Question

In the given figure, 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 is



(a) 18 cm (b) 13.5 cm (c) 12 cm (d) 9 cm [CBSE 2012]

Answer 1

In the given figure, DH and DK are tangents drawn to the circle from an external point D.

∴ DH = DK (Lengths of tangents drawn from an external point to a circle are equal)

FH and FM are tangents drawn to the circle from an external point F.

∴ FH = FM (Lengths of tangents drawn from an external point to a circle are equal)

EK and EM are tangents drawn to the circle from an external point E.

∴ EM = EK = 9 cm (Lengths of tangents drawn from an external point to a circle are equal)

Now,

Perimeter of ΔEDF = ED + DF + EF

= ED + (DH + FH) + EF

= ED + (DK + FM) + EF (DH = DK and FH = FM)

= (ED + DK) + (FM + EF)

= EK + EM

= 9 cm + 9 cm

= 18 cm

Thus, the perimeter of ΔEDF is 18 cm.

Hence, the corrrect answer is option A.