Question
In Fig., a circle touches the side DF of EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of EDF (in cm) is:
18
13.5
12
9
Solution
A.
18
It is known that the tangents from an external point to the circle are equal.
EK = EM, DK = DH and FM = FH .....(1)
Perimeter of EDF = ED + DF + FE
= (EK - DK) + (DH + HF) + EM - FM)
= (EK - DH) + (DH + HF) + (EM - FH) [Using (1)]
= EK + EM
= 2EK = 2(9 CM) = 18 CM
Hence, the perimeter of EDF is 18 cm.
