AC 是以 O 为圆心的圆的直径。$AC = \sqrt{2}r$ 且 $AB = BC$。 COB 是一个以 P 为圆心的半圆弧。 圆的未阴影部分占百分之几？

1. Find the radius of the large circle and the length of $CB$.

Let the radius of the large circle be $R$.

Since $AC$ is a diameter,
$2R=\sqrt{2}r$
$R=\frac{r}{\sqrt{2}}$

Since $AB=BC$, the equal chords subtend equal angles at the centre. Hence $\angle AOB=\angle BOC$.

As $AC$ is a diameter, $\angle AOC=180^\circ$, so
$\angle BOC=90^\circ.$

Therefore, in right-angled triangle $BOC$,
$CB^2=R^2+R^2=2R^2$
$CB=R\sqrt{2}=\frac{r}{\sqrt{2}}\times \sqrt{2}=r.$

2. Find the area of the semicircle with diameter $CB$.

The semicircular arc $COB$ has diameter $CB=r$.

So its radius is
$\frac{r}{2}.$

Area of this semicircle:
$\frac{1}{2}\pi\left(\frac{r}{2}\right)^2=\frac{\pi r^2}{8}.$

3. Find the area of the circular segment of the large circle between chord $CB$ and the minor arc $CB$.

For the large circle, $\angle BOC=90^\circ$.

Area of sector $BOC$:
$\frac{90}{360}\pi R^2=\frac{1}{4}\pi\left(\frac{r}{\sqrt{2}}\right)^2=\frac{\pi r^2}{8}.$

Area of triangle $BOC$:
$\frac{1}{2}R^2=\frac{1}{2}\left(\frac{r}{\sqrt{2}}\right)^2=\frac{r^2}{4}.$

Area of the circular segment:
$\frac{\pi r^2}{8}-\frac{r^2}{4}.$

4. Find the unshaded area.

Unshaded area
$=\text{area of semicircle }COB+\text{area of large circular segment}$
$=\frac{\pi r^2}{8}+\left(\frac{\pi r^2}{8}-\frac{r^2}{4}\right)$
$=\frac{\pi r^2}{4}-\frac{r^2}{4}$
$=\frac{r^2(\pi-1)}{4}.$

5. Express the unshaded area as a percentage of the whole large circle.

Area of the large circle:
$\pi R^2=\pi\left(\frac{r}{\sqrt{2}}\right)^2=\frac{\pi r^2}{2}.$

Percentage not shaded:
$\frac{\frac{r^2(\pi-1)}{4}}{\frac{\pi r^2}{2}}\times 100\%$
$=\frac{\pi-1}{2\pi}\times 100\%$
$=34.1\% \text{ correct to 3 s.f.}$