In Fig. 2, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, calculate the perimeter of ΔEDF (in cm)

Given: EK = 9 cm

By circle property we know, EK = EM.

[∵, EK, and EM are tangents from the same point onto the circle]

So, EK = EM = 9 cm

Also, KD = DH and HF = FM

[∵, KD, and DH are tangents to the circle from the same point; HF and FM are tangents to the circle from the same point]

We have to find the perimeter of the triangle EDF, which is given by

Perimeter of ∆EDF = ED + DF + EF

⇒ Perimeter of ∆EDF = (EK – KD) + (DH + HF) + (EM – FM) [∵, ED = EK – KD, DF = DH + HF and EF = EM – FM]

⇒ Perimeter of ∆EDF = (EK – KD) + (KD + FM) + (EK – FM) [∵, DH = KD, HF = FM and EM = EK]

⇒ Perimeter of ∆EDF = EK – KD + KD + FM + EK – FM

⇒ Perimeter of ∆EDF = (EK + EK) – KD + KD + FM – FM

⇒ Perimeter of ∆EDF = 2(EK) + 0

⇒ Perimeter of ∆EDF = 2(9) = 18

Hence, the perimeter of triangle EDF is 18 cm.