Past Exam Questions

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:

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