1.

In Fig. a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ΔEDF isA. 18 cm B. 13.5 cm C. 12 cm D. 9 cm

Answer»

Answer is A. 18 cm

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

Option A

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