Given, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm
Also ΔABC∼ΔDEF
∴ABDE=BCEF=ACDF
⇒46=BC9=AC12 ...(i)
On taking first two terms, we get,
46=BC9
⇒ BC=4×96=6 cm
=AC=6×129=8 cm (taking last two term of eq(i))
Now, perimeter of ΔABC=AB+BC+AC
=4+6+8=18 cm