Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.
Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.
You can use Pythagoras' theorem to calculate the fraction of the diameter for which the vertical line
to the semicircle is under half the radius.
$\displaystyle \displaystyle\left(\frac{r}{2}\right)^2+x^2=r^2\Ri ghtarrow\ x^2=r^2-\frac{r^2}{4}=\frac{4r^2-r^2}{4}=\frac{3r^2}{4}$
$\displaystyle \displaystyle\Rightarrow\ x=\frac{\sqrt{3}r}{2}$
This is the base length of two right-angled triangles both sides of the y-axis.
You only need to work with one of them.
To calculate the length of the line segment for which the perpendicular is under half the radius...
$\displaystyle \displaystyle\ r-\frac{\sqrt{3}r}{2}=r\left(1-\frac{\sqrt{3}}{2}\right)$
The probability is the ratio of that length to the radius.
The probability is the ratio of 2(r-x) to 2r,
because if the point is on either line segment of length r-x (on the diameter,
touching the circumference), then the perpendicular height to the semicircle
is less than half the radius.
You can also use trigonometry to find the ratio.
$\displaystyle \displaystyle\ probability=\frac{2(r-x)}{diameter}=\frac{2(r-x)}{2r}=\frac{r-x}{r}=\frac{r\left(1-\frac{\sqrt{3}}{2}\right)}{r}=\left(1-\frac{\sqrt{3}}{2}\right)$