Question

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. $\frac{5}{6}$
• B. $\frac{1}{16}$
• C. $\frac{1}{4}$
• D. $\frac{5}{12}$

Answer 1

The correct option is C $\frac{1}{4}$

If $l$ is the length of the chord, $r$ is the distance of the mid-point of the chord from the center of the circle and $a$ is radius of the given circle ,

then $r=a\cos \theta ,l=2a\sin \theta$

Given: $\frac{2}{3}2a<2a\sin \theta <\frac{5}{6}2a$

$\Rightarrow \frac{\sqrt{11}}{6}a<a\cos \theta <\frac{\sqrt{5}}{3}a\Rightarrow \frac{\sqrt{11}}{6}a<r<\frac{\sqrt{5}}{3}a$

$\therefore$ The given condition is satisfied if the mid-point of the chord lies within the region between the concentric circles of radius $\frac{\sqrt{11}}{6}$ and $\frac{\sqrt{5}}{3}$.

Hence the probability $=($ the area of the circular annulus$\left)/\right($ area of the given circle $)$

$=(\frac{5}{9}-\frac{11}{36})=\frac{1}{4}$