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

A. 18

B. 13.5

C. 12

D. 9

Answer

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Hint: EK and EM are tangents to the circle. Tangent segments to a circle from the same external point are congruent, (i.e. equal in lengths). Use this in finding the perimeter.

Complete step-by-step answer:

Given Data - A circle touches ED and EF at K and M respectively.

That makes EK and EM the tangents of the circle.

From the concepts of circles, we know that tangent segments to a circle from the same external point are congruent, (i.e. equal in lengths)

Given EK = 9cm

Therefore, EM = EK = 9cm (the exterior point is E)

Now, EK + EM = 18cm

We can write, EK = ED + DK and EM = EF + FM

⟹ED + DK + EF + FM = 18cm

Again from the exterior point D there are two tangents to the circle DK and DH and from exterior point F there are two tangents to the circle FH and FM.

Hence, DK = DH and FH = FM

⟹ED + DH + FH + EF = 18cm (FH = HF) - These are interchangeable

We can write DH + HF = DF

⟹ED + DF + EF = 18 cm

The perimeter of a triangle = sum of all sides

The perimeter of a ∆ EDF = ED + DF + EF

= 18 cm

Note: In such problems having a complete understanding of the concept of circles is very important. Here we used the concept of tangents, where tangent segments to a circle from the same external point are congruent (i.e. equal in length).

A tangent from a given point to a circle is a straight line which touches the circle at only one point.

Complete step-by-step answer:

Given Data - A circle touches ED and EF at K and M respectively.

That makes EK and EM the tangents of the circle.

From the concepts of circles, we know that tangent segments to a circle from the same external point are congruent, (i.e. equal in lengths)

Given EK = 9cm

Therefore, EM = EK = 9cm (the exterior point is E)

Now, EK + EM = 18cm

We can write, EK = ED + DK and EM = EF + FM

⟹ED + DK + EF + FM = 18cm

Again from the exterior point D there are two tangents to the circle DK and DH and from exterior point F there are two tangents to the circle FH and FM.

Hence, DK = DH and FH = FM

⟹ED + DH + FH + EF = 18cm (FH = HF) - These are interchangeable

We can write DH + HF = DF

⟹ED + DF + EF = 18 cm

The perimeter of a triangle = sum of all sides

The perimeter of a ∆ EDF = ED + DF + EF

= 18 cm

Note: In such problems having a complete understanding of the concept of circles is very important. Here we used the concept of tangents, where tangent segments to a circle from the same external point are congruent (i.e. equal in length).

A tangent from a given point to a circle is a straight line which touches the circle at only one point.

Last updated date: 25th Sep 2023

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