Past Exam Questions

(i) To find angle BOC, use the formula for the angle in radians: \(\theta = 2 \tan^{-1} \left( \frac{1}{2} \right)\). Calculating gives \(\theta = 0.9273\) radians.

(ii) First, find the radius OB using the Pythagorean theorem: \(OB = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\) cm.

Next, calculate the arc length BXC using \(s = r\theta\), where \(r = \sqrt{125}\) and \(\theta = 0.9273\). Thus, \(s = 11.2 \times 0.9273 = 10.4\) cm.

The perimeter of the shaded region is the arc length plus the side BC of the square: \(10.4 + 10 = 20.4\) cm.

(iii) The area of the sector BXC is given by \(\frac{1}{2} r^2 \theta\). Substituting the values, \(\frac{1}{2} \times 11.2^2 \times 0.9273 = 7.96\) cm².

(i) The area of sector OAB is given by \(\frac{1}{2} \times 8^2 \times \alpha = 32\alpha\).

The area of semicircle OAC is \(\frac{1}{2} \times \pi \times 4^2 = 8\pi\).

According to the problem, \(8\pi = 2 \times 32\alpha\).

Solving for \(\alpha\), we get \(\alpha = \frac{\pi}{8}\).

(ii) The perimeter of the complete figure consists of the straight lines OA and OB, and the arcs AB and AC.

The length of OA and OB is 8 cm each.

The length of arc AB is \(8 \times \frac{\pi}{8} = \pi\).

The length of arc AC is \(\frac{1}{2} \times 8\pi = 4\pi\).

Thus, the total perimeter is \(8 + 8 + \pi + 4\pi = 8 + 5\pi\).

The area of triangle \(ABC\) is calculated using the formula for the area of a right triangle:

\(\text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}\)

The angle \(A\) can be found using the tangent function:

\(\tan A = \frac{2\sqrt{3}}{2} = \sqrt{3}\)

Thus, \(A = \frac{\pi}{3}\) radians (or 60 degrees).

The area of the sector \(ACD\) is given by:

\(\text{Area of sector } ACD = \frac{1}{2} \times r^2 \times \theta = \frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}\)

The area of the shaded region \(BDC\) is the area of triangle \(ABC\) minus the area of sector \(ACD\):

\(\text{Shaded area } BDC = 2\sqrt{3} - \frac{2\pi}{3}\)

(i) From the diagram, we have OQ = x + OC = 20. Using trigonometry in the right triangle, we have:

\(\sin 0.6 = \frac{x}{OC} \Rightarrow OC = \frac{x}{\sin 0.6}\)

Substitute into the equation: \(x + \frac{x}{\sin 0.6} = 20\)

Solve for \(x\): \(x = 7.218\)

(ii) The area of the sector is \(\frac{1}{2} \times 20^2 \times 1.2\).

The area of the circle with centre C is \(\pi \times 7.218^2\).

Total area outside the circle: \(\frac{1}{2} \times 20^2 \times 1.2 - \pi \times 7.218^2 = 76.3 \text{ cm}^2\)

(iii) Angle PCR is \(\pi - 1.2\).

Arc PR is \(7.218 \times (\pi - 1.2) = 14.01 \text{ cm}\).

Using trigonometry, \(OP = OR = \frac{x}{\tan 0.6}\).

Perimeter of OPSR: \(14.01 + 2 \times \frac{7.218}{\tan 0.6} = 35.1 \text{ cm}\)

(i) To find \(AC\), note that \(AC = OA - OC\). Since \(OC = r \cos \theta\), we have:

\(AC = r - r \cos \theta\).

(ii) For the perimeter of the shaded region ABD:

1. Calculate arc AB: \(\text{arc } AB = \frac{4\pi}{3}\).

2. Calculate arc AD: \(\text{arc } AD = \frac{\pi}{2} \times \text{their } AC = \frac{\pi}{2} \times (4 - 4 \cos \frac{\pi}{3}) = \pi\).

3. Calculate BD: \(BD = 4 \sin \frac{\pi}{3} - \text{their } AC = 2\sqrt{3} - 2\).

4. Add these to find the perimeter: \(\text{Perimeter} = \frac{7\pi}{3} + 2\sqrt{3} - 2\).

(i) Since OBX is a right angle (90°), we use the cosine rule: \(\cos \theta = \frac{r}{2r} = \frac{1}{2}\). Therefore, \(\theta = \frac{1}{3}\pi\) radians.

(ii) The arc length AB is \(\frac{1}{3}r\pi\). The length BX is \(r\tan(\frac{1}{3}\pi) = r\sqrt{3}\). Thus, the perimeter P is \(r + \frac{1}{3}r\pi + r\sqrt{3}\).

(iii) The area of the shaded region is the area of triangle OBX minus the area of sector AOB. The area of triangle OBX is \(\frac{1}{2}r^2\sqrt{3}\) and the area of sector AOB is \(\frac{1}{6}r^2\pi\). Therefore, the area of the shaded region is \(\frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi\).

(i) To find the length of AB, use the formula for the chord length:

AB = 2 imes r imes ext{sin} rac{ heta}{2} = 2 imes 8 imes ext{sin} rac{2.4}{2} = 14.9 ext{ cm}.

(ii) The perimeter of the plate is the sum of the chord length and the arc length. The arc length is given by:

s = r heta = 8 imes (2 ext{π} - 2.4) = 31.1 ext{ cm}.

Thus, the perimeter is:

14.9 + 31.1 = 46.0 ext{ cm}.

(iii) The area of the plate is the area of the sector minus the area of the triangle. The area of the sector is:

rac{1}{2} imes r^2 imes heta = rac{1}{2} imes 8^2 imes (2 ext{π} - 2.4) = 124.3 ext{ cm}^2.

The area of the triangle is:

rac{1}{2} imes r^2 imes ext{sin} heta = rac{1}{2} imes 8^2 imes ext{sin} 2.4 = 21.6 ext{ cm}^2.

Thus, the area of the plate is:

124.3 + 21.6 = 146 ext{ cm}^2.

First, calculate the length \(AQ\) which is the radius of the arc. Since \(Q\) is the midpoint of \(BC\), and \(ABC\) is an equilateral triangle, \(AQ = \sqrt{3}\).

Next, calculate the area of triangle \(AQC\). The height from \(A\) to \(BC\) is \(\sqrt{3}\), so the area of \(\triangle AQC\) is \(\frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\).

Now, calculate the area of the sector \(APR\). The angle \(\angle BAC\) is \(60^\circ\) or \(\frac{\pi}{3}\) radians. The area of the sector is \(\frac{1}{2} \times (\sqrt{3})^2 \times \frac{\pi}{3} = \frac{\pi}{2}\).

The area of the shaded region is the area of \(\triangle AQC\) minus the area of the sector \(APR\):

\(\text{Shaded area} = \frac{\sqrt{3}}{2} - \frac{\pi}{2} = \sqrt{3} - \frac{\pi}{2}\)

(a) To find the perimeter of the shaded region, we first find the angle \(\angle APQ\) using the cosine rule:

\(\cos^{-1} \left( \frac{5}{6} \right) = 0.5857 \text{ radians}\)

The perimeter of the shaded region is given by:

\(4 \times r \times 0.5857 = 2.34r\)

Alternatively, it can be expressed as \(0.745\pi r\) or \(\frac{293}{125}r\).

(b) To find the area of the shaded region, we use the sector formula for \(\text{Sector APB}\):

\(\frac{1}{2}r^2 \times (2 \times 0.5857)\)

Using the appropriate formula for the area of a triangle and combining it with the sector, we find:

\(\text{Shaded area} = 0.250r^2\)

Alternatively, it can be expressed as \(0.0796\pi r^2\).

(i) The area of parallelogram ABCD can be found using the formula for the area of a parallelogram:

Area = AB × BD × sin(ABD)

Given AB = BD = 10 cm and ABD = 0.8 radians, the area is:

Area = 10 × 10 × sin(0.8) = 100 × sin(0.8) = 71.7

(ii) The area of the complete figure ABQCDP includes the area of parallelogram ABCD and the areas of sectors APD and BQC. Each sector has a radius of 10 cm and angle 0.8 radians.

Area of one sector = \(\frac{1}{2} \times 10^2 \times 0.8 = 40\)

Total area of sectors = 2 × 40 = 80

Total area of figure = Area of parallelogram + Total area of sectors = 71.7 + 8.3 = 80

(iii) The perimeter of the complete figure ABQCDP includes the sides AB, BC, CD, DA, and the arcs APD and BQC.

Length of one arc = 10 × 0.8 = 8

Total length of arcs = 2 × 8 = 16

Perimeter = AB + BC + CD + DA + Total length of arcs = 10 + 10 + 10 + 10 + 16 = 36

(i) The perpendicular distance from \(D\) to \(AX\) is \(6 \sin \frac{\pi}{3} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\).

The perpendicular distance from \(E\) to \(AX\) is \(10 \sin \theta\).

Equating these distances gives \(10 \sin \theta = 3\sqrt{3}\), so \(\sin \theta = \frac{3\sqrt{3}}{10}\).

Thus, \(\theta = \sin^{-1}\left(\frac{3\sqrt{3}}{10}\right)\).

(ii) The arc \(DX\) is \(6 \times \frac{1}{3}\pi = 2\pi\).

The arc \(EX\) is \(10 \times 0.5464 = 5.464\).

The horizontal steps are \(6 \cos \frac{\pi}{3}\) and \(10 \cos \theta\).

\(DE = 10 + 6 - 6 \cos \frac{\pi}{3} - 10 \cos \theta\).

The perimeter is \(\text{arc } DX + \text{arc } BX + DE = 16.20\).

(i) The perimeter of the plate consists of the arc AB and the sides OC, CB, and OA. The length of the arc AB is given by \(r\theta\). The side OC is \(r \sin \theta\) and CB is \(r \cos \theta\). The side OA is r. Therefore, the perimeter is:

\(r\theta + r \sin \theta + r \cos \theta + r = r(1 + \theta + \cos \theta + \sin \theta)\)

(ii) The area of the plate consists of the area of the sector OAB and the area of the triangle OCB. The area of the sector OAB is \(\frac{1}{2} r^2 \theta\). For r = 10 and \(\theta = \frac{1}{5}\pi\), this becomes:

\(\frac{1}{2} \times 10^2 \times \frac{\pi}{5} = 31.42\)

The area of the triangle OCB is \(\frac{1}{2} \times OC \times CB\). With OC = \(10 \sin \frac{\pi}{5}\) and CB = \(10 \cos \frac{\pi}{5}\), the area is:

\(\frac{1}{2} \times 10 \cos \frac{\pi}{5} \times 10 \sin \frac{\pi}{5} = 23.78\)

Thus, the total area of the plate is:

\(31.42 + 23.78 = 55.2\)

(i) Since AX is tangent to the circle at A, \(\angle OAX = \frac{\pi}{2}\). Using the tangent formula, \(AX = 6 \tan \frac{\pi}{3} = 6 \sqrt{3}\).

(ii) The area of triangle OAX is \(\frac{1}{2} \times 6 \times 6 \sqrt{3} = 18 \sqrt{3}\).

The area of sector OAB is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\).

Thus, the area of the shaded region is \(18 \sqrt{3} - 6\pi\).

(iii) The arc length AB is \(6 \times \frac{\pi}{3} = 2\pi\).

Using trigonometry, \(OX = \frac{6}{\cos \frac{\pi}{3}} = 12\), so \(BX = 6\).

The perimeter of the shaded region is \(6 \sqrt{3} + 2\pi + 6\).

(i) The length of AS is \(r \tan \theta\). The area of triangle OAB is \(r^2 \tan \theta\) or \(\frac{1}{2} r^2 \tan \theta\). The area of sector OPS is \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\). Therefore, the shaded area is \(r^2 (\tan \theta - \theta)\).

(ii) Given \(\theta = \frac{1}{3}\pi\) and \(r = 6\), we have \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\). Thus, \(AP = 6\). The length of AS is \(6 \tan \frac{\pi}{3} \Rightarrow AB = 12\sqrt{3}\). The arc PST is \(12 \times \frac{\pi}{3}\). Therefore, the perimeter is \(12 + 12\sqrt{3} + 4\pi\).

(i) To find the area of the shaded region BPDQ, we calculate the area of sector BPD and subtract the area of triangle BPD. The area of sector BPD is given by \(\frac{1}{2} \times 5^2 \times 1.2\). The area of triangle BPD is \(\frac{1}{2} \times 5^2 \times \sin 1.2\). Therefore, the area of the shaded region is:

\(2 \left[ \frac{1}{2} \times 5^2 \times 1.2 - \frac{1}{2} \times 5^2 \times \sin 1.2 \right] = 6.70\)

(ii) To find the length of PQ, we use the cosine rule in triangle APQ. The length of PQ is given by:

\(5 \cos 0.6\)

Then, subtract from 5:

\(5 - 5 \cos 0.6\)

Multiply by 2:

\(10(1 - \cos 0.6) = 1.75\)

(i) Use the Pythagorean theorem in triangle \(PQR\):

\(RS^2 = PR^2 - PQ^2 = 10^2 - 6^2\)

\(RS^2 = 100 - 36 = 64\)

\(RS = \sqrt{64} = 8 \text{ cm}\)

(ii) Use trigonometry to find \(\angle RPQ\):

\(\sin \theta = \frac{8}{10}\)

\(\theta = \arcsin\left(\frac{8}{10}\right) \approx 0.9273 \text{ radians}\)

(iii) The area of the shaded region is the area of the trapezium minus the areas of the two sectors:

Area of trapezium = \(40 \text{ cm}^2\)

Large sector area = \(\frac{1}{2} \times 8^2 \times 0.9273\)

Small sector angle = \(\pi - 0.9273\)

Small sector area = \(\frac{1}{2} \times 2^2 \times 2.214\)

Shaded area = \(40 - (\text{Large sector} + \text{Small sector}) = 5.90 \text{ cm}^2\)

(i) To find angle DOE, use the sine rule in triangle DOE:

\(\sin \frac{1}{2} \theta = \frac{6}{10}\)

Solving for \(\theta\), we find:

\(\theta = 1.287 \text{ radians}\)

(ii) To find the perimeter of the metal plate, calculate the arc length and add the straight edges:

Arc length of BOD = \(2 \times 10 \times (\pi - 1.287)\)

Perimeter = \(12 + 12 + 2 \times 10 \times (\pi - 1.287) = 61.1 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the area of the sectors and triangles:

Area of sector DOE = \(\frac{1}{2} \times 10^2 \times 1.287\)

Area of triangle DOE = \(\frac{1}{2} \times 10^2 \times \sin 1.287\)

Total area = \(\pi \times 10^2 - (2 \times \text{sectors} - 2 \times \text{triangles})\)

Area = 281 or 282 cm².

(i) To find the length of the arc BD, we first calculate the length of AB using the Pythagorean theorem:

\(AB = \sqrt{6^2 + 6^2} = \sqrt{72}\)

The angle BAD is 45° or \(\frac{\pi}{4}\) radians. The arc length s is given by \(s = r\theta\), where \(r = \sqrt{72}\) and \(\theta = \frac{\pi}{4}\):

\(s = \sqrt{72} \times \frac{\pi}{4} \approx 6.67 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OAB. The sector area is given by:

\(\text{Sector area} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 72 \times \frac{\pi}{4}\)

The area of triangle OAB is:

\(\text{Area of triangle} = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2\)

Thus, the shaded area is:

\(\text{Shaded area} = \frac{1}{2} \times 72 \times \frac{\pi}{4} - 18 \approx 10.3 \text{ cm}^2 \text{ or } 9\pi - 18 \text{ cm}^2\)

(a) The perimeter of the cross-section RASB is the sum of the arc lengths ARB and RBS. The arc length ARB is given by \(2.5 \times \frac{4\pi}{3}\) and the arc length RBS is given by \(2.24 \times \frac{5\pi}{6}\). Therefore, the perimeter is:

\(2.5 \times \frac{4\pi}{3} + 2.24 \times \frac{5\pi}{6} = 10.47 + 5.86 = 16.34\)

(b) The area of triangle AOB is \(\frac{1}{2} \times 2.5^2 \times \sin \frac{2\pi}{3} = 2.706\) and the area of triangle APB is \(\frac{1}{2} \times 2.24^2 \times \sin \frac{5\pi}{6} = 1.254\). The difference in area is:

\(2.706 - 1.254 = 1.45\)

(c) The area of the cross-section RASB is the sum of the areas of sectors OARB and PASB plus the area of quadrilateral OAPB. The area of sector OARB is \(\frac{1}{2} \times 2.5^2 \times \frac{4\pi}{3} = 13.09\) and the area of sector PASB is \(\frac{1}{2} \times 2.24^2 \times \frac{5\pi}{6} = 6.57\). Adding the difference in area from part (b), the total area is:

\(13.09 + 6.57 + 1.45 = 21.1\)

(i) The perimeter of R₁ is given by the sum of the two radii and the arc length: \(r + r + r\theta = 2r + r\theta\).

The length of the major arc AB is \(2\pi r - r\theta\).

Equating the two expressions: \(2r + r\theta = 2\pi r - r\theta\).

Solving for \(\theta\):

\(2r + r\theta = 2\pi r - r\theta\)

\(2r + r\theta + r\theta = 2\pi r\)

\(2r + 2r\theta = 2\pi r\)

\(2r(1 + \theta) = 2\pi r\)

\(1 + \theta = \pi\)

\(\theta = \pi - 1\)

(ii) The area of region R₁ is given by \(\frac{1}{2}r^2\theta\).

Given \(\theta = \pi - 1\) and area of R₁ is 30 cm²:

\(\frac{1}{2}r^2(\pi - 1) = 30\)

\(r^2 = \frac{60}{\pi - 1}\)

\(r = \sqrt{\frac{60}{\pi - 1}} \approx 5.29\)

The area of region R₂ is the area of the circle minus the area of R₁:

\(R_2 = \frac{1}{2}r^2(\pi + 1)\)

\(R_2 = \frac{1}{2} \times \frac{60}{\pi - 1} \times (\pi + 1)\)

\(R_2 \approx 58.0 \text{ cm}^2\)