Question

Let C be the semicircle of radius r > 0 with center at (0, 0) and lying above the x-axis. For each x in [−r, r], let L(x) be the length of the line from (x, 0) to the semicircle C and perpendicular to the x-axis. What is the probability that L(x) is less than r/2?

Answer 1

Note that the length from (x,0) to the semicircle C and perpendicular to the X-axis is nothing but the height 'y'.

And as we are interested in the upper half of the circle, so -

$y = \sqrt{1-r^2}$

So, the region L(x) < r/2 is nothing but the area of segment shaded below -

The angle subtended by the segment at the center is -

$\theta = 2 \times cos^{-1}({\frac{r/2}{r}}) = \frac{2 \pi}{3}$

Thus, area of segment = area of sector - area of triangle

  $\\ = (\frac{2 \pi / 3}{2 \pi}) \times \pi r^2 - \frac{1}{2} \times r \times r \times sin(120) \\ \\ = \frac{\pi r^2}{3} - \frac{\sqrt{3} r^2}{4}$

Thus, probability of y < l/2 is -

$P = \frac{Area \:\: of \:\: Segment }{Area \:\: of \:\: Semicircle} = \frac{ = \frac{\pi r^2}{3} - \frac{\sqrt{3} r^2}{4}}{0.5 \times \pi r^2} = \frac{\frac{\pi}{3} -\frac{\sqrt{3}}{4} }{0.5 \pi} \approx 0.3910$