The diagram shows a circle, centre O, radius 10 cm. AB is a chord of length 16 cm.

1. Form a right-angled triangle using the perpendicular bisector of the chord

Let $M$ be the midpoint of chord $AB$. Since the perpendicular from the centre of a circle to a chord bisects the chord,
$AM=\frac{16}{2}=8\text{ cm}.$

2. Use the sine ratio to find half the central angle, then double it

In right-angled triangle $OAM$,
$\sin \angle AOM=\frac{AM}{OA}=\frac{8}{10}=0.8.$
Hence
$\angle AOM=\sin^{-1}(0.8)=0.927295\ldots\text{ radians}.$
Therefore
$\angle AOB=2(0.927295\ldots)=1.85459\ldots=1.855\text{ radians} \quad \text{(to 4 s.f.)}.$

1. Find the angle subtended by the major arc of the shaded segment

The shaded segment is the larger segment, so it is bounded by the major arc $AB$.

$\text{major angle}=2\pi-1.855$
$=4.428\ldots\text{ radians}$

2. Calculate the length of the major arc

Using $\text{arc length}=r\theta$,

$\text{major arc length}=10(2\pi-1.855)$
$=44.2819\ldots\text{ cm}$

3. Add the chord length to get the perimeter of the shaded segment

$\text{Perimeter}=\text{major arc length}+AB$
$=44.2819\ldots+16$
$=60.2819\ldots$
$\approx 60.3\text{ cm}$

1. Find the area of the major sector corresponding to the shaded segment

From part (a), the minor angle is $1.855$ radians.

Major angle:
$2\pi-1.855$

Area of major sector:
$\frac12(10)^2(2\pi-1.855)$
$=50(2\pi-1.855)$

2. Add the area of triangle $AOB$ to obtain the shaded major segment

Area of triangle $AOB$:
$\frac12(10)(10)\sin 1.855=50\sin 1.855$

Area of shaded segment:
$50(2\pi-1.855)+50\sin 1.855$
$=269.4\ldots$
$\approx 269\text{ cm}^2$