(a) The area of sector ABC is given by \(\frac{1}{2} r^2 \theta\). Substituting \(\theta = \frac{1}{6} \pi\), the sector area is \(\frac{1}{2} r^2 \times \frac{1}{6} \pi = \frac{1}{12} \pi r^2\).

The triangle ABD has base AD and height BD. Using trigonometry, \(BD = r \sin \frac{\pi}{6} = \frac{1}{2} r\) and \(AD = r \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} r\).

The area of triangle ABD is \(\frac{1}{2} \times \frac{1}{2} r \times \frac{\sqrt{3}}{2} r = \frac{\sqrt{3}}{8} r^2\).

Therefore, the area of BCD is \(\frac{1}{12} \pi r^2 - \frac{\sqrt{3}}{8} r^2\).

(b) Given BD = \frac{\sqrt{3}}{2} r, use trigonometry to find angle BAC: \(\angle BAC = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}\).

Then, \(AD = r \cos \frac{\pi}{3} = \frac{1}{2} r\) and \(CD = \frac{1}{2} r\).

The length of arc BC is \(r \times \frac{\pi}{3}\).

Thus, the perimeter of BCD is \(\frac{\sqrt{3}}{2} r + \frac{1}{2} r + \frac{\pi}{3} r\).

(a) To find the perimeter of the shaded region, we first calculate the angles at A and B using trigonometry. We have:

\(\tan A = \frac{12}{5}\) or \(\cos A = \frac{5}{13}\) or \(\sin A = \frac{12}{13}\).

This gives \(A = 1.176\) radians and \(B = 0.3948\) radians.

The length of DE is given as 4 cm.

The arc lengths are calculated as:

Arc on circle A: \(5 \times 1.176 = 5.880\) cm

Arc on circle B: \(8 \times 0.3948 = 3.158\) cm

Thus, the perimeter of the shaded region is \(5.880 + 3.158 + 4 = 13.0\) cm.

(b) To find the area of the shaded region, we calculate the area of triangle ADE and subtract the areas of the sectors:

Area of triangle ADE: \(\frac{1}{2} \times 5 \times 12 = 30\) cm²

Area of sector on circle A: \(\frac{1}{2} \times 5^2 \times 1.176 = 14.70\) cm²

Area of sector on circle B: \(\frac{1}{2} \times 8^2 \times 0.3948 = 12.63\) cm²

Thus, the area of the shaded region is \(30 - 14.70 - 12.63 = 2.67\) cm².

(a) To find angle XYC, we use the sine rule in the right triangle XYC:

\(\sin(\angle XYC) = \frac{9}{11}\)

Thus,

\(\angle XYC = \sin^{-1}\left(\frac{9}{11}\right) \approx 0.9582 \text{ radians}\)

(b) To find the perimeter of ABC, we calculate each segment:

1. Calculate XY using the Pythagorean theorem in triangle XYC:

\(XY = \sqrt{11^2 - 9^2} = \sqrt{40}\)

2. Calculate AB:

\(AB = 9 + 11 - XY = 20 - \sqrt{40}\)

3. Calculate arc AC:

\(\text{Arc } AC = 11 \times 0.9582 = 10.5402\)

4. Calculate arc BC:

\(\text{Arc } BC = 9 \times \frac{\pi}{2} \approx 14.1372\)

5. Add all segments to find the perimeter:

\(\text{Perimeter} = AB + \text{Arc } AC + \text{Arc } BC = (20 - \sqrt{40}) + 10.5402 + 14.1372 \approx 38.4\)

(a) To find \(\angle ABC\), we can use trigonometry. Since \(P\) is the foot of the perpendicular from \(C\) to \(AB\), \(AP = 12\) cm and \(BP = 3\) cm. Therefore, \(\tan \angle ABC = \frac{9}{3} = 3\). Using the inverse tangent function, \(\angle ABC = \tan^{-1}(3) \approx 1.249\) radians, which rounds to \(1.25\) radians to 3 significant figures.

(b) To find the area of the shaded region, we first find \(BC\) using the sine rule: \(BC = \frac{9}{\sin 1.25} \approx 9.4869\) cm.

The area of sector \(BCQ\) is \(\frac{1}{2} \times (BC)^2 \times \angle ABC = \frac{1}{2} \times 9.4869^2 \times 1.25 \approx 56.207\) square cm.

The area of triangle \(PBC\) is \(\frac{1}{2} \times 9 \times 3 = 13.5\) square cm.

Therefore, the area of the shaded region is \(56.207 - 13.5 = 42.7\) square cm (or \(42.8\) square cm depending on rounding).

(a) Recognize that at least one of the angles \(A, B, C\) is \(\frac{\pi}{3}\) (60°). Each arc is \(\frac{1}{6}\) of a full circle. The circumference of a full circle with radius 6 cm is \(12\pi\). Therefore, the length of each arc is \(\frac{1}{6} \times 12\pi = 2\pi\). Since there are two arcs, the total arc length is \(4\pi\). The perimeter is the sum of the straight line and the arcs: \(6 + 4\pi\).

(b) The area of each sector is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\). The area of the triangle formed by the radii and the line \(AB\) is \(\frac{1}{2} \times 6 \times 6 \times \sin\left(\frac{\pi}{3}\right) = 9\sqrt{3}\). The area of the plate is the sum of the areas of the sectors minus the area of the triangle: \(2 \times 6\pi - 9\sqrt{3} = 12\pi - 9\sqrt{3}\).