Answer: area \(153\text{ cm}^2\); perimeter \(62.2\text{ cm}\).

(a) Since \(PQ\) is perpendicular to \(AB\), it bisects the chord \(AB\). Therefore

\(AQ=\frac{14}{2}=7.\)

In right-angled triangle \(AOQ\), \(OA=10\) and \(AQ=7\). Hence

\(\sin\angle AOQ=\frac7{10}.\)

So

Since \(P\), \(O\) and \(Q\) are collinear,

\(\angle POA=\pi-\angle AOQ.\)

Thus

\(\angle POA=\pi-0.775397\ldots=2.366195\ldots\)

So \(\angle POA=2.366\) radians correct to 3 decimal places.

(b) The shaded area is made from the sector with angle \(2.366\) together with triangle \(AOP\).

The sector area is

\(\frac12(10)^2(2.366)=118.3.\)

The triangle area is

\(\frac12(10)^2\sin(2.366)=35.0.\)

Therefore the shaded area is

\(118.3+35.0=153.3\ldots\)

So the area is

\(153\text{ cm}^2\)

to 3 significant figures.

(c) The major arc \(PB\) has length

\(10(2.366)=23.66.\)

To find \(AP\), use the chord formula in triangle \(AOP\):

Therefore

The perimeter is the major arc \(PB\), plus \(AP\), plus the two radii making \(20\text{ cm}\). Hence

\(23.66+18.5+20=62.2\text{ cm}\)

to 3 significant figures.