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Question
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
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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