# In Fig. 10.98, a

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

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