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

EK=9cmAs length of tangents drawn from an external point to the circle are equal.∴EK=EM=9cmAlso,DH=DKandFH=FM... (1)EK=EM=9cm⇒ED+DK=9cm andEF+FM=9cm⇒ED+DH=9cm andEF+HF=9cm— (2)Perimeter ofΔEDF=ED+DF+EF=ED+DH+HF+EF=(9+9)cm (From equation (2))=18cmTherefore, the perimeter=18cm.