Answer: (a) perimeter \(80.7\) cm; (b) area \(340\) cm\(^2\) or \(341\) cm\(^2\), depending on rounding.

For a sector, use radians throughout with arc length \(s=r\theta\) and area \(A=\frac12r^2\theta\).

(a) The arc length formula is

\(s=r\theta\).

Here \(s=27\) and \(r=12\), so

\(27=12\theta\).

Thus

\(\theta=\frac{27}{12}=\frac94\).

This is the angle \(\angle AOB\).

Since \(C\), \(O\) and \(A\) lie on a straight line,

\(\angle COB=\pi-\theta\).

Triangle \(OCB\) is right-angled at \(B\), and \(OB=12\). Therefore

\(\tan(\pi-\theta)=\frac{CB}{12}\).

So

\(CB=12\tan\left(\pi-\frac94\right)\).

This gives

\(CB\approx14.86\).

The perimeter consists of \(OA\), the arc \(AB\), \(BC\), \(CD\), and \(DO\).

Since \(OA=DO=12\) and \(BC=CD\), the perimeter is

\(12+27+CB+CB+12\).

Hence

\(\text{perimeter}=24+27+2(14.86)\approx80.7\).

So the perimeter is \(80.7\) cm to 3 significant figures.

(b) The area of the sector \(OAB\) is

\(\frac12r^2\theta=\frac12(12)^2\left(\frac94\right)=162\).

The kite area is twice the area of triangle \(OCB\). Since \(OB\perp CB\),

\(\text{area of kite}=2\cdot\frac12\cdot12\cdot CB=12CB\).

Using \(CB\approx14.86\),

\(12CB\approx178.3\).

Therefore the total area is approximately

\(162+178.3=340.3\).

So the area is about \(340\) cm\(^2\) to 3 significant figures.