(i) - By Theorem- For quadrilateral circumscribing circle
AB + DC = AD + BC
∴ 6 + DC = 5 + 8
⇒ DC = 7 cm
(ii) - By Theorem- Tangents from external point are equal in lengths
∴ DR = DQ = 7 cm
and, CR = CP = 2 cm
∵ DC = DR + CR
∴ DC = 7 + 2 = 9 cm
(iii) By Theorem- Tangents from an external point are equal in lengths
∴ CR = CQ = 4 cm
and BQ = BP = 2 cm
Now, AB = AP + BP = AP + 2,
DC = DR + CR = DR + 4
and, BC = BQ + CQ = 2 + 4
By Theorem- For quadrilateral circumscribing circle
AB + DC = AD + BC
∴ (AP + 2) + (DR + 4) = 5 + (2 + 4)
⇒ AP + DR = 5 cm