Q. 94.5( 11 Votes )

In Fig. 2, a circ

Answer :

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.

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