Answer: \(P=2r+\frac{18}{r}\); stationary value \(12\text{ cm}\), a minimum.

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The area of a sector is \(\frac12r^2\alpha\). Since the area is \(9\),

\(\frac12r^2\alpha=9\), so \(r^2\alpha=18\).

The arc length is \(r\alpha\). From \(r^2\alpha=18\), we have \(r\alpha=\frac{18}{r}\).

The perimeter is the two radii plus the arc length, so

\(P=2r+\frac{18}{r}\).

Differentiate:

\(\frac{\mathrm dP}{\mathrm dr}=2-\frac{18}{r^2}.\)

For a stationary value, set this equal to zero:

\(2-\frac{18}{r^2}=0\), so \(r^2=9\). Since \(r\gt0\), \(r=3\).

Then \(P=2(3)+\frac{18}{3}=12\).

Also, \(\frac{\mathrm d^2P}{\mathrm dr^2}=\frac{36}{r^3}\), which is positive at \(r=3\). Therefore the stationary value is a minimum.

This gives the required answer: \(P=2r+\frac{18}{r}\); stationary value \(12\text{ cm}\), a minimum.