Since, length of two tangents drawn from an external point of circle are equal.
So, AP = AS
BP = BQ DR = DS
and RC = CQ
Adding all, we get
(AP + BP) + (DR + RC) = AS + BQ + DS + CQ
⇒ (AP + BP) + (DR + RC)
= (AS + DS) + (BQ + CQ)
⇒ AB + DC = AD + BC
⇒ 6 + 4 = AD + 7
⇒ 10 = AD + 7
AD = 10 – 7 = 3 cm
Hence, AD = 3 cm.