Past Exam Questions

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