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

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

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

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

Correct Answer: A

Solution :

\[AB=\frac{5}{6}r\Rightarrow CD=\frac{2}{3}r\] \[\Rightarrow \,\,\,\,\,OB=\sqrt{{{r}^{2}}-\frac{25}{36}{{r}^{2}}}\] \[\Rightarrow \,\,\,\,\,OB=\sqrt{\frac{11}{6}}r\]

Similarly, \[OD=\sqrt{{{r}^{2}}-\frac{4}{9}{{r}^{2}}}=\frac{\sqrt{5}}{3}r\]

Required probability\[=\frac{\pi \left( \frac{5}{9}{{r}^{2}}-\frac{11}{36}{{x}^{2}} \right)}{\pi {{r}^{2}}}=\frac{20-11}{36}=\frac{1}{4}\]