Answer: (a) \(a=15-\frac{5\pi}{2}\), (b) \(24.1\) cm.

Answer: (a) \(a=15-\frac{5\pi}{2}\), (b) \(24.1\) cm.

(a) The area of sector \(AOB\) is

Since \(CB=a\) and \(OB=15\),

\(OC=15-a.\)

The area of triangle \(AOC\) is

\(\frac12(15)(15-a)\sin\frac{\pi}{6}.\)

Since \(\sin\frac{\pi}{6}=\frac12\), this area is

\(\frac{15(15-a)}{4}.\)

The shaded region \(ACB\) is the sector minus triangle \(AOC\). The question says that the area of triangle \(AOC\) is equal to the shaded area, so

Therefore

\(\frac{30(15-a)}{4}=\frac{75\pi}{4}.\)

Hence

\(15-a=\frac{5\pi}{2},\)

so

\(a=15-\frac{5\pi}{2}.\)

(b) The perimeter of the shaded region is

\(AC+\text{arc }AB+CB.\)

The arc length \(AB\) is

\(15\cdot\frac{\pi}{6}=\frac{5\pi}{2}.\)

Also \(CB=15-\frac{5\pi}{2}\), so

\(\text{arc }AB+CB=15.\)

Now \(OA=15\), \(OC=\frac{5\pi}{2}\), and angle \(AOC=\frac{\pi}{6}\). By the cosine rule,

Thus

\(AC\approx9.1.\)

Therefore the perimeter is approximately

\(9.1+15=24.1\text{ cm}.\)