Given AB=4 cm,DE=6 cm and EF=9 cm and FD=12 cm
Also, `DeltaABC~DeltaDEF`
`therefore(AB)/(ED)=(BC)/(EF)=(AC)/(DF)`
`rArr 4/6=(BC)/9=(AC)/(12)`
On taking first two terms, we get
`4/6=(BC)/9`
`rArr BC=(4xx9)/6=6 cm`
`=AC=(6xx12)/9=8 cm`
NOw, perimeter of `DeltaABC=AB+BC+AC`
=4+6+8=18 cm