Answer :

Mark the touching points as P, Q, R and S as shown

As tangents from a point are of equal length we have

AQ = AR = a

BR = BS = b

CP = CS = c

DP = DQ = d

From figure

⇒ BC = BS + SC

⇒ 7 = b + c

⇒ b = 7 – c … BC is 7 cm given …(i)

Also,

⇒ DC = DP + PC

⇒ 4 = d + c

⇒ c = 4 – d … DC is 4 cm given …(ii)

And

⇒ AB = AR + RB

⇒ 6 = a + b … AB is 6 cm given

⇒ a = 6 – b

Using (i)

⇒ a = 6 – (7 – c)

Using (ii)

⇒ a = 6 – (7 – (4 – d))

⇒ a = 6 – (7 – 4 + d)

⇒ a = 6 – 7 + 4 – d

⇒ a + d = 3

⇒ AQ + QD = 3 …since AQ = a and QD = d

From figure AQ + QD = AD

⇒ AD = 3 cm

Hence AD is 3 cm

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Expertsview all courses

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation