In the diagram, OAB is a sector of a circle with centre O. Given that OA = 12 cm, ∠AOB = $\theta$ radians, OC = 8 cm and the length of the arc AB = 15 cm. (i) Calculate the value of $\theta$. Ans: ___ (ii) Find the area of the shaded region. Ans: ___ cm^2 (iii) Calculate the perimeter of the shaded region. Ans: ___ cm In the diagram, A, B, C and D are four points on a circle, centre O. DO and CB are parallel.
Find, giving reasons for each answer, (i) angle DAB, Ans: ___ ° (ii) angle DOB, Ans: ___ ° (iii) angle ODC. Ans: ___ °

1. Apply the arc length formula to find the angle in radians.

Arc length $AB = 15$ cm and radius $OA = 12$ cm.

Using $s = r\theta$,

$15 = 12\theta$

$\theta = \frac{15}{12} = 1.25$

1. Find the area of sector OAB using the angle from part (i).

Using $\theta=1.25$ radians and radius $12$ cm,
$\text{Area of sector }OAB=\frac12 r^2\theta=\frac12(12)^2(1.25)=90\text{ cm}^2.$

2. Find the area of triangle OAC.

Since $C$ lies on $OB$, $\angle AOC=\angle AOB=1.25$ radians.
$\text{Area of }\triangle OAC=\frac12(12)(8)\sin 1.25=45.55\ldots\text{ cm}^2.$

3. Subtract the triangle area from the sector area to get the shaded area.

$\text{Shaded area}=90-45.55\ldots=44.44\ldots\text{ cm}^2.$

1. Find the straight boundary length $CB$.

Since $O$, $C$ and $B$ are collinear,

$CB=OB-OC=12-8=4\text{ cm}$.

2. Find $AC$ using the cosine rule in triangle $AOC$.

In $\triangle AOC$, $OA=12$, $OC=8$ and $\angle AOC=1.25$ radians.

$AC^2=12^2+8^2-2(12)(8)\cos 1.25$

$AC=\sqrt{12^2+8^2-2(12)(8)\cos 1.25}$

$AC=12.143\ldots\text{ cm}$

3. Add all boundary lengths of the shaded region.

Perimeter of shaded region

$=\text{arc }AB+CB+AC$

$=15+4+12.143\ldots$

$=31.143\ldots\text{ cm}$

$\approx 31.1\text{ cm}$

1. Use the angle at the centre theorem for chord $DB$.

From part (i), $\angle DAB = 70^\circ$.

Since $\angle DAB$ is the angle at the circumference standing on arc $DB$, and $\angle DOB$ is the angle at the centre standing on the same arc,
$\angle DOB = 2\times \angle DAB = 2\times 70^\circ = 140^\circ.$

1. Use the previous central angle to get the angle at the circumference standing on the same chord $DB$.

From part (ii),
$\angle DOB = 140^\circ$.

Since the angle at the centre is twice the angle at the circumference standing on the same chord $DB$,
$\angle DAB=\frac{140^\circ}{2}=70^\circ.$

2. Use opposite angles of a cyclic quadrilateral and the parallel lines to find $\angle ODC$.

In cyclic quadrilateral $ABCD$,
$\angle DCB=180^\circ-\angle DAB=180^\circ-70^\circ=110^\circ.$

Since $DO\parallel CB$, the angle between $DC$ and $DO$ is supplementary to the angle between $DC$ and $CB$:
$\angle ODC=180^\circ-110^\circ=70^\circ.$