Question

The probablilty 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. $\frac{5}{12}$

Answer 1

Answer: (C)

Let $l$ be the length of the chord $AB$ of the given circle of radius $a$ and $r$ be the distance of the mid point $D$ 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 mid point 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
$=\frac{\pi {\left(\frac{\sqrt{5}}{3}a\right)}^2-\pi {\left(\frac{\sqrt{11}}{6}a\right)}^2}{\pi a^2}=\frac{1}{4}$