In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm, and CD = 4cm. Find AD.

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
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