Past Exam Questions

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:

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