question_answer

The probability that the length of a randomly chosen chord of a circle lies between \[\frac{2}{3}\]and \[\frac{5}{6}\] of its diameter is

A) \[\frac{5}{16}\]

B) \[\frac{1}{16}\]

C) \[\frac{1}{4}\]

D) \[\frac{5}{12}\]

Correct Answer: C

Solution :

Let \[l\] be the length of the chord \[AB\] of the given circle of radius a and r be the distance of the midpoint D of the chord from the Centre c, then \[r=a\cos \theta \,\]and \[l=2a\sin \theta \] according to given condition:

\[\frac{2}{3}(2a)<2a\,sin\theta <\frac{5}{6}(2a)\] \[\Rightarrow \frac{2}{3}<\sin \theta <\frac{5}{6}\Rightarrow \frac{\sqrt{11}}{6}<\cos \theta <\frac{\sqrt{5}}{3}\]\[\Rightarrow \frac{\sqrt{11}}{6}a<r<\frac{\sqrt{5}}{3}a\]

\[\therefore \]The given condition is satisfied if the midpoint   of the chord lies within the region between the concentric circles of radii\[\frac{\sqrt{11}}{6}\]a and \[\frac{\sqrt{5}}{3}a\]

Hence, the required probability \[=\frac{\pi {{\left( \frac{\sqrt{5}}{3}a \right)}^{2}}-\pi {{\left( \frac{\sqrt{11}}{6}a \right)}^{2}}}{\pi {{a}^{2}}}=\frac{1}{4}\]