A semicircle OABCDE with centre O has a radius of 4.5 cm. Chord BD has a length of 6 cm and the perimeter of minor sector OAB is 12.785 cm.

1. Find the arc length of minor arc $AB$ from the sector perimeter.

Perimeter of minor sector $OAB = OA + OB + \text{arc }AB$

Since $OA=OB=4.5$ cm,

$12.785 = 4.5+4.5+\text{arc }AB$

$\text{arc }AB = 12.785-9=3.785\text{ cm}$

2. Use the arc length formula to calculate the angle in radians.

Using $s=r\theta$,

$3.785=4.5\theta$

$\theta=\frac{3.785}{4.5}=0.841111\ldots$

So,

$\angle AOB=0.841\text{ rad}$

1. State the circle theorem that justifies the equality.

$OM \perp BD$ and $O$ is the centre of the circle. The perpendicular from the centre of a circle to a chord bisects the chord. Therefore, $M$ is the midpoint of $BD$, so $BM = MD$.

1. Find the angle of sector $OBC$ in radians.

From part (i),
$\angle AOB=0.841\text{ rad}.$
Since $AOC$ is a quarter turn,
$\angle AOC=\frac{\pi}{2}\text{ rad}.$
Hence
$\angle BOC=\frac{\pi}{2}-0.841=0.7298\text{ rad}.$

2. Find the area of sector $OBC$.

Using $\text{area of sector}=\dfrac12 r^2\theta$,
$\text{Area of sector }OBC=\frac12(4.5)^2(0.7298)$
$=7.39\text{ cm}^2.$

3. Subtract the area of triangle $OMB$ from the sector to get the shaded area.

From part (ii), $BM=MD$, so
$BM=\frac{6}{2}=3\text{ cm}.$
In right triangle $OMB$,
$OM=\sqrt{4.5^2-3^2}=\sqrt{11.25}.$
So
$\text{Area of }\triangle OMB=\frac12(3)(\sqrt{11.25})=5.03\text{ cm}^2.$
Therefore,
$\text{shaded area}=7.39-5.03=2.36\text{ cm}^2.$