Answer: (a) \(16.4\text{ cm}\) to 3 significant figures; (b) \(\displaystyle a^2\left(\pi-4\phi\cos^2\phi\right)\).

Answer: (a) \(16.4\text{ cm}\) to 3 significant figures; (b) \(\displaystyle a^2\left(\pi-4\phi\cos^{2}\phi\right)\).

(a) Since \(OA:OC=4:5\) and \(OC=6.5\),

\(OA=\frac45(6.5)=5.2.\)

The perimeter of the shaded region is made from the two arcs and the two straight radial gaps.

The outer arc length is

\(6.5\cdot\frac{3\pi}{8}.\)

The inner arc length is

\(5.2\cdot\frac{3\pi}{8}.\)

Each straight radial gap has length

\(6.5-5.2=1.3.\)

Therefore the perimeter is

\(6.5\cdot\frac{3\pi}{8}+5.2\cdot\frac{3\pi}{8}+2(1.3).\)

This is

\(\frac{35.1\pi}{8}+2.6\approx 16.4.\)

(b) In triangle \(OPR\), \(OP=OR=a\). Since \(\angle OPR=\phi\), the angle at \(R\) is also \(\phi\).

Therefore

\(\angle POR=\pi-2\phi.\)

By the cosine rule,

\(y^2=a^2+a^2-2a^2\cos(\pi-2\phi).\)

Since \(\cos(\pi-2\phi)=-\cos2\phi\),

\(y^2=2a^2+2a^2\cos2\phi=4a^2\cos^{2}\phi.\)

So

\(y=2a\cos\phi.\)

The unshaded sector \(PQR\) has centre \(R\), radius \(y\), and angle \(2\phi\), so its area is

\(\frac12y^2(2\phi)=y^2\phi.\)

The total shaded area is the area of the circle with centre \(O\), minus this sector:

\(\pi a^2-y^2\phi.\)

Substitute \(y^2=4a^2\cos^{2}\phi\):

\(\pi a^2-4a^2\phi\cos^{2}\phi =a^2\left(\pi-4\phi\cos^{2}\phi\right).\)