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A circle whose radius is of 10 cm is having a chord inside it which is RS at a distance of 6 cm from it. It is having another chord PQ inside it which intersects chord PQ at T such that TS = 1/6 of RT. Find the value of chord PQ?

A circle whose radius is of 10 cm is having a chord inside it which is RS at a distance of 6 cm from it. It is having another chord PQ inside it which intersects chord PQ at T such that TS = 1/6 of RT. Find the value of chord PQ? Correct Answer 16√3 cm

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⇒ In this figure, OM = 6cm, OS = 10cm

⇒ In ΔOMS, ∠OMS = 90°

⇒ According to Pythagoras theorem, OS2 = OM2 + MS2

⇒ MS = √(102 – 62) = 8 cm

⇒ RS = 16 cm

According to condition given in the problem, TS = 1/6 RT

⇒ TS = 1/4 RS

⇒ TS = 4cm

According to intersecting chord theorem, when there are two intersecting chords, the product of rectangle formed by the segments of one chord is equal to product of rectangle formed by the segments of other.

⇒ RT × TS = PT × TQ

⇒ PT × TQ = 12 × 4 = 48cm

⇒ By AM-GM inequality, (PT + TQ) /2 ≥ √(PT × TQ)

⇒ (PT + TQ) /2 ≥ √48

⇒ PQ ≥ 2√48 = 16√3

∴ Answer is 16√ 3 cm

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