Past Exam Questions

(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 \(\approx 35.0 \text{ or } 34.9 \text{ cm}^2\).