Question

The probability that length of a randomly selected chord of a circle lies between $\frac{1}{2}$ and $\frac{3}{2}$ of its radius is
(correct answer + 1, wrong answer - 0.25)

• A. $\frac{1}{2}$
• B. $\frac{9}{16}$
• C. $\frac{2}{3}$
• D. $\frac{4}{9}$

Answer 1

The correct option is A $\frac{1}{2}$


Let $O$ be the centre and $r$ be the radius of circle
$AB=\frac{3r}{2},CD=\frac{r}{2}$
Perpendicular distance of $AB$ from $O$ is,
$(OM{)}^2=r^2-{\left(\frac{3r}{4}\right)}^2\Rightarrow OM=\frac{\sqrt{7}}{4}r$
similarly, $(ON{)}^2=r^2-{\left(\frac{r}{4}\right)}^2\Rightarrow ON=\frac{\sqrt{15}}{4}r$

Now, if length of a randomly selected chord of a circle lies between $\frac{1}{2}$ and $\frac{3}{2}$ of it's radius, then mid-point of the chord should lie within the region between concentric circles of radii $\frac{\sqrt{7}}{4}r$ and $\frac{\sqrt{15}}{4}r.$
$\therefore P=\frac{\pi \left(\frac{15r^2}{16}\right)-\pi \left(\frac{7}{16}{r}^2\right)}{\pi r^2}$
$=\frac{8}{16}$
$=\frac{1}{2}$