+\underline{12\frac{√3}{4}(\frac{1}{9})^2}$
Is am I correct? Further, if my calculations are correct then, what will be the area after repeating the process indefinitely?
You are right, that should be 12 upfront coefficient, obvious printer's devil/error.
If we take constant $C= \sqrt{3} /4 $ then the total area of infinitely many diminishing Fractal Koch snowflake triangles is given by:
$$ C*(1+ 3(1/3)^2 + 12 (1/9)^2 + 27 (1/27)^2+...+...)$$ $$= C*3/2 $$
finite area after evaluating the sum of infinite series, whereas the perimeter length diverges.