(a) To find angle BAD, use the formula for the area of a sector: \(\frac{1}{2} \times r^2 \times \theta = 10\). Here, \(r = 4\) cm and \(\theta\) is the angle in radians.

\(\frac{1}{2} \times 4^2 \times \theta = 10\)

\(8 \theta = 10\)

\(\theta = \frac{10}{8} = 1.25\) radians

(b) To find the perimeter of the shaded region, calculate the arc length BD and the lengths BC and CD.

Arc BD = \(4 \times 1.25 = 5\) cm

Using \(BC = 4 \tan(1.25)\), calculate \(BC\).

\(BC \approx 12.04\) cm

Using \(CD = \frac{4}{\cos(1.25)} - 4\), calculate \(CD\).

\(CD \approx 8.69\) cm

Perimeter = Arc BD + BC + CD = \(5 + 12.04 + 8.69 = 25.7\) cm

(a) By symmetry, there are six sectors around the diagram that make up a complete circle. Therefore, \(6 \times \angle PAQ = 2\pi\), so \(\angle PAQ = \frac{2\pi}{6} = \frac{\pi}{3}\) radians.

(b) Each straight section of rope has length 40 cm. Each curved section around each pipe has length \(r\theta = 20 \times \frac{\pi}{3}\). Total length = \(6 \times (40 + \frac{20\pi}{3}) = 240 + 40\pi \approx 366\) cm.

(c) The area of one equilateral triangle formed by the centers is \(\frac{1}{2} \times 40 \times 40 \times \sin(\frac{\pi}{3}) = 400 \sqrt{3}\). Total area of hexagon = \(6 \times 400 \sqrt{3} = 2400 \sqrt{3}\) cm\(^2\).

(d) The area of the complete region enclosed by the rope includes the hexagon and the sectors. Total area = \(2400 \sqrt{3} + 4800 + 400\pi \approx 7200\) cm\(^2\).

(a) Each arc is part of a circle with a central angle of 60 degrees (since ABC is equilateral). The length of one arc is \(\frac{60}{360} \times 2\pi \times 2 = \frac{2\pi}{3}\). The perimeter of the coin is the sum of the lengths of the three arcs: \(3 \times \frac{2\pi}{3} = 2\pi\).

(b) The area of one sector is \(\frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}\). The area of the equilateral triangle is \(\frac{1}{2} \times 2 \times 2 \times \sin\left(\frac{\pi}{3}\right) = \sqrt{3}\). The area of the coin is the area of three sectors minus the area of the triangle: \(3 \times \frac{2\pi}{3} - \sqrt{3} = 2\pi - 2\sqrt{3}\).

(a) To show that angle PCS = \(\frac{2}{3} \pi\) radians, consider the triangle PCQ. Since PQ = 4 cm, the triangle PCQ is equilateral, making angle PCQ = \(\frac{\pi}{3}\). Therefore, angle PCS = \(\pi - \frac{\pi}{6} - \frac{\pi}{3} = \frac{2}{3} \pi\).

(b) The perimeter of the plate consists of the lengths of arcs PS and QR, and the circumferences of the semicircles PQ and RS. The length of each arc is \(\frac{2\pi}{3} \times 4 = \frac{8\pi}{3}\). The circumference of each semicircle is \(2 \times 2 = 4\). Therefore, the total perimeter is \(2 \times \frac{8\pi}{3} + 2 \times 2 = \frac{28\pi}{3}\).

(c) The area of the plate is calculated by subtracting the areas of the removed segments from the area of the original circle. The area of sector CPQ is \(\frac{1}{2} \times 4^2 \times \frac{\pi}{3} = \frac{8\pi}{3}\). The area of the segment is \(\frac{8\pi}{3} - \frac{1}{2} \times 4^2 \times \sin\left(\frac{\pi}{3}\right) = \frac{8\pi}{3} - 4\sqrt{3}\). The area of the small semicircle is \(\pi \times 2^2 = 4\pi\). Therefore, the area of the plate is \(\pi \times 4^2 - 2 \times 4\pi - 2 \times \left(\frac{8\pi}{3} - 4\sqrt{3}\right) = \frac{20\pi}{3} + 8\sqrt{3}\).

(a) To find k when \angle BAC = \frac{1}{6}\pi radians:

The area of triangle \Delta ADE is given by:

\(\frac{1}{2} (ka)^2 \sin \frac{\pi}{6} = \frac{1}{4} k^2 a^2\)

The area of sector ABC is:

\(\frac{1}{2} a^2 \frac{\pi}{6}\)

Since DE divides the sector into two equal areas:

\(2 \times \frac{1}{4} k^2 a^2 = \frac{1}{2} a^2 \frac{\pi}{6}\)

Simplifying gives:

\(k^2 = \frac{\pi}{6}\)

\(k = \sqrt{\frac{\pi}{6}} \approx 0.7236\)

(b) For the general case with \angle BAC = \theta:

The area of triangle \Delta ADE is:

\(\frac{1}{2} (ka)^2 \sin \theta\)

Equating twice the area of \Delta ADE to the area of sector ABC:

\(2 \times \frac{1}{2} (ka)^2 \sin \theta = \frac{1}{2} a^2 \theta\)

\(k^2 = \frac{\theta}{2 \sin \theta}\)

Given \(\frac{\theta}{\sin \theta} > 1\), it follows:

\(k^2 > \frac{1}{2}\)

Thus, \(\frac{1}{\sqrt{2}} < k < 1\)