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Answer: \(3\pi\) cm (about 9.42 cm)
The area of a sector is related to its arc length \(l\) by
\[A = \frac{1}{2}\,r\,l\]
Substituting \(A = 54\pi\ \mathrm{cm^2}\) and \(r = 36\) cm:
\[54\pi = \frac{1}{2}\times36\times l = 18\,l\]
\[l = \frac{54\pi}{18} = 3\pi\ \text{cm} \approx 9.42\ \text{cm}\]
Check using the angle: \(\dfrac{\theta}{360^\circ}\times\pi(36)^{2} = 54\pi\) gives \(\theta = 15^\circ\), and then \(l = \dfrac{15}{360}\times2\pi\times36 = 3\pi\) cm.
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