$\mathrm{One} \mathrm{side} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle} \mathrm{is} 24\mathrm{cm}. \mathrm{The} \mathrm{midpoints} \mathrm{of} \mathrm{its} \mathrm{sides} \mathrm{are} \mathrm{joined} \mathrm{to} \mathrm{form} \mathrm{another}$

 $\mathrm{triangle} \mathrm{whose} \mathrm{midpoints}, \mathrm{in} \mathrm{turn} \mathrm{are} \mathrm{joined} \mathrm{to} \mathrm{form} \mathrm{still} \mathrm{another} \mathrm{triangle}. \mathrm{This} \mathrm{process} \mathrm{continues} \mathrm{indefinitely},$

 $\mathrm{then} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{perimeters} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{triangles} \mathrm{is}$

$Let ‘a’ be the side length of an equilateral triangle, then AB=BC=CA=a$

          $D, E, F are the midpoints of BC, CA, AB respectively EF=FD=DE=$ $\frac{a}{2}$

          $and H,I,J are midpoints of EF and FD and DE respectively$

$\therefore \mathrm{IJ}=\mathrm{JH}=\mathrm{HI}=\frac{a}{4}$

$\text{ Similarly, }\mathrm{KL}=\mathrm{ML}=\mathrm{KM}=\frac{a}{8},\dots .$

$\therefore P= Sum of perimeters$

      $=3\left(a+\frac{a}{2}+\frac{a}{4}+\frac{a}{8}+\mathrm{.......}\right)$

      $$\begin{gathered} =3\left(\frac{a}{1-\frac{1}{2}}\right) \\ =6\mathrm{a} \\ =6\times 24 \left(\therefore a=24\right) \\ =144 \mathrm{cm} \end{gathered}$$