In Diagram I below, ABCDE is a regular pentagon, centre O. $OA = OB = 4\text{ cm}$. (Diagram I and Diagram II are shown — see figure_description.)

1. Find one central angle of the regular pentagon

Since $ABCDE$ is a regular pentagon, the 5 angles at the centre are equal.

$\angle AOB = \frac{360^\circ}{5} = 72^\circ$

1. Express the pentagon as 5 congruent triangles with central angle $72^\circ$

Since ABCDE is a regular pentagon, it can be divided into 5 congruent triangles from the centre $O$.

From part (a),

$\angle AOB = 72^\circ$.

Also, $OA=OB=4\text{ cm}$.

2. Calculate the total area using the area formula for a triangle

Area of pentagon

$=5\left(\frac12 \times 4 \times 4 \times \sin 72^\circ\right)$

$=40\sin72^\circ$

$=38.042\ldots$

$\approx 38.0\text{ cm}^2$

1. Find angle BAD using the interior angle of a regular pentagon

Since $ABCDE$ is a regular pentagon,
$\angle BAE=\frac{(5-2)\times180^\circ}{5}=108^\circ.$
In a regular pentagon, diagonal $AD$ is parallel to side $BC$, so
$\angle DAE=\angle CBA=108^\circ.$
Hence
$\angle BAD=\angle BAE-\angle DAE=108^\circ-72^\circ=36^\circ.$

2. Use isosceles triangle ABD to find angle ABD

The arc $AB$ has centre $D$ and radius $AD$, so $DA=DB$.

Therefore triangle $ABD$ is isosceles and
$\angle DAB=\angle ABD.$
Thus
$\angle ABD=36^\circ.$

1. Identify the included angle and equal radii in triangle $OBD$

Since $ABCDE$ is a regular pentagon with centre $O$, $OB=OD=4\text{ cm}$.

The angle from one vertex to the next at the centre is $72^\circ$, so

$\angle BOD=2\times72^\circ=144^\circ.$

2. Apply the cosine rule to find $BD$

In $\triangle BOD$,

$BD^2=OB^2+OD^2-2(OB)(OD)\cos\angle BOD$

$BD^2=4^2+4^2-2(4)(4)\cos144^\circ$

$BD=\sqrt{32-32\cos144^\circ}$

$BD=7.608\ldots$

Therefore, $BD\approx7.61\text{ cm}$.

1. Find the angle at the centre of the arc

From part (c)(i), $\angle ABD=36^\circ$.

Since the arc $AB$ has centre $D$ and radius $AD$, $AD=BD$.

So triangle $ABD$ is isosceles and
$\angle BAD=\angle ABD=36^\circ.$

Hence
$\angle ADB=180^\circ-36^\circ-36^\circ=108^\circ.$

2. Calculate the area of sector $ADB$

Using $BD\approx 7.61\text{ cm}$ from part (c)(ii),
$\text{Area of sector }ADB=\frac{108}{360}\pi(7.61)^2.$

$\text{Area of sector }ADB\approx 54.6\text{ cm}^2.$

3. Subtract the area of triangle $ADB$ to find the shaded segment

$\text{Area of }\triangle ADB=\frac12(7.61)(7.61)\sin108^\circ.$

$\text{Area of }\triangle ADB\approx 27.5\text{ cm}^2.$

Therefore,
$\text{Area of shaded segment}=54.6-27.5=27.1\text{ cm}^2.$

Using more exact calculator values before rounding gives
$\text{Area of shaded segment}\approx 27.0\text{ cm}^2.$

1. Recognise that the face is made up of the pentagon and 5 equal shaded segments

Since the design is based on a regular pentagon, there are $5$ congruent circular segments, one on each side of the pentagon.

From earlier parts,
$\text{area of pentagon } ABCDE = 38.0\text{ cm}^2$
and
$\text{area of one shaded segment} = 27.0\text{ cm}^2.$

2. Add the pentagon area and the areas of the 5 segments

$\text{Area of face} = 38.0 + 5(27.0)$
$= 38.0 + 135.0$
$= 173\text{ cm}^2$