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Let AC = 4k and BC = 3k. Since C is the midpoint of AB (as AB is a diameter and C is where PQ perpendicularly intersects AB): AC + BC = AB Therefore, 4k + 3k = AB: 7k = AB Given AB = 2 × radius = 14 cm: 7k =14 k=2 Hence: AC =4k =4 × 2 = 8 cm, BC=3k=3×2=6cm PC = CQ (AB is the diameter of given circle and PQ is the perpendicular on AB) As we know, PC × CQ = AC × CB PC × PC = 8 × 6 PC2 = 48 PC = √48 PC = 4√3 Now, PQ = 2×PC PQ = 2 × 4√3 PQ = 8√3
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