In the diagram, BA and BC are tangents to the circle with centre O. BO meets the circle at D, OA = 10 cm and BC = 28 cm.

1. Use the tangent-radius property and Pythagoras' theorem to find $BO$.

Since $BC$ is a tangent and $OC$ is a radius, $OC \perp BC$.

In right-angled triangle $BOC$,
$BO^2 = BC^2 + OC^2$
$BO^2 = 28^2 + 10^2 = 884$
$BO = \sqrt{884} = 29.732\ldots$

2. Subtract the radius from $BO$ to find $BD$.

Since $D$ is on the circle and $O$ is the centre,
$OD = 10$

Hence,
$BD = BO - OD$
$BD = 29.732\ldots - 10 = 19.732\ldots$
$BD = 19.7\text{ cm} \quad \text{(3 s.f.)}$

1. Find the angle between one radius and $BO$ using the right-angled triangle $AOB$.

Since $BA$ is a tangent, $OA \perp BA$.

In right-angled triangle $AOB$,
$\tan \angle AOB=\frac{AB}{OA}=\frac{28}{10}.$
$\angle AOB=\tan^{-1}\left(\frac{28}{10}\right)=70.346^\circ.$

2. Find the central angle of the sector $AOC$.

The two tangents from $B$ are equal, so $BO$ bisects $\angle AOC$.

Hence
$\angle AOC=2(70.346^\circ)=140.692^\circ.$

3. Find the area of quadrilateral $ABCO$.

Quadrilateral $ABCO$ is made up of two congruent right-angled triangles, $\triangle AOB$ and $\triangle COB$.

$\text{Area of }ABCO=2\left(\frac12 \times 10 \times 28\right)=280\text{ cm}^2.$

4. Subtract the sector area from the quadrilateral area to get the shaded area.

$\text{Area of sector }AOC=\frac{140.692}{360}\times \pi \times 10^2=122.776\text{ cm}^2.$

$\text{Shaded area}=280-122.776=157.224\text{ cm}^2.$

$\text{Shaded area}\approx 157\text{ cm}^2.$