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In Fig 2, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ΔEDF (in cm) is:

#### Options

A. 18

B. 13.5

C. 12

D. 9

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#### Solution

EK = 9 cm

As length of tangents drawn from an external point to the circle are equal.

∴EK = EM = 9 cm

Also, DH = DK and FH = FM … (i)

EK = EM = 9 cm

⇒ ED + DK = 9 cm and EF + FM = 9 cm

⇒ ED + DH = 9 cm and EF + HF = 9 cm [From equation (i)] … (ii)

Perimeter of ΔEDF = ED + DF + EF

= ED + DH + HF + EF

= (9 + 9) cm [From equation (ii)]

= 18 cm

Hence, the correct option is A.

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