Q. 363.7( 3 Votes )

In Fig. 10.98, a

Answer :

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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