Answer: (a) \(2\pi r+4x+2x\theta\); (b) \(\pi r^2-x^2\theta\); (c) \(r^2(\pi-\theta)\).

Answer: (a) \(2\pi r+4x+2x\theta\); (b) \(\pi r^2-x^2\theta\); (c) \(r^2(\pi-\theta)\).

(a) The perimeter of the shaded region includes the full circumference of the large circle, which is

\(2\pi r.\)

The two unshaded sectors each contribute two radii of length \(x\), so the total straight-line contribution is

\(4x.\)

Each small sector also contributes an arc of length \(x\theta\). There are two such arcs, so their total length is

\(2x\theta.\)

Hence the perimeter is

\(2\pi r+4x+2x\theta.\)

(b) The area of the large circle is

\(\pi r^2.\)

The area of one small sector is

\(\frac12x^2\theta.\)

There are two such sectors, so the total unshaded area is

\(x^2\theta.\)

Therefore the shaded area is

\(\pi r^2-x^2\theta.\)

(c) Since \(r\) and \(\theta\) are constant, the shaded area

\(\pi r^2-x^2\theta\)

is least when \(x\) is as large as possible. This gives \(x=r\), so the least possible area is

\(\pi r^2-r^2\theta=r^2(\pi-\theta).\)