Let a be the side length of equilateral triangle, than `AB=BC=CA=a`
`therefore D,E,F` are the mid-points of BC,CA and AB, respectively.
`therefore EF=FD=DE=(a)/(2)`
and H,I,J are the mid-points of EF, FD and DE, respectively.
`therefore IJ=JH=HI=(a)/(4)`
Similarly, `KL=ML=KM=(a)/(8),"..... "`
P=Sum of perimetes `=3(a+(a)/(2)+(a)/(4)+(a)/(8)+"......")`
`=3((a)/(1-(1)/(2)))=6a=6xx24=144 " cm " " "[ therefore a=24 " cm "]`
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