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


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