Given:

EK = 9 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By above property,

EM = EK = 9 cm (tangent from E)

DK = DH (tangent from D)

FM = FH (tangent from F)

Now,

Perimeter of ∆EDF = ED + DF + FE

⇒ Perimeter of ∆EDF = (EK – KD) + (DH + HF) + (EM – MF)

[∵ED = EK – KD

DF = DH + HF

FE = EM – MF]

⇒ Perimeter of ∆EDF = EK – KD + DH + HF + EM – MF

⇒ Perimeter of ∆EDF = EK – DH + DH + HF + EM – HF [∵DK = DH and FM = FH]

⇒ Perimeter of ∆EDF = EK + EM

⇒ Perimeter of ∆EDF = 9 cm + 9 cm

⇒ Perimeter of ∆EDF = 18 cm

Hence, Perimeter of ∆EDF = 18 cm