Part 1
$d^2 + \left( \dfrac{r}{2} \right)^2 = r^2$
$d = \sqrt{r^2 - \dfrac{r^2}{4}}$
$d = \dfrac{\sqrt{3}r}{2}$
$P = \dfrac{d}{r} = \dfrac{\sqrt{3}r/2}{r}$
$P = \dfrac{\sqrt{3}}{2} \approx 0.866$ ← [ C ] answer for part 1
Part 2
$P = \dfrac{\pi d^2}{\pi r^2} = \dfrac{(\sqrt{3}r/2)^2}{r^2}$
$P = \dfrac{3}{4} = 0.75$ ← [ D ] answer for part 2
Part 3
If one end of the chord is at A, the other end must be on arc BDC.
$P = \dfrac{4}{6}$
$P = 0.667$ ← [ B ] answer for part 3
