Answer: (a) \(\dfrac{\pi}{3}\). (b) \(2\sqrt3-\dfrac{2\pi}{3}\).

(a) In the circle with centre \(O\), the radius is \(2\) and the chord \(AB\) has length \(2\sqrt3\).

Let

\(\angle AOB=\theta\).

Using the chord formula \(\text{chord}=2r\sin\left(\frac{\theta}{2}\right)\),

\(2\sqrt3=2(2)\sin\left(\frac{\theta}{2}\right)\).

So

\(\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt3}{2}\).

Hence

\(\frac{\theta}{2}=\frac{\pi}{3}\),

so

\(\theta=\frac{2\pi}{3}\).

The point \(Q\) lies on the circle and \(AQ=BQ\), so \(Q\) is on the perpendicular bisector of \(AB\). The angle \(AQB\) is an angle at the circumference subtending the chord \(AB\). Therefore it is half the central angle subtending the same chord:

\(\angle AQB=\frac12\angle AOB=\frac12\cdot\frac{2\pi}{3}=\frac{\pi}{3}\).

(b) Since \(Q\) lies on the original circle and \(AQ=BQ\), we have

\(AQ=BQ=2\sqrt3\).

First find the area of the circular segment of the circle with centre \(O\), cut off by chord \(AB\). This segment has radius \(2\) and angle \(\frac{2\pi}{3}\), so its area is

\(\frac12(2)^2\left(\frac{2\pi}{3}\right)-\frac12(2)^2\sin\left(\frac{2\pi}{3}\right)\).

This simplifies to

\(\frac{4\pi}{3}-\sqrt3\).

Next find the area of the circular segment of the circle with centre \(Q\), cut off by the same chord \(AB\). This circle has radius \(2\sqrt3\) and angle \(\frac{\pi}{3}\), so this segment has area

\(\frac12(2\sqrt3)^2\left(\frac{\pi}{3}\right)-\frac12(2\sqrt3)^2\sin\left(\frac{\pi}{3}\right)\).

This simplifies to

\(2\pi-3\sqrt3\).

The shaded region is the difference between these two segments, so its area is

\(\left(\frac{4\pi}{3}-\sqrt3\right)-\left(2\pi-3\sqrt3\right)\).

Therefore

\(\text{area}=2\sqrt3-\frac{2\pi}{3}\).

Answer: \(\angle AOB=2.46\text{ rad}\), perimeter \(=37.5\text{ cm}\), and area \(=111\text{ cm}^2\) approximately.

Since \(AC\) and \(BD\) are each \(12\text{ cm}\) and are bisected by \(O\), each radius of the circle is

\(OA=OB=OC=OD=6\text{ cm}\).

Let the smaller angle \(\angle BOC\) be \(\theta\). The chord \(BC\) has length \(4\text{ cm}\), so

\(4=2(6)\sin\left(\frac{\theta}{2}\right)\).

Thus

\(\sin\left(\frac{\theta}{2}\right)=\frac13\),

and

\(\theta=2\sin^{-1}\left(\frac13\right)=0.6796\ldots\text{ rad}\).

Because \(A\) is opposite \(C\) and \(B\) is opposite \(D\), the obtuse angle \(AOB\) is

\(\angle AOB=\pi-\theta=\pi-2\sin^{-1}\left(\frac13\right)=2.4619\ldots\text{ rad}\).

Hence

\(\angle AOB=2.46\text{ rad}\quad\text{to 3 significant figures.}\)

The perimeter consists of the two straight sides \(AD\) and \(BC\), each of length \(4\), and two equal arcs, each with radius \(6\) and angle \(\angle AOB\). Therefore

\(\text{perimeter}=4+4+2(6)(2.4619\ldots)=37.54\ldots\text{ cm}\).

So the perimeter is

\(37.5\text{ cm}\quad\text{to 3 significant figures.}\)

For the area, it is helpful to view the shaded region as the area of the whole circle minus two identical small circular segments cut off by the chords \(AD\) and \(BC\).

For one of these small segments, the central angle is \(\theta=0.6796\ldots\). Its area is

\(\frac12(6)^2\theta-\frac12(6)^2\sin\theta=18\theta-18\sin\theta\).

So the shaded area is

\(36\pi-2(18\theta-18\sin\theta)=36\pi-36\theta+36\sin\theta\).

Using \(\theta=2\sin^{-1}\left(\frac13\right)\),

\(\text{area}=111.26\ldots\text{ cm}^2\).

Hence the area is

\(111\text{ cm}^2\quad\text{to 3 significant figures.}\)

Answer: \(\theta=1.85\text{ rad}\), perimeter \(=18.5\text{ cm}\), area \(=22.4\text{ cm}^2\) approximately.

(a) In the isosceles triangle, the two equal sides are \(5\text{ cm}\) and the base is \(8\text{ cm}\). Bisecting the angle \(\theta\) splits the triangle into two right-angled triangles.

Each right-angled triangle has hypotenuse \(5\) and opposite side \(4\) for the angle \(\dfrac\theta2\). Therefore

\(\sin\dfrac\theta2=\dfrac45.\)

Hence

\(\dfrac\theta2=\sin^{-1}\left(\dfrac45\right),\)

so

\(\theta=2\sin^{-1}\left(\dfrac45\right)=1.854\ldots\text{ rad}.\)

Thus

\(\theta=1.85\text{ rad}\)

to 3 significant figures.

(b)(i) Each arc has radius \(5\text{ cm}\) and angle \(\theta\) radians. The length of one arc is

\(r\theta=5\theta.\)

There are two equal arcs, so the perimeter is

\(2(5\theta)=10\theta.\)

Using \(\theta=1.854\ldots\),

\(\text{perimeter}=10(1.854\ldots)=18.54\ldots\text{ cm}.\)

Therefore the perimeter is

\(18.5\text{ cm}\)

to 3 significant figures.

(b)(ii) The area of one circular segment is the area of the sector minus the area of the isosceles triangle:

\(\dfrac12r^2\theta-\dfrac12r^2\sin\theta=\dfrac12r^2(\theta-\sin\theta).\)

Here \(r=5\), and the shape consists of two identical segments. So

\(\text{area}=2\cdot\dfrac12(5)^2(\theta-\sin\theta)=25(\theta-\sin\theta).\)

Using \(\theta=1.854\ldots\),

\(\text{area}=25(1.854\ldots-\sin1.854\ldots)=22.36\ldots\text{ cm}^2.\)

Therefore the area is

\(22.4\text{ cm}^2\)

to 3 significant figures.