The distance between chord RS and the center of a circle is 3 cm. Chor

Look at the attached figures..

Fig I : This gives us the placement of chord RS. where the center of RS is at N and NO=3.
\(\triangle{NOS}\) gives \(NS=\sqrt{5^2-3^2}=\sqrt{16}=4\)

Fig II : This gives us the placement of chord PQ where ST=2, and it shows that PQ can be made in different ways.
However we are looking for the LEAST length of PQ, so it will be when we take point T to be the closest distance to the center. Join OT, and OT will be perpendicular to PQ.


Fig III :
a) Join OT
\(\triangle{NOT}\) gives \(OT=\sqrt{2^2+3^2}=\sqrt{13}\).
b) Join OQ
\(\triangle{OQT}\) gives \(TQ=\sqrt{5^2-13}=\sqrt{12}=2\sqrt{3}\)
So, \(PQ=2*TQ=2*2\sqrt{3}=4\sqrt{3}\)
Here a rule that helps you but NOT tested on GMAT is product of two portions on either side of intersection of two chords is same, that is RT*ST=PT*QT..
\(RT*ST=PT*QT........6*2=PT^2.......PT=\sqrt{12}.......PQ=4\sqrt{3}\)

C

minustark

∠R = ∠Q and ∠P = ∠S {Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.}

Hence, triangle PRT is similar to SQT

\(\frac{PT}{ST} = \frac{RT}{QT}\)

\(PT*QT = RT*ST\)

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