Past Exam Questions

(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:

\[ \tan(\angle BDC) = \frac{4}{3} \]

Thus, \( \angle BDC = 0.927 \) radians.

Then, \( \angle ADC = \pi - 0.927 = 2.214 \) radians.

The arc length \( \widehat{AC} \) is given by:

\[ \widehat{AC} = 5 \times 2.214 = 11.07\ \text{cm} \]

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

\[ AC = \sqrt{8^2 + 4^2} = 8.94\ \text{cm} \]

Therefore, the perimeter of the shaded region is:

\[ 11.07 + 8.94 = 20.0\ \text{cm} \]

(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:

\[ \frac{1}{2} \times 5^2 \times 2.214 = 27.7\ \text{cm}^2 \]

The area of triangle \( ADC \) is:

\[ \frac{1}{2} \times 5 \times 4 = 10\ \text{cm}^2 \]

Thus, the area of the shaded region is:

\[ 27.7 - 10 = 17.7\ \text{cm}^2 \]

(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 \(21.4 \text{ cm}\).