One side of an equilateral triangle is 18 cm.

According to the midpoint theorem, the sides of each triangle formed by joining the midpoints of an equilateral triangle are half of the sides of the equilateral triangle. In other words, the triangles formed are equilateral triangles with sides 18 cm, 9 cm, 4.5 cm, 2.25 cm, ...

$($ i $)$ Sum of the perimeters of all the triangles, $\mathrm{P}=3 \times 18+3 \times 9+3 \times 4.5+3 \times 2.25+\ldots$$\infty$

$\Rightarrow \mathrm{P}=3 \times(18+9+4.5+2.25+\ldots \infty)$

It is a G.P. with $a=18$ and $r=\frac{1}{2}$.

$\therefore P=3 \times\left(\frac{18}{1-\frac{1}{2}}\right)$

$\Rightarrow \mathrm{P}=3 \times 36=108 \mathrm{~cm}$

(ii) Sum of the areas of all the triangles, $\mathrm{A}=\frac{\sqrt{3}}{4}(18)^{2}+\frac{\sqrt{3}}{4}(9)^{2}+\frac{\sqrt{3}}{4}(4.5)^{2}+\ldots \infty$

$\Rightarrow \mathrm{A}=\frac{\sqrt{3}}{4}\left((18)^{2}+(9)^{2}+(4.5)^{2}+\ldots \infty\right)$

$\therefore A=\frac{\sqrt{3}}{4}\left(\frac{(18)^{2}}{1-\frac{1}{4}}\right)$

$\Rightarrow \mathrm{A}=\frac{\sqrt{3}}{3} \times 324$

$\Rightarrow \mathrm{A}=108 \sqrt{3} \mathrm{~cm}^{2}$