(a) Since the total angle around point O is 2π radians, and angle AOB is 2.8 radians, angle ACO is given by:

\(\angle ACO = \pi - 2.8 = 0.7 \text{ radians}\)

(b) Using the sine rule in triangle ACO, we have:

\(\frac{R}{\sin(2.8)} = \frac{r}{\sin(0.7)}\)

Solving for R gives:

\(R = \frac{r \cdot \sin(2.8)}{\sin(0.7)} \approx 1.53r\)

(c) The area of sector OAB is:

\(\text{Area of sector } OAB = \frac{1}{2} r^2 \times 2.8 = 1.4r^2\)

The area of sector CAB is:

\(\text{Area of sector } CAB = \frac{1}{2} R^2 \times 2 \times 0.7 = 1.638r^2\)

The area of triangle OAC is:

\(\text{Area of } \triangle OAC = \frac{1}{2} r^2 \sin(\pi - 2.8) = 0.4927r^2\)

The area of the shaded region is:

\(1.4r^2 - (1.638r^2 - 0.985r^2) = 0.747r^2 \text{ to } 0.748r^2\)

(a) To find the perimeter of the shaded segment, we first determine the radius of the circle. The perimeter of the sector is given by the sum of the arc length and twice the radius:

\(2r + 8 = 20\)

Solving for \(r\), we get \(r = 6\) cm.

The angle \(\angle AOB\) in radians is given by:

\(\frac{8}{6} = \frac{4}{3} \text{ radians}\)

Using the cosine rule in triangle \(OAB\), the length of \(AB\) is:

\(AB = \sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos \left( \frac{4}{3} \right)}\)

Calculating this gives \(AB \approx 7.42\) cm.

Thus, the perimeter of the shaded segment is:

\(7.42 + 8 = 15.4 \text{ cm}\)

(b) To find the area of the shaded segment, we calculate the area of the sector and subtract the area of triangle \(OAB\).

The area of the sector is:

\(\frac{1}{2} \times 6^2 \times \frac{4}{3}\)

The area of triangle \(OAB\) is:

\(\frac{1}{2} \times 6 \times 6 \times \sin \left( \frac{4}{3} \right)\)

Calculating these gives the sector area as \(24\) cm² and the triangle area as \(17.49\) cm².

Thus, the area of the shaded segment is:

\(24 - 17.49 = 6.51 \text{ cm}^2\)

First, find the length of BC using the tangent of angle 60 degrees (\(\frac{\pi}{3}\) radians):

\(\tan 60 = \frac{BC}{6}\)

\(BC = 6\sqrt{3}\)

Next, calculate the area of triangle OBC:

\(\text{Area of } \triangle OBC = \frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

Calculate the area of sector AOC:

\(\text{Area of sector } AOC = \frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\)

Finally, find the area of the shaded region by subtracting the area of the sector from the area of the triangle:

\(\text{Area of shaded region} = 18\sqrt{3} - 6\pi\)

(i) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OCD.

The area of the sector is given by \(\frac{1}{2} r^2 \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the area of the sector is \(\frac{1}{2} \times 6^2 \times 0.8 = 14.4 \text{ cm}^2\).

The area of triangle OCD is given by \(\frac{1}{2} ab \sin C\), where \(a = b = 10 \text{ cm}\) and \(C = 0.8 \text{ radians}\). Thus, the area of the triangle is \(\frac{1}{2} \times 10^2 \times \sin 0.8 = 35.9 \text{ cm}^2\).

The shaded area is \(14.4 - 35.9 = 21.5 \text{ cm}^2\).

(ii) To find the perimeter of the shaded region, we calculate the arc length and the length of CD.

The arc length is given by \(s = r \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the arc length is \(6 \times 0.8 = 4.8 \text{ cm}\).

The length of CD can be found using the cosine rule or \(2 \times 10 \times \sin 0.4\). Thus, \(CD = 7.8 \text{ cm}\).

The perimeter of the shaded region is \(8 + 4.8 + 7.8 = 20.6 \text{ cm}\).

(i) The perimeter of the sector is given by the sum of the two radii and the arc length: \(2r + r\theta = 20\). Solving for \(\theta\), we have:

\(20 = 2r + r\theta\)

\(r\theta = 20 - 2r\)

\(\theta = \frac{20}{r} - 2\)

(ii) The area of the sector is given by \(\frac{1}{2}r^2\theta\). Substituting \(\theta = \frac{20}{r} - 2\), we get:

\(A = \frac{1}{2}r^2\left(\frac{20}{r} - 2\right)\)

\(A = \frac{1}{2}(20r - 2r^2)\)

\(A = 10r - r^2\)

(iii) Using the cosine rule in triangle \(OPQ\), where \(r = 8\) and \(\theta = 0.5\) radians:

\(PQ^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos(0.5)\)

\(PQ^2 = 64 + 64 - 128 \times \cos(0.5)\)

\(PQ = \sqrt{128 - 128 \times \cos(0.5)}\)

Alternatively, using the sine rule:

\(PQ = 2 \times 8 \times \sin(0.25)\)

\(PQ \approx 3.96\) (allow 3.95)

(i) For \(\theta = 1\), angle BOC = \(\pi - \theta\). The area of sector BOC is given by \(\frac{1}{2} r^2 (\pi - \theta)\). Substituting \(r = 8\) and \(\theta = 1\), the area is \(\frac{1}{2} \times 8^2 \times (\pi - 1) = 32(\pi - 1)\) or approximately 68.5.

(ii) The perimeter of sector AOB is \(8 + 8 + 8\theta\) and the perimeter of sector BOC is \(8 + 8 + 8(\pi - \theta)\). Setting \(8 + 8 + 8\theta = \frac{1}{2}(8 + 8 + 8(\pi - \theta))\), solve for \(\theta\):

\(16 + 8\theta = 8 + 8 + 4\pi - 4\theta\)

\(12\theta = 4\pi - 16\)

\(\theta = \frac{1}{3}(\pi - 2)\) or approximately 0.381.

(iii) For \(\theta = \frac{1}{3}\pi\), AB = 8 cm. Using the cosine rule or trigonometry, BC = \(2 \times 8 \sin \frac{\pi}{3} = 8\sqrt{3}\). The perimeter of triangle ABC is \(8 + 8 + 8\sqrt{3} = 24 + 8\sqrt{3}\) cm.

(i) To find the area of the shaded region, we need to calculate the area of triangle OQR and subtract the area of sector OPQ.

The length QR is given by QR = r \tan \theta.

The area of triangle OQR is \(\frac{1}{2} \times r \times r \tan \theta = \frac{1}{2}r^2 \tan \theta\).

The area of sector OPQ is \(\frac{1}{2}r^2 \theta\).

Thus, the area of the shaded region is \(\frac{1}{2}r^2 \tan \theta - \frac{1}{2}r^2 \theta = \frac{1}{2}r^2(\tan \theta - \theta)\).

(ii) Given θ = 0.8 and r = 15, we calculate the perimeter of the shaded region.

The arc length PQ is \(r \theta = 15 \times 0.8 = 12 \text{ cm}\).

The length OR is \(\frac{r}{\cos \theta} = \frac{15}{\cos 0.8} \approx 21.53 \text{ cm}\).

The perimeter of the shaded region is \(r \tan \theta + \text{arc } PQ + (r - r \cos \theta)\).

Substituting the values, we get \(15 \tan 0.8 + 12 + (15 - 15 \cos 0.8) \approx 34.0 \text{ cm}\).

(i) To find angle AOB, use the sine rule in the right triangle formed by the radius and half of AB. The sine of half the angle is given by:

\(\sin(\frac{1}{2} \text{angle}) = \frac{16}{20}\)

Solving for the angle, we find:

\(\text{Required angle} = 1.855 \text{ radians}\)

(ii) The area of sector AXBO is calculated using the formula:

\(\frac{1}{2} r^2 \theta\)

Substituting the values, we get:

\(\frac{1}{2} \times 20^2 \times 1.855 = 371 \text{ cm}^2\)

(iii) The area of the new cross-section is found by subtracting the area of the rectangle and the sector from the area of the circle, and then adding the area of the triangle:

\(\pi r^2 - l \times b - \frac{1}{2} r^2 \theta + \frac{1}{2} bh\)

Substituting the values, we get:

\(\pi \times 20^2 - 32 \times 18 - \frac{1}{2} \times 20^2 \times 1.855 + \frac{1}{2} \times 32 \times 18 = 502 \text{ cm}^2\) (accept 501)

(a) To find the perimeter of the shaded region, we first need to find the length of AC. Using the cosine rule:

\(AC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(1.8)\)

\(AC = 9.40 \text{ cm}\)

Next, calculate the angle CAB:

\(\text{Angle } CAB = \frac{1}{2}(\pi - 1.8)\)

\(\text{Angle } CAB = 0.6708 \text{ radians}\)

Now, find the length of arc CD:

\(\text{Arc } CD = 9.40 \times 0.6708 = 6.306 \text{ cm}\)

The perimeter of the shaded region is:

\(6 + 3.40 + 6.306 = 15.7 \text{ cm}\)

(b) To find the area of the shaded region, calculate the area of sector ADC and subtract the area of triangle ABC:

\(\text{Area of sector } ADC = \frac{1}{2} \times 9.40^2 \times 0.6708\)

\(\text{Area of triangle } ABC = \frac{1}{2} \times 6^2 \times \sin(1.8)\)

\(\text{Area of shaded region} = 29.64 - 17.53 = 12.1 \text{ cm}^2\)

(a) To find angle BOC, we use the formula for arc length: \(s = r\theta\). Given the arc length AB is 2 cm and radius is 10 cm, we have:

\(\theta = \frac{s}{r} = \frac{2}{10} = 0.2 \text{ radians}\)

Angle BOC is the sum of angle AOC and angle AOB:

\(\angle BOC = \angle AOC + \angle AOB = \frac{1}{6} \pi + 0.2 \approx 0.724 \text{ radians}\)

(b) To find the area of the shaded region BPC, we first find the lengths of BP and OP using trigonometry in triangle OBP:

\(BP = 10 \sin \left( \frac{5\pi + 6}{30} \right) \approx 6.6208\)

\(OP = 10 \cos \left( \frac{5\pi + 6}{30} \right) \approx 7.494\)

The area of triangle OBP is:

\(\text{Area of } \triangle OBP = \frac{1}{2} \times 10 \times 10 \times \sin \left( \frac{5\pi + 6}{30} \right) \approx 24.809 \text{ cm}^2\)

The area of sector BOC is:

\(\text{Area of sector BOC} = \frac{1}{2} \times 10^2 \times \left( \frac{5\pi + 6}{30} \right) \approx 36.1799 \text{ cm}^2\)

The area of the shaded region BPC is:

\(\text{Area of region BPC} = \text{Area of sector BOC} - \text{Area of } \triangle OBP \approx 11.4 \text{ cm}^2\)

(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: \(2 \times 6\pi - 9\sqrt{3} = 12\pi - 9\sqrt{3}\).

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

(a) To find angle BAO, use the cosine rule in triangle AOB:

\(\cos BAO = \frac{8^2 + 12^2 - 8^2}{2 \times 8 \times 12}\)

\(\cos BAO = \frac{6}{8}\)

Thus, \(BAO = 0.723\) radians.

(b) The area of the shaded region is the area of sector ABC minus the area of triangle AOB.

Area of sector ABC:

\(\frac{1}{2} \times 12^2 \times 0.7227 = 52.0\)

Area of triangle AOB:

\(\frac{1}{2} \times 8 \times 12 \times \sin(0.7227) = 31.7\)

Shaded area:

\(52.0 - 31.7 = 20.3\)

(c) The perimeter of the shaded region is the sum of arc BC and line segments OC and AB.

Arc BC:

\(12 \times 0.7227 = 8.67\)

Perimeter:

\(8 + 4 + 8.67 = 20.7\)

(a) The area of triangle ABC can be calculated using the formula for the area of a triangle: \(\frac{1}{2} r^2 \sin(2θ)\).

The area of sector ABD is \(\frac{1}{2} r^2 θ\).

Therefore, the area of the shaded region is the area of triangle ABC minus the area of sector ABD: \(\frac{1}{2} r^2 \sin(2θ) - \frac{1}{2} r^2 θ = \frac{1}{2} r^2 (\sin 2θ - θ)\).

(b) The arc BD is given by \(rθ = 10 \times 0.6 = 6\) cm.

The length AC is \(2r \cos θ = 2 \times 10 \times \cos 0.6 = 16.506\) cm.

The length DC is \(AC - r = 16.506 - 10 = 6.506\) cm.

The perimeter of the shaded region is the sum of the arc BD, and the lengths DC and CD: \(6 + 6.506 + 10 = 22.5\) cm.

(a) Using the relationship \(\sin \theta = \frac{r}{OC} \\), we have \(OC = \frac{r}{\sin \theta} \\). Therefore, \(CD = r + \frac{r}{\sin \theta} \\).

(b) The radius of arc AB is \(4 + \frac{4}{\sin \frac{\pi}{6}} = 4 + 8 = 12 \\). The arc length \(AB = 12 \times \frac{2\pi}{6} = 4\pi \\). Thus, the perimeter is \(24 + 4\pi \\).

(c) The area of \(\triangle FOC = \frac{1}{2} \times 4 \times OC \times \sin \frac{\pi}{3} = 8\sqrt{3} \\). The area of sector \(FOE = \frac{1}{2} \times \frac{2\pi}{3} \times 4^2 = \frac{16\pi}{3} \\). Therefore, the shaded area is \(16\sqrt{3} - \frac{16\pi}{3} \\).

First, calculate the angle \(\angle POA\) using the cosine rule:

\(\cos \angle POA = \frac{5}{13}\)

\(\angle POA = \cos^{-1} \left( \frac{5}{13} \right) \approx 1.17 \text{ radians}\) (or 67.4°)

The reflex angle \(\angle AOB = 2\pi - 2 \times 1.17 \approx 3.93 \text{ radians}\).

The length of the major arc AB is:

\(5 \times 3.93 = 19.65 \text{ cm}\)

Calculate the length of AP (or BP) using the Pythagorean theorem:

\(AP = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}\)

The total length of the cord is:

\(19.65 + 12 + 12 = 43.7 \text{ cm}\)

(a) To find \(BC\), use the cosine rule in triangle \(OBC\):

\(BC^2 = r^2 + (2r)^2 - 2 \cdot r \cdot 2r \cdot \cos\left(\frac{\pi}{6}\right)\)

\(BC^2 = r^2 + 4r^2 - 4r^2 \cdot \frac{\sqrt{3}}{2}\)

\(BC^2 = 5r^2 - 2r^2\sqrt{3}\)

\(BC = r\sqrt{5 - 2\sqrt{3}}\)

(b) The perimeter of the shaded region is the arc length of \(AB\) plus \(AC\) and \(BC\):

Arc length \(AB = \frac{2\pi r}{6}\)

\(AC = r\)

Perimeter = \(\frac{2\pi r}{6} + r + r\sqrt{5 - 2\sqrt{3}}\)

(c) The area of the shaded region is the area of the sector minus the area of triangle \(OAB\):

Sector area = \(\frac{1}{2} \cdot 4r^2 \cdot \frac{\pi}{6}\)

Triangle area = \(\frac{1}{2} \cdot r \cdot 2r \cdot \sin\left(\frac{\pi}{6}\right)\)

Shaded area = \(r^2 \left( \frac{\pi}{3} - \frac{1}{2} \right)\)

First, find the angle \\ \angle AOB \\ using the arc length formula: \\ \text{Arc length} = r \theta \\ where \\ r = 6 \\ and \\ \text{Arc length} = 15. \\ Thus, \\ \theta = \frac{15}{6} = 2.5 \\text{ radians}.

The angle \\ \angle BOC = \pi - 2.5 \\text{ radians}.

Using the cosine rule in triangle \\ \triangle BOC, \\ BC = 6(\pi - 2.5) = 3.850 \\text{ cm}.

Using the sine rule, \\ \sin(\pi - 2.5) = \frac{BX}{6}, \\ so \\ BX = 6 \sin(\pi - 2.5) = 3.59 \\text{ cm}.

To find \\ OX, \\ use either \\ OX = 6 \cos(\pi - 2.5) \\ or Pythagoras' theorem: \\ OX = 4.807 \\text{ cm}.

Finally, \\ XC = 6 - OX = 1.193 \\text{ cm}.

The perimeter \\ P = BX + XC + BC = 3.59 + 1.193 + 3.850 = 8.63 \\text{ cm}.

First, calculate the length of OC using the cosine of angle AOB:

\(OC = 6 \cos 0.8 = 4.18 \text{ cm}\).

Next, find the area of sector OCD:

\(\text{Area of sector } OCD = \frac{1}{2} \times (4.18)^2 \times 0.8\).

Then, calculate the area of triangle OCA:

\(\Delta OCA = \frac{1}{2} \times 6 \times 4.18 \times \sin 0.8\).

Finally, find the required area of the shaded region by subtracting the area of sector OCD from the area of triangle OCA:

\(\text{Required area} = \Delta OCA - \text{sector } OCD\).

The calculated area is approximately 2.01 cm².

(i) Since arc OC is part of a circle with centre A, the angle CAO is \(\frac{\pi}{3}\) radians.

(ii) The area of sector AOC is given by \(\frac{1}{2} r^2 \times \frac{\pi}{3}\).

The area of triangle ABC is \(\frac{1}{2} (r)(2r) \sin\left(\frac{\pi}{3}\right)\).

Using \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), the area of triangle ABC becomes \(\frac{1}{2} (2r^2) \frac{\sqrt{3}}{2} = \frac{r^2 \sqrt{3}}{2}\).

The area of the shaded region is the area of triangle ABC minus the area of sector AOC, which is \(\frac{r^2 \sqrt{3}}{2} - \frac{1}{2} r^2 \frac{\pi}{3}\).

(a) The angle \(ACB\) is given as \(k\pi\) radians. From the mark scheme, \(k = \frac{2}{3}\).

(b) The perimeter of the shaded motif is the circumference of the circle with radius \(r\), which is \(2\pi r\).

(c) To find the area of the shaded motif:

- The area of the major sector \(OAB\) is \(\frac{1}{2} r^2 \times \frac{4\pi}{3} = \frac{2\pi r^2}{3}\).

- The area of the minor sector \(CAOB\) is \(\frac{1}{2} r^2 \times \frac{\pi}{3} = \frac{\pi r^2}{6}\).

- The area of triangle \(ACB\) is \(\frac{1}{2} r^2 \sin \frac{\pi}{3} = \frac{r^2 \sqrt{3}}{4}\).

- The area of the shaded motif is the area of the major sector minus the area of the minor sector and the triangle: \(\frac{2\pi r^2}{3} - \left( \frac{\pi r^2}{6} + \frac{r^2 \sqrt{3}}{4} \right) = \frac{\pi r^2}{3} + \frac{r^2 \sqrt{3}}{2}\).

(i) The arc length AB is given by \(2r\theta\). The tangent segments AT and BT are equal, and each is \(r \tan \theta\) because \(\tan \theta = \frac{AT}{r}\). Therefore, the perimeter \(P\) of the shaded region is:

\(P = 2r\theta + 2r \tan \theta\)

(ii) For r = 5 and \theta = 1.2, the area of triangle AOT is \(\frac{1}{2} \times 5 \times 5 \times \tan 1.2\). The area of sector AOB is \(\frac{1}{2} \times 5^2 \times 2.4\). The shaded area is twice the area of triangle AOT minus the area of sector AOB:

\(\text{Area} = 2 \times \left( \frac{1}{2} \times 5 \times 5 \times \tan 1.2 \right) - \frac{1}{2} \times 25 \times 2.4\)

Calculating gives:

\(\text{Area} = 34.3 \text{ cm}^2\)

(i) The length of the arc AC is given by the formula \(r \theta = 6\), where \(r = OA\) and \(\theta = \frac{3}{8} \pi\). Solving for \(r\), we have:

\(OA \times \frac{3}{8} \pi = 6\)

\(OA = \frac{6 \times 8}{3 \pi} = \frac{16}{\pi} \approx 5.093 \text{ cm}\)

(ii) The perimeter of the shaded region is the sum of the lengths of AB, BC, and the arc AC. Since AB and BC are tangents, \(AB = BC\). Using the tangent formula:

\(AB = OA \times \tan\left(\frac{3}{16} \pi\right)\)

\(AB = 5.093 \times \tan\left(\frac{3}{16} \pi\right) \approx 3.403 \text{ cm}\)

The perimeter is:

\(2 \times 3.403 + 6 = 12.8 \text{ cm}\)

(iii) The area of the shaded region is the difference between the area of triangle OABC and the area of sector OAC. The area of triangle OABC is:

\(\text{Area of } OABC = \frac{1}{2} \times OA \times AB = \frac{1}{2} \times 5.093 \times 3.403 \approx 8.665 \text{ cm}^2\)

The area of sector OAC is:

\(\text{Area of sector } OAC = \frac{1}{2} \times (5.093)^2 \times \frac{3}{8} \pi \approx 6.615 \text{ cm}^2\)

The shaded area is:

\(8.665 - 6.615 = 2.05 \text{ cm}^2\)

(i) To find the length of AD, we use the angle EAD = ACD = \frac{3\pi}{10} radians. Using the sine rule in triangle ACD, we have:

\(AD = 8 \sin\left(\frac{3\pi}{10}\right)\)

Calculating this gives \(AD \approx 6.47\) cm.

(ii) To find the area of the shaded region, we calculate the area of sector ADE and subtract the area of triangle ADC.

The area of sector ADE is:

\(\frac{1}{2} \times (\text{their } AD)^2 \times \frac{\pi}{5}\)

The area of triangle ADC is:

\(\frac{1}{2} \times 8 \times (\text{their } AD) \times \sin\left(\frac{\pi}{5}\right)\)

Calculating these areas and subtracting gives the shaded area \(\approx 35.0 \text{ or } 34.9 \text{ cm}^2\).

The perimeter of sector \(AOC\) is given by the arc length plus the radii, which is \(2r + r\theta\).

The angle \(COB\) is \(\pi - \theta\) because the total angle in a semicircle is \(\pi\) radians.

The perimeter of sector \(BOC\) is \(2r + r(\pi - \theta)\).

According to the problem, the perimeter of \(BOC\) is twice the perimeter of \(AOC\):

\(2r + r(\pi - \theta) = 2(2r + r\theta)\)

Simplifying, we have:

\(2r + r\pi - r\theta = 4r + 2r\theta\)

\(r\pi - r\theta = 2r\theta + 2r\)

\(r\pi = 3r\theta + 2r\)

\(\pi = 3\theta + 2\)

\(\theta = \frac{\pi - 2}{3}\)

Calculating \(\theta\) gives \(\theta \approx 0.38\) to 2 significant figures.

To find the perimeter of the sector, we need to express it in terms of the radius \(r\) and the area \(A\).

The area of a sector is given by \(A = \frac{1}{2} r^2 \theta\), where \(\theta\) is the angle in radians.

Rearranging for \(\theta\), we have:

\(\theta = \frac{2A}{r^2}\)

The perimeter \(P\) of the sector is the sum of the two radii and the arc length:

\(P = r + r + r\theta\)

Substituting \(\theta\) gives:

\(P = 2r + r \left( \frac{2A}{r^2} \right)\)

Simplifying, we get:

\(P = 2r + \frac{2A}{r}\)

1. Calculate angle CBA using the sine rule: \(\angle CBA = \sin^{-1}\left(\frac{7}{8}\right) \approx 1.0654 \text{ radians}\).

2. Calculate the area of sector BCYD: \(\frac{1}{2} \times 8^2 \times 2 \times 1.0654 \approx 68.19 \text{ cm}^2\).

3. Calculate the area of triangle BCD: \(7 \times \sqrt{8^2 - 7^2} \approx 27.11 \text{ cm}^2\).

4. Calculate the area of the semicircle CXD: \(\frac{1}{2} \pi \times 7^2 \approx 76.97 \text{ cm}^2\).

5. Total area enclosed by the arcs: \(68.19 - 27.11 + 76.97 \approx 118.0 \text{ to } 118.1 \text{ cm}^2\).

(i) The angle BAC in radians is given by the formula for arc length: \(\theta = \frac{\text{arc length}}{\text{radius}}\). Here, \(\theta = \frac{4}{5} = 0.8\) radians.

(ii) To find the area of the shaded region BDC, we calculate:

1. The length of BD: \(BD = 5 \sin(0.8) \approx 3.587\) cm.

2. The length of DC: \(DC = 5 - 5 \cos(0.8) \approx 1.516\) cm.

3. The area of sector BAC: \(\frac{1}{2} \times 5^2 \times 0.8 = 10\) cm².

4. The area of segment BC: \(\frac{1}{2} \times 5^2 \times (0.8 - \sin(0.8)) \approx 1.03\) cm².

5. The area of triangle BDC: \(\frac{1}{2} \times BD \times DC \approx 2.719\) cm².

6. The area of the trapezium: \(\frac{1}{2} \times (5 + DC) \times BD \approx 11.69\) cm².

7. The shaded area BDC: \(\text{Trapezium} - \text{Sector} + \text{Segment} = 11.69 - 10 + 1.03 = 1.69\) cm².

(i) To find angle CBY, use the cosine rule in triangle ACB:

\(\cos(\angle ABC) = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC} = \frac{12^2 + 8^2 - 8^2}{2 \times 12 \times 8} = \frac{80}{192} = \frac{5}{12}\)

\(\angle ABC = \cos^{-1}\left(\frac{5}{12}\right) \approx 1.696 \text{ radians}\)

Since CBY is a straight line, \(\angle CBY = \pi - 2 \times \angle ABC\)

\(\angle CBY = \pi - 2 \times 1.696 \approx 1.445 \text{ radians}\)

(ii) The perimeter of the shaded region consists of arcs CY and XC:

Arc CY is part of a circle with radius 8 ext{ cm} and angle CBY:

\(\text{Arc } CY = 8 \times 1.445 = 11.56 \text{ cm}\)

To find arc XC, use angle BAC:

\(\angle BAC = \frac{\pi - \angle ABC}{2} \approx 0.7227 \text{ radians}\)

Arc XC is part of a circle with radius 12 ext{ cm}:

\(\text{Arc } XC = 12 \times 0.7227 = 8.673 \text{ cm}\)

The perimeter of the shaded region is:

\(11.56 + 8.673 + 4 = 24.2 \text{ cm}\)

First, calculate angle OAB:

\(\angle OAB = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}\)

Calculate the area of sector CAB:

\(\text{Area of sector } CAB = \frac{1}{2} \times \left(\frac{3\pi}{10}\right) \times 5^2 = 11.78\)

Calculate OA using the sine rule:

\(OA = \frac{5}{\sin \frac{\pi}{5}} = 8.507\)

Calculate the area of sector COD:

\(\text{Area of sector } COD = \frac{1}{2} \times (3.507)^2 \times \frac{\pi}{5} = 3.86\)

Calculate the area of triangle OAB:

\(\Delta OAB = \frac{1}{2} \times 5 \times 8.507 \times \sin \frac{3\pi}{10} = 17.20 \text{ or } 17.21\)

Calculate the shaded area:

\(\text{Shaded area} = 17.20 \text{ or } 17.21 - 11.78 - 3.86 = 1.56 \text{ or } 1.57\)

First, find the angle ∠AOC using the arc length formula:

Arc length = radius × angle in radians.

Given arc length = 6 cm and radius = 5 cm, we have:

6 = 5 × ∠AOC

∠AOC = \frac{6}{5} \text{ radians} \approx 1.2 \text{ radians}

Next, find the length of AB using the tangent function:

AB = 5 \times \tan(1.2) \approx 12.86 \text{ cm}

Calculate the area of triangle OAB:

\text{Area of } \triangle OAB = \frac{1}{2} \times 5 \times 12.86 \approx 32.15 \text{ cm}^2

Calculate the area of sector OAC:

\text{Area of sector OAC} = \frac{1}{2} \times 5^2 \times 1.2 \approx 15 \text{ cm}^2

Finally, find the area of the shaded region:

\text{Shaded region} = 32.15 - 15 = 17.2 \text{ cm}^2

(a) To find the area of the shaded region, we first need to find the radius r and angle θ.

We have the equations:

\(\frac{1}{2} r^2 \theta = 76.8\)

\(r \theta = 9.6\)

From the second equation, \(\theta = \frac{9.6}{r}\).

Substitute \(\theta\) in the first equation:

\(\frac{1}{2} r^2 \left( \frac{9.6}{r} \right) = 76.8\)

\(\frac{1}{2} r \times 9.6 = 76.8\)

\(4.8r = 76.8\)

\(r = 16\)

Now, substitute \(r = 16\) back to find \(\theta\):

\(\theta = \frac{9.6}{16} = 0.6 \text{ radians}\)

The area of triangle \(\Delta OAB\) is:

\(\frac{1}{2} \times 16^2 \times \sin(0.6) \approx 4.53 \text{ cm}^2\)

The area of the shaded region is:

\(76.8 - 72.27 = 4.53 \text{ cm}^2\)

(b) To find the perimeter of the shaded region, we need the length of AB and the arc length.

Using the cosine rule or sine rule, we find:

\(AB = 2 \times 16 \times \sin(0.3) \approx 9.46 \text{ cm}\)

The perimeter is:

\(9.6 + 9.46 = 19.1 \text{ cm}\)

(i) The area of sector AOB is given by \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\).

The area of the triangle AOT is \(\frac{1}{2} \times r \times r \tan \theta = \frac{1}{2} r^2 \tan \theta\).

Since the area of the sector is equal to the area of the shaded region, we have:

\(r^2 \theta = \frac{1}{2} r^2 \tan \theta\)

\(2\theta = \tan \theta\)

(ii) Given \(r = 8\) cm and the length of the minor arc AB is 19.2 cm, we find \(\theta\) using the arc length formula:

\(r \times 2\theta = 19.2\)

\(8 \times 2\theta = 19.2\)

\(\theta = 1.2\) radians

The area of sector AOB is \(\frac{1}{2} \times 8^2 \times 2 \times 1.2 = 76.8\) cm².

The area of the kite (or quadrilateral) is \(165\) cm².

The area of the shaded region is \(165 - 76.8 = 88.2\) cm², approximately \(87.8\) cm².

(i) To find the area of the shaded region, we need to calculate the area of triangle AOT and subtract the area of sector AOB.

The area of triangle AOT is given by:

\(\frac{1}{2} \times AT \times r\)

Using trigonometry, \(\tan \theta = \frac{AT}{r}\), so \(AT = r \tan \theta\).

Thus, the area of triangle AOT is:

\(\frac{1}{2} r^2 \tan \theta\)

The area of sector AOB is:

\(\frac{1}{2} r^2 \theta\)

Therefore, the area of the shaded region is:

\(\frac{1}{2} r^2 \tan \theta - \frac{1}{2} r^2 \theta\)

(ii) For r = 3 and \theta = 1.2, we find the perimeter of the shaded region.

First, calculate AT:

\(\tan \theta = \frac{AT}{3} \Rightarrow AT = 3 \tan 1.2 \approx 7.716\)

The arc length of AB is:

\(r \theta = 3 \times 1.2 = 3.6\)

Using the Pythagorean theorem or trigonometry, find OT:

\(OT = \frac{3}{\cos 1.2} \approx 8.279\)

The perimeter of the shaded region is:

\(AT + \text{arc length} + OT - r = 7.716 + 3.6 + 8.279 - 3 = 16.6\)

(i) To find the length of PQ, we use the sine rule in triangle POQ:

\(\frac{PQ}{2} = 10 \times \sin 1.1\)

\(PQ = 2 \times 10 \times \sin 1.1 = 17.8 \text{ cm}\)

(ii) First, find angle OPQ:

\(\angle OPQ = \frac{\pi}{2} - 1.1 \approx 0.471 \text{ radians}\)

Next, calculate the arc length QR:

\(\text{Arc } QR = 17.8 \times \left(\frac{\pi}{2} - 1.1\right) \approx 8.39 \text{ cm}\)

Finally, find the perimeter of the shaded region:

\(\text{Perimeter} = 17.8 + 10 + 10 + 8.39 = 26.2 \text{ cm}\)

(i) To find angle ABP, use the sine rule in triangle ABP:

\(\sin^{-1}\left(\frac{3}{5}\right) = 0.6435\) radians.

(ii) The area of a sector is given by \(\frac{1}{2} r^2 \theta\).

For sector BAP, \(r = 5\) and \(\theta = 0.6435\):

\(\text{Area} = \frac{1}{2} \times 5^2 \times 0.6435 = 8.04\).

For sector DAQ, \(r = 3\) and \(\theta = \frac{\pi}{2}\):

\(\text{Area} = \frac{1}{2} \times 3^2 \times \frac{\pi}{2} = 7.07\).

(iii) The area of the shaded region can be calculated by subtracting the area of triangle BPC from the sum of the areas of the sectors:

Area of triangle BPC = \(\frac{1}{2} \times 3 \times 4 = 6\).

Shaded area = \(8.04 + 7.07 - (15 - 6) = 6.11\).

(i) To find the length of the arc DC, we first need to determine the radius of the circle with centre B. Using the Pythagorean theorem, we find \(r = \sqrt{72}\). The angle at the centre for the arc DC is \(\frac{\pi}{4}\) radians. The length of the arc is given by \(s = r \theta\), so \(s = \sqrt{72} \times \frac{\pi}{4} = \frac{3\sqrt{2}}{2} \pi \approx 6.66 \text{ cm}\).

(ii) The area of sector BDC is \(\frac{1}{2} \times 72 \times \frac{\pi}{4} = 9\pi\). The area of region Q is \(9\pi - 18\). The area of region P is \(\frac{1}{4} \pi \times 6^2 - \text{area of Q} = 18\). The ratio is \(\frac{18}{9\pi - 18} \approx 1.75\).

(i) To find angle BAC, use the cosine rule in triangle ABC:

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

where a = 16, b = 10, c = 10. Substituting the values:

\(\cos A = \frac{10^2 + 10^2 - 16^2}{2 \times 10 \times 10} = \frac{100 + 100 - 256}{200} = \frac{-56}{200} = -0.28\)

Thus,

\(A = \cos^{-1}(-0.28) \approx 0.6435 \text{ radians}\)

(ii) To find the area of the shaded region, calculate the area of triangle ABC and subtract it from the sum of the areas of the sectors:

Area of triangle ABC:

\(\text{Area} = \frac{1}{2} \times 16 \times 10 \times \sin(0.6435) = 48\)

Area of one sector (e.g., sector ABE):

\(\text{Area} = \frac{1}{2} \times 10^2 \times 0.6435 = 32.175\)

Total area of two sectors:

\(2 \times 32.175 = 64.35\)

Shaded area:

\(\text{Shaded area} = 64.35 - 48 = 16.35 \approx 16.4\)

(i) To find angle ABC, use the sine function in the right triangle ABC:

\(\sin \angle ABC = \frac{AC}{BC} = \frac{8}{10}\)

\(\angle ABC = \arcsin\left(\frac{8}{10}\right) \approx 0.927 \text{ radians}\)

(ii) First, find AB using the Pythagorean theorem:

\(AB = \sqrt{BC^2 - AC^2} = \sqrt{10^2 - 8^2} = 6 \text{ cm}\)

Calculate the area of triangle BCD:

\(\text{Area of } \triangle BCD = \frac{1}{2} \times 8 \times 6 = 48 \text{ cm}^2\)

Calculate the area of sector BCD:

\(\text{Area of sector } BCD = \frac{1}{2} \times 10^2 \times 2 \times 0.927 = 92.73 \text{ cm}^2\)

Calculate the area of segment BCD:

\(\text{Area of segment } BCD = 92.73 - 48 = 44.73 \text{ cm}^2\)

Calculate the area of the semicircle with radius 8 cm:

\(\text{Area of semicircle} = \frac{1}{2} \times \pi \times 8^2 = 100.53 \text{ cm}^2\)

Finally, find the shaded area:

\(\text{Shaded area} = 100.53 - 44.73 = 55.8 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we need to consider the lengths of AB, the arc AB, and the sides AD and DC of the rectangle.

The length of AB can be found using the formula: \(AB = 2r\sin\theta\).

The length of the arc AB is given by: \(2r\theta\).

The perimeter P of the shaded region is: \(P = AD + DC + AB + \text{arc } AB = 2r + 2r\theta + 2r\sin\theta\).

(ii) For r = 5 and θ = \(\frac{1}{6} \pi\), we calculate the area of the shaded region.

The area of sector AOB is \(\frac{1}{2} r^2 (2\theta) = \frac{25\pi}{6}\) or 13.1.

The area of triangle AOB is \(\frac{1}{2} \times 2r\sin\theta \times r\cos\theta = \frac{25\sqrt{3}}{4}\) or 10.8.

The area of rectangle ABCD is \(r \times 2r\sin\theta = 25\).

The area of the shaded region is: \(25 - (\frac{25\pi}{6} - \frac{25\sqrt{3}}{4})\) or 22.7.

(i) Letting \(M\) be the midpoint of \(AB\), we have \(OM = 8\) cm using the Pythagorean theorem. Therefore, \(XM = 2\) cm. Using trigonometry, \(\tan AXM = \frac{6}{2}\), so \(AXB = 2 \times \tan^{-1}(3) = 2.498\) radians.

(ii) The length \(AX = \sqrt{6^2 + 2^2} = \sqrt{40}\). The arc \(AYB = r\theta = \sqrt{40} \times 2.498\). The perimeter is \(12 + \text{arc} = 27.8\) cm.

(iii) The area of sector \(AXBY = \frac{1}{2} \times (\sqrt{40})^2 \times 2.498\). The area of triangle \(AXB = \frac{1}{2} \times 12 \times 2\). Subtract these to find the shaded area: 38.0 cm\(^2\).

(i) To find \(\angle CBD\), we first find \(\angle ABC\). Since \(\angle CAB = \frac{2}{7} \pi\), and \(\triangle ABC\) is isosceles with \(AB = AC\), the remaining angle \(\angle ABC\) is:

\(\angle ABC = \pi - \angle CAB = \pi - \frac{2}{7} \pi = \frac{5}{7} \pi\)

Since \(ABD\) is a straight line, \(\angle CBD = \pi - \angle ABC = \pi - \frac{5}{7} \pi = \frac{9}{14} \pi\).

(ii) To find the perimeter of the shaded region, we need to calculate the lengths of \(BC\), arc \(CD\), and arc \(CB\).

Using the sine rule in \(\triangle ABC\):

\(\frac{\sin \frac{\pi}{7}}{8} = \frac{\sin \frac{2\pi}{7}}{BC}\)

Solving for \(BC\), we find \(BC \approx 6.94 \text{ cm}\).

The length of arc \(CD\) is given by:

\(\text{arc } CD = BC \times \frac{9\pi}{14} \approx 6.94 \times \frac{9\pi}{14} \approx 14.02 \text{ cm}\)

The length of arc \(CB\) is given by:

\(\text{arc } CB = 8 \times \frac{2\pi}{7} \approx 7.18 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(6.94 + 14.02 + 7.18 = 28.1 \text{ cm}\)

(a) To find the arc length \(AB\), we first need to find the length of \(OA\). Using the cosine rule in triangle \(OAP\):

\(OA^2 = OP^2 + AP^2 - 2 \cdot OP \cdot AP \cdot \cos(\theta)\)

\(OA^2 = x^2 + x^2 - 2x^2 \cos(\theta)\)

\(OA^2 = 2x^2(1 - \cos(\theta))\)

\(OA = x \sqrt{2(1 - \cos(\theta))}\)

Using the identity \(1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})\), we have:

\(OA = x \sqrt{4\sin^2(\frac{\theta}{2})} = 2x \sin(\frac{\theta}{2})\)

Since \(\sin(\frac{\theta}{2}) = \cos(\frac{\pi}{2} - \frac{\theta}{2})\), we have:

\(OA = 2x \cos(\frac{\pi}{2} - \frac{\theta}{2})\)

Thus, \(OA = 2x \cos \theta\).

The arc length \(AB\) is given by \(\theta \times OA\):

\(AB = \theta \times 2x \cos \theta = 2x\theta \cos \theta\)

(b) The area of the sector \(OAB\) is:

\(\text{Sector area} = \frac{1}{2} \times (2x \cos \theta)^2 \times \theta\)

\(= 2x^2 \cos^2 \theta \times \theta\)

The area of triangle \(OAP\) is:

\(\text{Triangle area} = \frac{1}{2} \times x \times x \times \sin(\pi - 2\theta)\)

\(= \frac{1}{2} x^2 \sin(2\theta)\)

The area of the shaded region \(APB\) is:

\(\text{Area of } APB = \text{Sector area} - \text{Triangle area}\)

\(= 2x^2 \cos^2 \theta \times \theta - \frac{1}{2} x^2 \sin(2\theta)\)

\(= x^2 \left( 2\cos^2 \theta - \frac{1}{2} \sin 2\theta \right)\)

(i) To find the radius \(OE\) of the arc \(CED\), we use the cosine rule in the triangle \(OEC\):

\(\cos(0.9) = \frac{OE}{6}\)

Solving for \(OE\), we get:

\(OE = 6 \cos(0.9) = 3.73 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the large sector and subtract the area of the small sector:

The angle of the large sector is \(2\pi - 1.8\).

Area of the large sector:

\(\frac{1}{2} \times 6^2 \times (2\pi - 1.8) = 80.7 \text{ cm}^2\)

Area of the small sector:

\(\frac{1}{2} \times 3.73^2 \times 1.8 = 12.52 \text{ cm}^2\)

Total area of the shaded region:

\(80.7 + 12.52 = 93.2 \text{ cm}^2\)

(i) To find the radius of the semicircle, use the cosine rule in triangle BOD:

\(BD = \sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos(1.2)} = \sqrt{200 - 200 \cos(1.2)} = 11.3 \text{ cm}\)

Since BD is the diameter of the semicircle, the radius \(r\) is:

\(r = \frac{BD}{2} = \frac{11.3}{2} = 5.646 \text{ cm}\)

(ii) The perimeter of the metal plate consists of the major arc of the circle and the semicircle:

The major arc length is given by:

\(10 \times (2\pi - 1.2) = 10 \times 5.083 = 50.832 \text{ cm}\)

The semicircle perimeter is:

\(\pi \times 5.646 = 17.737 \text{ cm}\)

Thus, the total perimeter is:

\(50.832 + 17.737 = 68.6 \text{ cm}\)

(iii) The area of the metal plate consists of the area of the major sector, the semicircle, and triangle OBD:

The area of the major sector is:

\(\frac{1}{2} \times 10^2 \times (2\pi - 1.2) = 254.159 \text{ cm}^2\)

The area of triangle OBD is:

\(\frac{1}{2} \times 10 \times 10 \times \sin(1.2) = 46.602 \text{ cm}^2\)

The area of the semicircle is:

\(\frac{1}{2} \times \pi \times (5.646)^2 = 50.1 \text{ cm}^2\)

Thus, the total area is:

\(254.159 + 46.602 + 50.1 = 351 \text{ cm}^2\)

(i) The perimeter of the shaded region ABDC is given by the sum of the lengths of arcs AB and CD, and the radii OA and OB. The length of arc AB is \(2r\alpha\) and the length of arc CD is \(r\alpha\). Therefore, the perimeter is:

\(2r\alpha + r\alpha + 2r = 4.4r\)

Simplifying, we get:

\(3r\alpha + 2r = 4.4r\)

\(3r\alpha = 2.4r\)

\(\alpha = 0.8\)

(ii) The area of the shaded region is the difference between the area of sector OAB and sector OCD. The area of sector OAB is \(\frac{1}{2}(2r)^2\alpha\) and the area of sector OCD is \(\frac{1}{2}r^2\alpha\). Therefore, the area of the shaded region is:

\(\frac{1}{2}(2r)^2 \times 0.8 - \frac{1}{2}r^2 \times 0.8 = 30\)

\(2r^2 \times 0.8 - 0.5r^2 \times 0.8 = 30\)

\((3/2)r^2 \times 0.8 = 30\)

\(r^2 \times 1.2 = 30\)

\(r^2 = 25\)

\(r = 5\)

First, calculate the angles in triangle ABC:

\(\angle BAC = \sin^{-1}(3/5) \approx 36.87^\circ\)

\(\angle ABC = \sin^{-1}(4/5) \approx 53.13^\circ\)

\(\angle ACB = \pi/2\) (right angle)

Calculate the area of triangle ABC:

\(\Delta ABC = \frac{1}{2} \times 4 \times 3 = 6 \, \text{cm}^2\)

Calculate the areas of the sectors:

\(\text{Sector AEF} = \frac{1}{2} \times 3^2 \times 0.6435\)

\(\text{Sector BEG} = \frac{1}{2} \times 2^2 \times 0.9273\)

\(\text{Sector CFG} = \frac{1}{2} \times 1^2 \times 1.5708\)

Sum of sectors = \(\frac{1}{2} [3^2 \times 0.6435 + 2^2 \times 0.9273 + 1^2 \times 1.5708]\)

Sum of sectors = 5.536 \, \text{cm}^2

Area of shaded region EFG:

\(\text{Area of EFG} = \Delta ABC - \text{Sum of sectors} = 6 - 5.536 = 0.464 \, \text{cm}^2\)

(i) To find the perimeter of the shaded region PQT, we need to calculate the lengths of PT, QT, and the arc PQ.

The length of PT is given by the tangent formula: \(PT = r \tan \alpha\).

The length of QT is \(QT = OT - OQ = \frac{r}{\cos \alpha} - r\).

The length of the arc PQ is \(r \alpha\).

Therefore, the perimeter of the shaded region PQT is \(r \tan \alpha + \frac{r}{\cos \alpha} - r + r \alpha\).

(ii) For \(\alpha = \frac{1}{3} \pi\) and \(r = 10\):

The area of triangle POT is \(\frac{1}{2} \times 10 \times 10 \times \tan \frac{\pi}{3} = 50 \sqrt{3}\).

The area of sector POQ is \(\frac{1}{2} \times 10^2 \times \frac{1}{3} \pi = \frac{50 \pi}{3}\).

The area of the shaded region is \(50 \sqrt{3} - \frac{50 \pi}{3} \approx 34\) (to 2 significant figures).

(i) To find the perimeter of the shaded region, we need to calculate the lengths of CD, BD, and the arc CB.

The length of CD is given by \(CD = r \cos \theta\).

The length of BD is given by \(BD = r - r \sin \theta\).

The length of the arc CB is given by \(r \left( \frac{1}{2} \pi - \theta \right)\).

Thus, the perimeter \(P\) is:

\(P = r \cos \theta + r - r \sin \theta + r \left( \frac{1}{2} \pi - \theta \right)\)

(ii) To find the area of the shaded region when r = 5 cm and θ = 0.6, we calculate the area of the sector and subtract the areas of triangle AOC and sector COD.

The area of sector AOB is \(\frac{1}{2} \times 5^2 \times \left( \frac{1}{2} \pi - 0.6 \right) = 12.135\).

The area of triangle AOC is \(\frac{1}{2} \times 5 \times \cos 0.6 \times 5 \times \sin 0.6 = 5.825\).

The area of the shaded region is:

\(12.135 - 5.825 = 6.31\)

(a)(i) Since \(AX\) is a tangent, \(\angle BAO = \angle OBA = \frac{\pi}{2} - \alpha\). Therefore, \(\angle AOB = \pi - (\frac{\pi}{2} - \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha\).

(a)(ii) The area of the sector \(OAB\) is \(\frac{1}{2} r^2 (2\alpha)\). The area of triangle \(OAB\) is \(\frac{1}{2} r^2 \sin 2\alpha\). Thus, the area of the shaded segment is \(\frac{1}{2} r^2 (2\alpha) - \frac{1}{2} r^2 \sin 2\alpha\).

(b) Use \(\alpha = \frac{\pi}{6}\) and \(r = 4\). The area of one segment is \(\frac{1}{2} (4)^2 \left( \frac{\pi}{3} \right) - \frac{1}{2} (4)^2 \sin \frac{\pi}{3} = \frac{8\pi}{3} - 4\sqrt{3}\).

The area of triangle \(ABC\) is \(\frac{1}{2} (4)^2 \sin \frac{\pi}{3} = 4\sqrt{3}\).

The area of the shaded region is \(3 \left( \frac{8\pi}{3} - 4\sqrt{3} \right) - 4\sqrt{3} = 16\sqrt{3} - 8\pi\).

(i) The area of sector OCD is given by \(\frac{1}{2} (2r)^2 \theta = 2r^2 \theta\).

The area of sectors OAB and OEF is \(2 \times \frac{1}{2} r^2 (\pi - \theta) = r^2 (\pi - \theta)\).

Thus, the total area of the metal plate is \(2r^2 \theta + r^2 (\pi - \theta) = r^2 (\pi + \theta)\).

(ii) The arc length CD is \(2r \theta\).

The arc lengths AB and EF are \(2 \times r (\pi - \theta) = 2r (\pi - \theta)\).

The straight edges AO, OE, BO, and OF sum to \(4r\).

Thus, the total perimeter is \(2r \theta + 2r (\pi - \theta) + 4r = 2\pi r + 4r\), which is independent of \(\theta\).

(i) To find the length of OB, use the cosine rule in the right-angled triangle OAB:

\(OB = \frac{OA}{\cos(\angle AOB)} = \frac{6}{\cos(0.6)} = 7.270 \text{ cm}\)

(ii) To find the perimeter of the metal plate, calculate the lengths of AB and the arc BC:

\(AB = OA \times \tan(\angle AOB) = 6 \times \tan(0.6) = 4.1 \text{ cm}\)

The arc length BC is given by:

\(s = r \theta = 7.27 \times (\frac{\pi}{2} - 0.6) = 7.06 \text{ cm}\)

Thus, the perimeter is:

\(6 + 7.27 + 7.06 + 4.1 = 24.4 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the areas of triangle OAB and sector OBC:

The area of triangle OAB is:

\(\frac{1}{2} \times 6 \times 7.27 \times \sin(0.6) = 12.31 \text{ cm}^2\)

The area of sector OBC is:

\(\frac{1}{2} \times 7.27^2 \times (\frac{\pi}{2} - 0.6) = 25.65 \text{ cm}^2\)

Thus, the total area is:

\(12.31 + 25.65 = 38.0 \text{ cm}^2\)

(i) To find the radius of the larger circle, consider the right triangle ABC where AB = r and AC = r. By the Pythagorean theorem:

\(BC^2 = AB^2 + AC^2 = r^2 + r^2 = 2r^2\)

\(BC = \sqrt{2r^2} = r\sqrt{2}\)

Thus, the radius of the larger circle is \(r\sqrt{2}\).

(ii) To find the area of the shaded region:

1. Calculate the area of sector BCFD of the larger circle:

\(\text{Area of sector BCFD} = \frac{1}{4} \pi (r\sqrt{2})^2 = \frac{1}{2} \pi r^2\)

2. Calculate the area of triangle BCAD:

\(\text{Area of } \triangle BCAD = \frac{1}{2} (2r)r = r^2\)

3. Calculate the area of segment CFDA:

\(\text{Area of segment CFDA} = \frac{1}{2} \pi r^2 - r^2\)

4. Calculate the area of semicircle CADE:

\(\text{Area of semicircle CADE} = \frac{1}{2} \pi r^2\)

5. Calculate the shaded area:

\(\text{Shaded area} = \frac{1}{2} \pi r^2 - \left( \frac{1}{2} \pi r^2 - r^2 \right)\)

\(= r^2\)

Thus, the area of the shaded region is \(r^2\).

(i) To show \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\), consider the area of triangle \(\triangle OAC\). The area is \(\frac{1}{2} r^2 \sin \alpha \cos \alpha\). Since \(AC\) divides the sector into two regions of equal area, the area of \(\triangle OAC\) is half the area of sector \(OAB\), which is \(\frac{1}{2} \times \frac{1}{2} r^2 \alpha\). Equating these gives \(\frac{1}{2} r^2 \sin \alpha \cos \alpha = \frac{1}{2} \times \frac{1}{2} r^2 \alpha\), simplifying to \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\).

(ii) The perimeter of \(\triangle OAC\) is \(r + r \sin \alpha + r \cos \alpha = 2.4r\). The perimeter of \(ACB\) is \(r\alpha + r \sin \alpha + r - r \cos \alpha = 2.18r\) or \(2.17r\). The ratio is \(\frac{2.4}{2.18} : 1 = 1.1 : 1\).

(iii) Given \(\alpha = 0.9477\) radians, convert to degrees: \(0.9477 \times \frac{180}{\pi} \approx 54.3°\).

The radius of the semicircle is \(\frac{1}{2} AB = r \sin \theta\).

The area of the semicircle is \(\frac{1}{2} \pi r^2 \sin^2 \theta = A_1\).

The shaded area is the area of the semicircle minus the area of the segment.

Shaded area = \(A_1 - \frac{1}{2} r^2 \theta + \frac{1}{2} r^2 \sin 2\theta\).

(i) To find the perimeter of the shaded region, we need to calculate the lengths of AB, BC, and the arc AC.

The length of AB or CB is given by \(\frac{3}{\tan \frac{\pi}{6}} = 3 \tan \frac{\pi}{3} = 3\sqrt{3}\).

The length of the arc AC is \(3 \times \frac{2\pi}{3} = 2\pi\).

Therefore, the perimeter of the shaded region is \(3\sqrt{3} + 3\sqrt{3} + 2\pi = 6\sqrt{3} + 2\pi\).

(ii) To find the area of the shaded region, we calculate the area of triangle OABC and subtract the area of sector OADC.

The area of triangle OABC is \(2 \times \frac{1}{2} \times 3 \times 3\sqrt{3} = 9\sqrt{3}\).

The area of sector OADC is \(\frac{1}{2} \times 3^2 \times \frac{2\pi}{3} = 3\pi\).

Thus, the area of the shaded region is \(9\sqrt{3} - 3\pi\).

(i) To find angle BAO, use the tangent function: \(\tan \theta = \frac{5}{12}\). Solving for \(\theta\), we get \(\theta = \tan^{-1}\left(\frac{5}{12}\right) \approx 0.3948\) radians.

(ii) The area of triangle AOB is \(\frac{1}{2} \times 12 \times 5 = 30\) cm².

The shaded area is the sum of two sectors minus the triangle area. The area of the sector with center A is \(\frac{1}{2} \times 12^2 \times 0.3948\).

The other angle in the triangle is \(\frac{\pi}{2} - 0.3948\). The area of the sector with center B is \(\frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right)\).

Thus, the shaded area is:

\(\frac{1}{2} \times 12^2 \times 0.3948 + \frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right) - 30\)

\(= 28.43 + 14.70 - 30 = 13.1\) cm².

(i) The arc length of AB is given by \(4\alpha\) since the radius is 4 cm and the angle is \(\alpha\) radians.

The arc length of DC is \((4 \cos \alpha) \alpha\) because the radius of the arc DC is \(4 \cos \alpha\).

The length AC (or DB) is \(4 - 4 \cos \alpha\) since AD is perpendicular to OB.

Thus, the perimeter of the shaded region ABDC is \(4\alpha \cos \alpha + 4\alpha + 8 - 8\cos \alpha\).

(ii) For \(\alpha = \frac{1}{6}\pi\), the length OD is \(4 \cos \frac{\pi}{6} = 2\sqrt{3}\).

The area of the sector AOB is \(\frac{1}{2} \times 4^2 \times \frac{\pi}{6}\).

The area of the sector ODC is \(-\frac{1}{2} (2\sqrt{3})^2 \times \frac{\pi}{6}\).

The shaded area is \(\frac{\pi}{3}\), which can be expressed as \(k\pi\) where \(k = \frac{1}{3}\).

(i) The area of triangle AOB is \(\frac{1}{2} r^2 \sin \theta\).

The area of the sector AOB is \(\frac{1}{2} r^2 \theta\).

The area of the segment AXB is \(\frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\).

Setting the areas equal: \(\frac{1}{2} r^2 \sin \theta = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\).

Simplifying gives \(2 \sin \theta = \theta\), so \(p = 2\).

(ii) The chord length AB is \(2 \times 8 \sin \frac{2.4}{2} = 8 \sin 1.2 \times 2 \approx 14.9\).

The arc length AXB is \(2.4 \times 8 = 19.2\).

The perimeter of the segment AXB is the sum of these: \(14.9 + 19.2 = 34.1\).

To find the area of the shaded region, we need to calculate the area of triangle ABC and subtract the area of sector ADE.

The area of triangle ABC is given by:

\(\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 4 \times 4 \tan \alpha = 8 \tan \alpha\)

The area of sector ADE is given by:

\(\text{Area of sector } ADE = \frac{1}{2} \times 2^2 \times \alpha = 2 \alpha\)

Therefore, the area of the shaded region is:

\(8 \tan \alpha - 2 \alpha\)

(i) To find \(\theta\), first calculate the slant height of the cone using the Pythagorean theorem: \(l = \sqrt{6^2 + 8^2} = 10\) cm.

The circumference of the base of the cone is \(2\pi \times 6 = 12\pi\) cm.

The arc length of the sector is equal to the circumference of the base: \(10\theta = 12\pi\).

Solving for \(\theta\), we get \(\theta = \frac{12\pi}{10} = 1.2\pi\) or approximately 3.77 radians.

(ii) The area of the sector (paper) is given by \(\frac{1}{2} r^2 \theta\), where \(r = 10\) cm and \(\theta = 1.2\pi\).

Thus, the area is \(\frac{1}{2} \times 10^2 \times 1.2\pi = 60\pi\) cm\(^2\) or approximately 188.5 cm\(^2\).

(a) To find the perimeter of the shaded region, we first need to find the angle \\ \angle ADC \\ using trigonometry. We have:

\\ \tan BDC = \frac{4}{3} \\

Thus, \\ BDC = 0.927 \\ radians.

Then, \\ ADC = \pi - 0.927 = 2.214 \\ radians.

The arc length \\ AC \\ is given by:

\\ AC = 5 \times 2.214 = 11.07 \\ cm.

The length \\ AC \\ in the triangle is:

\\ AC = \sqrt{8^2 + 4^2} = 8.94 \\ cm.

Therefore, the perimeter of the shaded region is:

\\ 11.07 + 8.94 = 20.0 \\ cm.

(b) To find the area of the shaded region, we calculate the area of sector \\ ACD \\ and subtract the area of triangle \\ ADC \\ from it.

The area of sector \\ ACD \\ is:

\\ \frac{1}{2} \times 5^2 \times 2.214 = 27.7 \\ cm².

The area of triangle \\ ADC \\ is:

\\ \frac{1}{2} \times 5 \times 4 = 10 \\ cm².

Thus, the area of the shaded region is:

\\ 27.7 - 10 = 17.7 \\ cm².

(i) The perimeter of the metal plate consists of the arc lengths and the straight edges. The arc length of OABCD is \(2r(2\pi - \alpha)\) and the arc length of OAED is \(r\alpha\). The straight edges are \(2r\) and \(r\). Therefore, the perimeter is:

\(2r(2\pi - \alpha) + r\alpha + 2r + r = 2\pi r + r\alpha + 2r\)

(ii) The area of the metal plate is the sum of the areas of the two sectors. The area of sector OABCD is \(\frac{1}{2}(2r)^2\alpha\) and the area of sector OAED is \(\pi r^2 - \frac{1}{2}r^2\alpha\). Therefore, the total area is:

\(\frac{1}{2}(2r)^2\alpha + \pi r^2 - \frac{1}{2}r^2\alpha = \frac{3r^2\alpha}{2} + \pi r^2\)

(iii) Given that the shaded and unshaded pieces are equal in area, we equate the areas:

\(\pi r^2 - \frac{1}{2}r^2\alpha = \frac{1}{2}(2r)^2\alpha\)

Solving for \(\alpha\):

\(\pi r^2 - \frac{1}{2}r^2\alpha = 2r^2\alpha\)

\(\pi r^2 = \frac{5}{2}r^2\alpha\)

\(\alpha = \frac{2}{5}\pi\)

(i) The area of circle C is \(\frac{1}{2} \times 3^2 \pi = \frac{9}{2} \pi\).

The area of the sector ORS is \(\frac{1}{2} \times 9^2 \theta - \frac{1}{2} \times 3^2 \theta\).

Equating the areas: \(\frac{9}{2} \pi = \frac{1}{2} \times 9^2 \theta - \frac{1}{2} \times 3^2 \theta\).

Simplifying gives \(\theta = \frac{1}{4}\pi\).

(ii) The perimeter of the shaded region includes the arc RS and the straight lines PS and PR.

The arc length RS is \(9 \times \frac{1}{4}\pi\).

The straight lines PS and PR are both 6 cm.

Thus, the perimeter is \(6 + 6 + 9 \times \frac{1}{4}\pi = 12 + 3\pi\) or approximately \(21.4 \text{ cm}\).

(i) To find angle BOC, use the formula for the angle in radians: \(\theta = 2 \tan^{-1} \left( \frac{1}{2} \right)\). Calculating gives \(\theta = 0.9273\) radians.

(ii) First, find the radius OB using the Pythagorean theorem: \(OB = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\) cm.

Next, calculate the arc length BXC using \(s = r\theta\), where \(r = \sqrt{125}\) and \(\theta = 0.9273\). Thus, \(s = 11.2 \times 0.9273 = 10.4\) cm.

The perimeter of the shaded region is the arc length plus the side BC of the square: \(10.4 + 10 = 20.4\) cm.

(iii) The area of the sector BXC is given by \(\frac{1}{2} r^2 \theta\). Substituting the values, \(\frac{1}{2} \times 11.2^2 \times 0.9273 = 7.96\) cm².

(i) The area of sector OAB is given by \(\frac{1}{2} \times 8^2 \times \alpha = 32\alpha\).

The area of semicircle OAC is \(\frac{1}{2} \times \pi \times 4^2 = 8\pi\).

According to the problem, \(8\pi = 2 \times 32\alpha\).

Solving for \(\alpha\), we get \(\alpha = \frac{\pi}{8}\).

(ii) The perimeter of the complete figure consists of the straight lines OA and OB, and the arcs AB and AC.

The length of OA and OB is 8 cm each.

The length of arc AB is \(8 \times \frac{\pi}{8} = \pi\).

The length of arc AC is \(\frac{1}{2} \times 8\pi = 4\pi\).

Thus, the total perimeter is \(8 + 8 + \pi + 4\pi = 8 + 5\pi\).

The area of triangle \(ABC\) is calculated using the formula for the area of a right triangle:

\(\text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}\)

The angle \(A\) can be found using the tangent function:

\(\tan A = \frac{2\sqrt{3}}{2} = \sqrt{3}\)

Thus, \(A = \frac{\pi}{3}\) radians (or 60 degrees).

The area of the sector \(ACD\) is given by:

\(\text{Area of sector } ACD = \frac{1}{2} \times r^2 \times \theta = \frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}\)

The area of the shaded region \(BDC\) is the area of triangle \(ABC\) minus the area of sector \(ACD\):

\(\text{Shaded area } BDC = 2\sqrt{3} - \frac{2\pi}{3}\)

(i) From the diagram, we have OQ = x + OC = 20. Using trigonometry in the right triangle, we have:

\(\sin 0.6 = \frac{x}{OC} \Rightarrow OC = \frac{x}{\sin 0.6}\)

Substitute into the equation: \(x + \frac{x}{\sin 0.6} = 20\)

Solve for \(x\): \(x = 7.218\)

(ii) The area of the sector is \(\frac{1}{2} \times 20^2 \times 1.2\).

The area of the circle with centre C is \(\pi \times 7.218^2\).

Total area outside the circle: \(\frac{1}{2} \times 20^2 \times 1.2 - \pi \times 7.218^2 = 76.3 \text{ cm}^2\)

(iii) Angle PCR is \(\pi - 1.2\).

Arc PR is \(7.218 \times (\pi - 1.2) = 14.01 \text{ cm}\).

Using trigonometry, \(OP = OR = \frac{x}{\tan 0.6}\).

Perimeter of OPSR: \(14.01 + 2 \times \frac{7.218}{\tan 0.6} = 35.1 \text{ cm}\)

(i) To find \(AC\), note that \(AC = OA - OC\). Since \(OC = r \cos \theta\), we have:

\(AC = r - r \cos \theta\).

(ii) For the perimeter of the shaded region ABD:

1. Calculate arc AB: \(\text{arc } AB = \frac{4\pi}{3}\).

2. Calculate arc AD: \(\text{arc } AD = \frac{\pi}{2} \times \text{their } AC = \frac{\pi}{2} \times (4 - 4 \cos \frac{\pi}{3}) = \pi\).

3. Calculate BD: \(BD = 4 \sin \frac{\pi}{3} - \text{their } AC = 2\sqrt{3} - 2\).

4. Add these to find the perimeter: \(\text{Perimeter} = \frac{7\pi}{3} + 2\sqrt{3} - 2\).

(i) Since OBX is a right angle (90°), we use the cosine rule: \(\cos \theta = \frac{r}{2r} = \frac{1}{2}\). Therefore, \(\theta = \frac{1}{3}\pi\) radians.

(ii) The arc length AB is \(\frac{1}{3}r\pi\). The length BX is \(r\tan(\frac{1}{3}\pi) = r\sqrt{3}\). Thus, the perimeter P is \(r + \frac{1}{3}r\pi + r\sqrt{3}\).

(iii) The area of the shaded region is the area of triangle OBX minus the area of sector AOB. The area of triangle OBX is \(\frac{1}{2}r^2\sqrt{3}\) and the area of sector AOB is \(\frac{1}{6}r^2\pi\). Therefore, the area of the shaded region is \(\frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi\).

(i) To find the length of AB, use the formula for the chord length:

AB = 2 imes r imes ext{sin} rac{ heta}{2} = 2 imes 8 imes ext{sin} rac{2.4}{2} = 14.9 ext{ cm}.

(ii) The perimeter of the plate is the sum of the chord length and the arc length. The arc length is given by:

s = r heta = 8 imes (2 ext{π} - 2.4) = 31.1 ext{ cm}.

Thus, the perimeter is:

14.9 + 31.1 = 46.0 ext{ cm}.

(iii) The area of the plate is the area of the sector minus the area of the triangle. The area of the sector is:

rac{1}{2} imes r^2 imes heta = rac{1}{2} imes 8^2 imes (2 ext{π} - 2.4) = 124.3 ext{ cm}^2.

The area of the triangle is:

rac{1}{2} imes r^2 imes ext{sin} heta = rac{1}{2} imes 8^2 imes ext{sin} 2.4 = 21.6 ext{ cm}^2.

Thus, the area of the plate is:

124.3 + 21.6 = 146 ext{ cm}^2.

First, calculate the length \(AQ\) which is the radius of the arc. Since \(Q\) is the midpoint of \(BC\), and \(ABC\) is an equilateral triangle, \(AQ = \sqrt{3}\).

Next, calculate the area of triangle \(AQC\). The height from \(A\) to \(BC\) is \(\sqrt{3}\), so the area of \(\triangle AQC\) is \(\frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\).

Now, calculate the area of the sector \(APR\). The angle \(\angle BAC\) is \(60^\circ\) or \(\frac{\pi}{3}\) radians. The area of the sector is \(\frac{1}{2} \times (\sqrt{3})^2 \times \frac{\pi}{3} = \frac{\pi}{2}\).

The area of the shaded region is the area of \(\triangle AQC\) minus the area of the sector \(APR\):

\(\text{Shaded area} = \frac{\sqrt{3}}{2} - \frac{\pi}{2} = \sqrt{3} - \frac{\pi}{2}\)

(a) To find the perimeter of the shaded region, we first find the angle \(\angle APQ\) using the cosine rule:

\(\cos^{-1} \left( \frac{5}{6} \right) = 0.5857 \text{ radians}\)

The perimeter of the shaded region is given by:

\(4 \times r \times 0.5857 = 2.34r\)

Alternatively, it can be expressed as \(0.745\pi r\) or \(\frac{293}{125}r\).

(b) To find the area of the shaded region, we use the sector formula for \(\text{Sector APB}\):

\(\frac{1}{2}r^2 \times (2 \times 0.5857)\)

Using the appropriate formula for the area of a triangle and combining it with the sector, we find:

\(\text{Shaded area} = 0.250r^2\)

Alternatively, it can be expressed as \(0.0796\pi r^2\).

(i) The area of parallelogram ABCD can be found using the formula for the area of a parallelogram:

Area = AB × BD × sin(ABD)

Given AB = BD = 10 cm and ABD = 0.8 radians, the area is:

Area = 10 × 10 × sin(0.8) = 100 × sin(0.8) = 71.7

(ii) The area of the complete figure ABQCDP includes the area of parallelogram ABCD and the areas of sectors APD and BQC. Each sector has a radius of 10 cm and angle 0.8 radians.

Area of one sector = \(\frac{1}{2} \times 10^2 \times 0.8 = 40\)

Total area of sectors = 2 × 40 = 80

Total area of figure = Area of parallelogram + Total area of sectors = 71.7 + 8.3 = 80

(iii) The perimeter of the complete figure ABQCDP includes the sides AB, BC, CD, DA, and the arcs APD and BQC.

Length of one arc = 10 × 0.8 = 8

Total length of arcs = 2 × 8 = 16

Perimeter = AB + BC + CD + DA + Total length of arcs = 10 + 10 + 10 + 10 + 16 = 36

(i) The perpendicular distance from \(D\) to \(AX\) is \(6 \sin \frac{\pi}{3} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\).

The perpendicular distance from \(E\) to \(AX\) is \(10 \sin \theta\).

Equating these distances gives \(10 \sin \theta = 3\sqrt{3}\), so \(\sin \theta = \frac{3\sqrt{3}}{10}\).

Thus, \(\theta = \sin^{-1}\left(\frac{3\sqrt{3}}{10}\right)\).

(ii) The arc \(DX\) is \(6 \times \frac{1}{3}\pi = 2\pi\).

The arc \(EX\) is \(10 \times 0.5464 = 5.464\).

The horizontal steps are \(6 \cos \frac{\pi}{3}\) and \(10 \cos \theta\).

\(DE = 10 + 6 - 6 \cos \frac{\pi}{3} - 10 \cos \theta\).

The perimeter is \(\text{arc } DX + \text{arc } BX + DE = 16.20\).

(i) The perimeter of the plate consists of the arc AB and the sides OC, CB, and OA. The length of the arc AB is given by \(r\theta\). The side OC is \(r \sin \theta\) and CB is \(r \cos \theta\). The side OA is r. Therefore, the perimeter is:

\(r\theta + r \sin \theta + r \cos \theta + r = r(1 + \theta + \cos \theta + \sin \theta)\)

(ii) The area of the plate consists of the area of the sector OAB and the area of the triangle OCB. The area of the sector OAB is \(\frac{1}{2} r^2 \theta\). For r = 10 and \(\theta = \frac{1}{5}\pi\), this becomes:

\(\frac{1}{2} \times 10^2 \times \frac{\pi}{5} = 31.42\)

The area of the triangle OCB is \(\frac{1}{2} \times OC \times CB\). With OC = \(10 \sin \frac{\pi}{5}\) and CB = \(10 \cos \frac{\pi}{5}\), the area is:

\(\frac{1}{2} \times 10 \cos \frac{\pi}{5} \times 10 \sin \frac{\pi}{5} = 23.78\)

Thus, the total area of the plate is:

\(31.42 + 23.78 = 55.2\)

(i) Since AX is tangent to the circle at A, \(\angle OAX = \frac{\pi}{2}\). Using the tangent formula, \(AX = 6 \tan \frac{\pi}{3} = 6 \sqrt{3}\).

(ii) The area of triangle OAX is \(\frac{1}{2} \times 6 \times 6 \sqrt{3} = 18 \sqrt{3}\).

The area of sector OAB is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\).

Thus, the area of the shaded region is \(18 \sqrt{3} - 6\pi\).

(iii) The arc length AB is \(6 \times \frac{\pi}{3} = 2\pi\).

Using trigonometry, \(OX = \frac{6}{\cos \frac{\pi}{3}} = 12\), so \(BX = 6\).

The perimeter of the shaded region is \(6 \sqrt{3} + 2\pi + 6\).

(i) The length of AS is \(r \tan \theta\). The area of triangle OAB is \(r^2 \tan \theta\) or \(\frac{1}{2} r^2 \tan \theta\). The area of sector OPS is \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\). Therefore, the shaded area is \(r^2 (\tan \theta - \theta)\).

(ii) Given \(\theta = \frac{1}{3}\pi\) and \(r = 6\), we have \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\). Thus, \(AP = 6\). The length of AS is \(6 \tan \frac{\pi}{3} \Rightarrow AB = 12\sqrt{3}\). The arc PST is \(12 \times \frac{\pi}{3}\). Therefore, the perimeter is \(12 + 12\sqrt{3} + 4\pi\).

(i) To find the area of the shaded region BPDQ, we calculate the area of sector BPD and subtract the area of triangle BPD. The area of sector BPD is given by \(\frac{1}{2} \times 5^2 \times 1.2\). The area of triangle BPD is \(\frac{1}{2} \times 5^2 \times \sin 1.2\). Therefore, the area of the shaded region is:

\(2 \left[ \frac{1}{2} \times 5^2 \times 1.2 - \frac{1}{2} \times 5^2 \times \sin 1.2 \right] = 6.70\)

(ii) To find the length of PQ, we use the cosine rule in triangle APQ. The length of PQ is given by:

\(5 \cos 0.6\)

Then, subtract from 5:

\(5 - 5 \cos 0.6\)

Multiply by 2:

\(10(1 - \cos 0.6) = 1.75\)

(i) Use the Pythagorean theorem in triangle \(PQR\):

\(RS^2 = PR^2 - PQ^2 = 10^2 - 6^2\)

\(RS^2 = 100 - 36 = 64\)

\(RS = \sqrt{64} = 8 \text{ cm}\)

(ii) Use trigonometry to find \(\angle RPQ\):

\(\sin \theta = \frac{8}{10}\)

\(\theta = \arcsin\left(\frac{8}{10}\right) \approx 0.9273 \text{ radians}\)

(iii) The area of the shaded region is the area of the trapezium minus the areas of the two sectors:

Area of trapezium = \(40 \text{ cm}^2\)

Large sector area = \(\frac{1}{2} \times 8^2 \times 0.9273\)

Small sector angle = \(\pi - 0.9273\)

Small sector area = \(\frac{1}{2} \times 2^2 \times 2.214\)

Shaded area = \(40 - (\text{Large sector} + \text{Small sector}) = 5.90 \text{ cm}^2\)

(i) To find angle DOE, use the sine rule in triangle DOE:

\(\sin \frac{1}{2} \theta = \frac{6}{10}\)

Solving for \(\theta\), we find:

\(\theta = 1.287 \text{ radians}\)

(ii) To find the perimeter of the metal plate, calculate the arc length and add the straight edges:

Arc length of BOD = \(2 \times 10 \times (\pi - 1.287)\)

Perimeter = \(12 + 12 + 2 \times 10 \times (\pi - 1.287) = 61.1 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the area of the sectors and triangles:

Area of sector DOE = \(\frac{1}{2} \times 10^2 \times 1.287\)

Area of triangle DOE = \(\frac{1}{2} \times 10^2 \times \sin 1.287\)

Total area = \(\pi \times 10^2 - (2 \times \text{sectors} - 2 \times \text{triangles})\)

Area = 281 or 282 cm².

(i) To find the length of the arc BD, we first calculate the length of AB using the Pythagorean theorem:

\(AB = \sqrt{6^2 + 6^2} = \sqrt{72}\)

The angle BAD is 45° or \(\frac{\pi}{4}\) radians. The arc length s is given by \(s = r\theta\), where \(r = \sqrt{72}\) and \(\theta = \frac{\pi}{4}\):

\(s = \sqrt{72} \times \frac{\pi}{4} \approx 6.67 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OAB. The sector area is given by:

\(\text{Sector area} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 72 \times \frac{\pi}{4}\)

The area of triangle OAB is:

\(\text{Area of triangle} = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2\)

Thus, the shaded area is:

\(\text{Shaded area} = \frac{1}{2} \times 72 \times \frac{\pi}{4} - 18 \approx 10.3 \text{ cm}^2 \text{ or } 9\pi - 18 \text{ cm}^2\)

(a) The perimeter of the cross-section RASB is the sum of the arc lengths ARB and RBS. The arc length ARB is given by \(2.5 \times \frac{4\pi}{3}\) and the arc length RBS is given by \(2.24 \times \frac{5\pi}{6}\). Therefore, the perimeter is:

\(2.5 \times \frac{4\pi}{3} + 2.24 \times \frac{5\pi}{6} = 10.47 + 5.86 = 16.34\)

(b) The area of triangle AOB is \(\frac{1}{2} \times 2.5^2 \times \sin \frac{2\pi}{3} = 2.706\) and the area of triangle APB is \(\frac{1}{2} \times 2.24^2 \times \sin \frac{5\pi}{6} = 1.254\). The difference in area is:

\(2.706 - 1.254 = 1.45\)

(c) The area of the cross-section RASB is the sum of the areas of sectors OARB and PASB plus the area of quadrilateral OAPB. The area of sector OARB is \(\frac{1}{2} \times 2.5^2 \times \frac{4\pi}{3} = 13.09\) and the area of sector PASB is \(\frac{1}{2} \times 2.24^2 \times \frac{5\pi}{6} = 6.57\). Adding the difference in area from part (b), the total area is:

\(13.09 + 6.57 + 1.45 = 21.1\)

(i) The perimeter of R₁ is given by the sum of the two radii and the arc length: \(r + r + r\theta = 2r + r\theta\).

The length of the major arc AB is \(2\pi r - r\theta\).

Equating the two expressions: \(2r + r\theta = 2\pi r - r\theta\).

Solving for \(\theta\):

\(2r + r\theta = 2\pi r - r\theta\)

\(2r + r\theta + r\theta = 2\pi r\)

\(2r + 2r\theta = 2\pi r\)

\(2r(1 + \theta) = 2\pi r\)

\(1 + \theta = \pi\)

\(\theta = \pi - 1\)

(ii) The area of region R₁ is given by \(\frac{1}{2}r^2\theta\).

Given \(\theta = \pi - 1\) and area of R₁ is 30 cm²:

\(\frac{1}{2}r^2(\pi - 1) = 30\)

\(r^2 = \frac{60}{\pi - 1}\)

\(r = \sqrt{\frac{60}{\pi - 1}} \approx 5.29\)

The area of region R₂ is the area of the circle minus the area of R₁:

\(R_2 = \frac{1}{2}r^2(\pi + 1)\)

\(R_2 = \frac{1}{2} \times \frac{60}{\pi - 1} \times (\pi + 1)\)

\(R_2 \approx 58.0 \text{ cm}^2\)

(i) To find the angle \(\angle POQ\) in radians, use the formula for arc length: \(s = r\theta\). Given \(s = 9\) cm and \(r = 5\) cm, we have:

\(9 = 5\theta\)

\(\theta = \frac{9}{5} = 1.8 \text{ rad}\)

(ii) To find the length of \(PT\), consider triangle \(POT\). The angle \(\angle POT\) is half of \(\angle POQ\), so \(\angle POT = \frac{1.8}{2} = 0.9 \text{ rad}\). Using the tangent function:

\(PT = 5 \tan(0.9) \approx 6.30 \text{ cm}\)

(iii) To find the area of the shaded region, first calculate the area of sector \(POQ\):

\(\text{Area of sector} = \frac{1}{2} \times 5^2 \times 1.8 = 22.5 \text{ cm}^2\)

Next, calculate the area of triangle \(POT\):

\(\text{Area of } POT = \frac{1}{2} \times 5 \times 6.30 = 15.75 \text{ cm}^2\)

The shaded area is twice the area of triangle \(POT\) minus the area of the sector:

\(\text{Shaded area} = 2 \times 15.75 - 22.5 = 9.00 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we first use the Pythagorean theorem in triangle OPT to find OT:

\(OT = \sqrt{OP^2 + PT^2} = \sqrt{5^2 + 12^2} = 13 \text{ cm}\)

Next, find QT:

\(QT = OT - OQ = 13 - 5 = 8 \text{ cm}\)

Calculate the angle POQ using the tangent function:

\(\angle POQ = \tan^{-1}\left(\frac{12}{5}\right) = 1.176 \text{ radians}\)

Find the arc length S:

\(S = r\theta = 5 \times 1.176 = 5.88 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(5.88 + 12 + 8 = 25.9 \text{ cm}\)

(ii) To find the area of the shaded region, calculate the area of sector POQ:

\(\text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.176 = 14.7 \text{ cm}^2\)

Calculate the area of triangle OPT:

\(\text{Area of triangle} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)

The area of the shaded region is:

\(30 - 14.7 = 15.3 \text{ cm}^2\)

(i) The area of sector AOB is \(\frac{1}{2}r^2\theta\).

In \(\triangle OAX\), \(AX = r\sin\theta\) and \(OX = r\cos\theta\).

The area of \(\triangle OAX\) is \(\frac{1}{2} \times OX \times AX = \frac{1}{2}r^2\sin\theta\cos\theta\).

Thus, the area of the shaded region AXB is:

\(A = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)\)

(ii) Given \(r = 12\) and \(\theta = \frac{1}{6}\pi\):

\(AX = 12\sin\frac{1}{6}\pi = 6\)

\(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\)

\(BX = OB - OX = 12 - 6\sqrt{3}\)

The length of arc AB is \(r\theta = 12 \times \frac{1}{6}\pi = 2\pi\).

Therefore, the perimeter of the shaded region AXB is:

\(P = AX + BX + \text{arc } AB = 6 + (12 - 6\sqrt{3}) + 2\pi = 18 - 6\sqrt{3} + 2\pi\)

To find the length of AX, we use the tangent of the angle \(\frac{1}{6} \pi\) (half of \(\frac{1}{3} \pi\)) in the right triangle formed by O, A, and X.

\(\tan \frac{1}{6} \pi = \frac{AX}{12}\)

\(\tan \frac{1}{6} \pi = \frac{\sqrt{3}}{3}\)

Thus, \(\frac{\sqrt{3}}{3} = \frac{AX}{12}\)

Solving for \(AX\), we get:

\(AX = 12 \times \frac{\sqrt{3}}{3} = 4\sqrt{3}\)

First, calculate the length of OC and AC using trigonometry. Since \(\angle AOB = \frac{1}{3} \pi\), we have:

\(OC = 12 \cos\left(\frac{1}{3} \pi\right) = 12 \times \frac{1}{2} = 6\)

\(AC = 12 \sin\left(\frac{1}{3} \pi\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\)

The area of the sector AOB is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{1}{3} \pi = 24\pi\)

The area of rectangle OCDB is:

\(12 \times 6\sqrt{3} = 72\sqrt{3}\)

The area of triangle OAC is:

\(\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

The area of the shaded region is the area of the rectangle minus the area of the sector and the triangle:

\(72\sqrt{3} - (24\pi + 18\sqrt{3}) = 54\sqrt{3} - 24\pi\)

Thus, the values of a and b are 54 and 24, respectively.

(i) To find angle AOB, use the tangent formula in the right triangle formed by the radius and the tangent: \(\tan(\frac{x}{2}) = \frac{15}{8} = 1.875\). Solving for \(\frac{x}{2}\), we get \(\frac{x}{2} = 1.081\). Therefore, \(x = 2.16\) radians.

(ii) The perimeter of the shaded region is the sum of the lengths of AT, BT, and the arc AB. The arc length is given by \(r\theta = 8 \times 2.16 = 17.3\) cm. Thus, the perimeter is \(15 + 15 + 17.3 = 47.3\) cm.

(iii) The area of sector AOB is \(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times 2.16 = 69.1\) cm². The area of triangle AOT is \(2 \times \frac{1}{2} \times 8 \times 15 = 120\) cm². The shaded area is \(120 - 69.1 = 50.9\) cm².

The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).

For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).

At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:

\(2 \frac{dy}{dx} + 6 = 0\)

\(\frac{dy}{dx} = -\frac{6}{2} = -3\)

The gradient of the tangent at \(P\) is \(-3\).

The angle \(\theta\) between the line and the tangent is given by:

\(\tan \theta = \left| \frac{-3 - (-2)}{1 + (-3)(-2)} \right| = \left| \frac{-1}{1 + 6} \right| = \frac{1}{7}\)

\(\theta = \tan^{-1} \left( \frac{1}{7} \right) \approx 8.1^\circ\)

(i) To find BD, use the formula for the perpendicular from the center to the chord: BD = 9\sin(\pi - 2.4). Calculating gives:

\(BD = 9\sin(\pi - 2.4) = 6.08 \text{ cm}\)

(ii) To find the perimeter of the shaded region, calculate the arc length AB and add the lengths of BD, OD, and OA:

Arc AB = 9 \times 2.4 = 21.6 \text{ cm}

Using Pythagoras or trigonometry, OD = 9\cos(\pi - 2.4) = 6.64 \text{ cm}

Perimeter = 21.6 + 6.08 + 9 + 6.64 = 43.3 \text{ cm}

(iii) To find the area of the shaded region, calculate the area of the sector and subtract the area of triangle OBD:

Area of sector = \(\frac{1}{2} \times 9^2 \times 2.4\)

Area of triangle OBD = \(\frac{1}{2} \times 6.08 \times 6.64\)

Total area = \(\frac{1}{2} \times 81 \times 2.4 - \frac{1}{2} \times 6.08 \times 6.64 = 117 \text{ cm}^2\)