A, B and C are points on the circle centre O and AB = BC. P is the midpoint of chord AB and Q is the midpoint of chord BC.

1. Establish that both triangles are right-angled.

Since $P$ is the midpoint of chord $AB$, $OP \perp AB$.

Since $Q$ is the midpoint of chord $BC$, $OQ \perp BC$.

Hence
$\angle OPA=\angle OQC=90^\circ.$

2. Show one pair of corresponding sides is equal using radii.

$OA=OC$
because both are radii of the same circle.

3. Show the other required pair of corresponding sides is equal, then conclude congruence.

Since $P$ and $Q$ are midpoints,
$AP=\frac12 AB, \qquad CQ=\frac12 BC.$
Given $AB=BC$,
$AP=CQ.$

Therefore,
$\triangle OAP \cong \triangle OCQ$
by RHS congruence.

1. Find the central angle subtended by each equal chord.

Since $AB=BC$, the equal chords subtend equal angles at the centre.

So
$\angle AOB=\angle BOC=\frac12\angle AOC=\frac12\times \frac{7\pi}{9}=\frac{7\pi}{18}.$

2. Find the area of one sector cut off by a chord.

For one sector, with radius $6$ cm and angle $\frac{7\pi}{18}$,
$\text{sector area}=\frac12 r^2\theta$
$=\frac12(6)^2\left(\frac{7\pi}{18}\right)=7\pi.$

3. Find the area of one isosceles triangle inside the sector.

For one triangle, for example $\triangle AOB$,
$\text{triangle area}=\frac12 ab\sin C$
$=\frac12(6)(6)\sin\left(\frac{7\pi}{18}\right)$
$=18\sin\left(\frac{7\pi}{18}\right).$

4. Calculate the total shaded area from the two identical circular segments.

Area of one shaded segment is
$7\pi-18\sin\left(\frac{7\pi}{18}\right).$

There are two identical shaded segments, so
$\text{shaded area}=2\left[7\pi-18\sin\left(\frac{7\pi}{18}\right)\right]$
$=14\pi-36\sin\left(\frac{7\pi}{18}\right)$
$=10.153\ldots$
$\approx 10.2\text{ cm}^2.$