Circle PQR is inscribed in a quadrilateral ABCD. The circle touches side AD at
point S. AP = 8 cm, QC = 3 cm and DC = 6 cm. The length of side AD is?
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Answer:
Let the circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at points P, Q, R and S respectively.
Since the lengths of the tangents drawn from an external point are equal. So
DR = DS (tangents on circle from point D)
CR = CQ (tangents on circle from point C)
BP = BQ (tangents on circle from point B)
AP = AS (tangents on circle from point A)
Adding all these equations
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
AD = 8 cm + 6 cm - 9 cm = 5 cm
Step-by-step explanation:
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