该图（未按比例绘制）显示了两个以 O 为圆心的同心圆。AB 和 AD 分别是较小圆在 B 和 D 处的切线。AC 是较大圆的直径。

1. State equality of tangent lengths from point A

$AB=AD$ since tangents drawn from the same external point $A$ to the smaller circle are equal.

2. State equality of tangent lengths from point C

$BC=DC$ since tangents drawn from the same external point $C$ to the smaller circle are equal.

3. Use the common side and conclude congruence

$AC$ is common to both triangles.

Therefore, the three corresponding sides are equal:
$AB=AD,\quad BC=DC,\quad AC=AC.$
Hence $\triangle ABC \cong \triangle ADC$ by SSS.

1. Identify the right angle and the known side in triangle AOB

Since $AB$ is a tangent to the smaller circle at $B$, $OB \perp AB$.

So $\angle ABO = 90^\circ$ and $AO = 8$ cm.

2. Find the length of OB using sine in the right-angled triangle

In right-angled triangle $AOB$,
$\sin 42.5^\circ = \frac{OB}{AO}$
$OB = 8\sin 42.5^\circ$

3. Calculate the area of triangle AOB

Using $AB$ as the base and $OB$ as the height, first note that
$AB = 8\cos 42.5^\circ.$

Hence
$\text{Area of }\triangle AOB = \frac12 \times AB \times OB$
$= \frac12 \times (8\cos 42.5^\circ) \times (8\sin 42.5^\circ)$
$= 18.386\ldots$
$\approx 18.4\text{ cm}^2.$

1. Find the radius of the smaller circle and the central angle of the sector

Since $AB$ is a tangent, $OB\perp AB$.

In right-angled $\triangle AOB$,
$OB=8\sin 42.5^\circ=5.406\ldots\text{ cm}.$
Also,
$\angle AOB=90^\circ-42.5^\circ=47.5^\circ.$
By symmetry, $\angle AOD=47.5^\circ$, so
$\angle BOD=47.5^\circ+47.5^\circ=95^\circ.$

2. Find the area of quadrilateral $ABOD$

From part (i),
$\text{area of }\triangle AOB=18.4\text{ cm}^2.$
By symmetry,
$\text{area of quadrilateral }ABOD=2(18.4)=36.8\text{ cm}^2.$

3. Subtract the sector area from the quadrilateral area

Area of sector $BOD$ is
$\frac{95}{360}\pi(5.406\ldots)^2=24.2\ldots\text{ cm}^2.$

Hence shaded area
$=36.8-24.2\ldots=12.6\text{ cm}^2\quad\text{(3 s.f.)}.$