Exams › SSC CGL (Prelims) › General

In a circle, chords PQ and RS intersect at T such that \(PT:TQ = 3:4\) and \(RT:TS = 6:5\). If \(PQ = p\) and \(RS = q\), which of the following is true?

1. \(p^2/q^2 = 147/242\)
2. \(pq = 7/11\)
3. \(p^2/q^2 = 245/242\)
4. \(pq = 12/11\)

Correct answer: \(p^2/q^2 = 245/242\)

Solution

By the intersecting chords theorem, \(PT\cdot TQ = RT\cdot TS\). Let \(PT=3x, TQ=4x\) and \(RT=6y, TS=5y\), so \(12x^2=30y^2\), giving \(x^2/y^2=5/2\). Now \(p=PQ=7x\) and \(q=RS=11y\), hence \(p^2/q^2 = 49x^2/121y^2 = 49\cdot(5/2)/121 = 245/242\).

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