O Level Elementary Mathematics Practice Paper 3

Answer Key and Marking Scheme

TuitionGoWhere Practice Paper - Elementary Mathematics O-Level Topic: Geometry & Trigonometry (Version 3)

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Section A

1.

• Concept: Pythagoras' Theorem.
• Working: $AC^2 = AB^2 + BC^2$ $AC^2 = 12^2 + 5^2 = 144 + 25 = 169$ $AC = \sqrt{169} = 13$
• Answer: 13 cm
• Marks: [2] (1 for substitution, 1 for answer)

2.

• Concept: Inverse trigonometric ratios.
• Working: $x = \tan^{-1}(0.5)$ $x \approx 26.565...$
• Answer: 26.6
• Marks: [2] (1 for correct inverse operation, 1 for correct rounding)

3.

• Concept: Angle at centre is twice angle at circumference.
• Working: The angle at the centre $\angle AOC = 130^\circ$. The angle at the circumference $\angle ABC$ subtends the same arc $AC$. $\angle ABC = \frac{1}{2} \times \angle AOC$ $\angle ABC = \frac{1}{2} \times 130^\circ = 65^\circ$
• Answer: 65
• Marks: [2] (1 for stating relationship, 1 for answer)

4.

• Concept: Cosine ratio in right-angled triangle.
• Working: Let $\theta$ be the angle with the ground. Adjacent side = 2.5 m, Hypotenuse = 6 m. $\cos \theta = \frac{2.5}{6}$ $\theta = \cos^{-1}\left(\frac{2.5}{6}\right)$ $\theta \approx 65.375...^\circ$
• Answer: 65.4
• Marks: [2] (1 for correct ratio, 1 for answer)

5.

• Concept: Parallel lines and angles (Zig-zag theorem).
• Working: Draw a line through $E$ parallel to $AB$ and $CD$. The angle at $E$ is split into two parts: alternate interior to $\angle BAE$ and alternate interior to $\angle CDE$. $\angle AED = \angle BAE + \angle CDE$ $\angle AED = 40^\circ + 35^\circ = 75^\circ$
• Answer: 75
• Marks: [2] (1 for method/reasoning, 1 for answer)

6.

• Concept: Volume of a cylinder.
• Working: $V = \pi r^2 h$ $500 = \pi r^2 (10)$ $r^2 = \frac{500}{10\pi} = \frac{50}{\pi}$ $r = \sqrt{\frac{50}{\pi}} \approx 3.989...$
• Answer: 3.99
• Marks: [3] (1 for formula, 1 for substitution/rearrangement, 1 for answer)

7.

• Concept: Cosine Rule.
• Working: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)$ $PR^2 = 8^2 + 10^2 - 2(8)(10)\cos(60^\circ)$ $PR^2 = 64 + 100 - 160(0.5)$ $PR^2 = 164 - 80 = 84$ $PR = \sqrt{84} \approx 9.165...$
• Answer: 9.17
• Marks: [3] (1 for formula, 1 for substitution, 1 for answer)

8.

• Concept: Interior angles of regular polygons.
• Working: Sum of interior angles $= (n-2) \times 180^\circ$. For hexagon, $n=6$: $(6-2) \times 180 = 720^\circ$. One interior angle $= \frac{720}{6} = 120^\circ$. Alternatively: Exterior angle $= \frac{360}{6} = 60^\circ$. Interior $= 180 - 60 = 120^\circ$.
• Answer: 120
• Marks: [2] (1 for method, 1 for answer)

9.

• Concept: Trigonometric identities / Pythagorean theorem in trig.
• Working: $\sin \theta = \frac{3}{5}$. Imagine a right triangle with opposite 3, hypotenuse 5. Adjacent side $= \sqrt{5^2 - 3^2} = \sqrt{25-9} = \sqrt{16} = 4$. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}$.
• Answer: $\frac{4}{5}$ or 0.8
• Marks: [2] (1 for finding adjacent side, 1 for ratio)

10.

• Concept: Total Surface Area of a Cone.
• Working: TSA $= \pi r^2 + \pi r l$ $r = 3, l = 5$. TSA $= \pi(3^2) + \pi(3)(5) = 9\pi + 15\pi = 24\pi$.
• Answer: $24\pi$
• Marks: [3] (1 for base area, 1 for curved surface area, 1 for total)

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Section B

11. (a)

• Concept: Cosine Rule in $\triangle ABD$.
• Working: $BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(60^\circ)$ $BD^2 = 7^2 + 8^2 - 2(7)(8)(0.5)$ $BD^2 = 49 + 64 - 56 = 57$ $BD = \sqrt{57} \approx 7.55$
• Answer: 7.55 cm
• Marks: [3]

(b)

• Concept: Cosine Rule in $\triangle BCD$.
• Working: Sides are $BC=5, CD=6, BD=\sqrt{57}$. $BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD)$ $57 = 5^2 + 6^2 - 2(5)(6)\cos(\angle BCD)$ $57 = 25 + 36 - 60\cos(\angle BCD)$ $57 = 61 - 60\cos(\angle BCD)$ $60\cos(\angle BCD) = 61 - 57 = 4$ $\cos(\angle BCD) = \frac{4}{60} = \frac{1}{15}$ $\angle BCD = \cos^{-1}\left(\frac{1}{15}\right) \approx 86.2^\circ$
• Answer: 86.2
• Marks: [3]

12. (a)

• Concept: Trigonometry in right-angled $\triangle STB$.
• Working: $\tan(45^\circ) = \frac{h}{TB} \Rightarrow 1 = \frac{h}{TB} \Rightarrow TB = h$.
• Answer: $h$
• Marks: [1]

(b)

• Concept: Trigonometry in right-angled $\triangle STA$.
• Working: $\tan(30^\circ) = \frac{h}{TA} \Rightarrow TA = \frac{h}{\tan(30^\circ)} = h\sqrt{3}$ or $\frac{h}{1/\sqrt{3}}$.
• Answer: $h\sqrt{3}$ or $\frac{h}{\tan 30^\circ}$
• Marks: [1]

(c)

• Concept: Forming and solving equations.
• Working: $TA - TB = AB = 20$. $h\sqrt{3} - h = 20$ $h(\sqrt{3} - 1) = 20$ $h = \frac{20}{\sqrt{3} - 1}$ $h \approx \frac{20}{1.732 - 1} = \frac{20}{0.732} \approx 27.32$
• Answer: 27.3 m
• Marks: [4] (1 for equation, 1 for rearrangement, 1 for calculation, 1 for accuracy)

13. (a)

• Concept: Tangent-Radius theorem.
• Working: Radius is perpendicular to tangent at point of contact.
• Answer: 90
• Marks: [1]

(b)

• Concept: Angles in $\triangle OAP$.
• Working: Sum of angles in $\triangle OAP = 180^\circ$. $\angle AOP = 180 - 90 - 40 = 50^\circ$.
• Answer: 50
• Marks: [2]

(c)

• Concept: Angle at centre vs circumference.
• Working: $\angle AOP$ is the angle at centre subtending arc $AB$? No, subtending arc $AC$? Wait, $PBC$ is a line through centre. So $\angle AOP$ is the angle at centre subtending arc $AB$? No, $A$ and $B$ are on the circle. Angle at centre $\angle AOB$? No, $B$ is on the line $POC$. The angle at the centre subtending arc $AC$ is $\angle AOC$. $\angle AOC = 180 - \angle AOP = 180 - 50 = 130^\circ$. Angle at circumference $\angle ABC$? No, question asks for $\angle ACB$. $\angle ACB$ subtends arc $AB$? No. Let's look at $\triangle AOC$. It is isosceles ($OA=OC$). $\angle AOC = 130^\circ$. $\angle OCA = \angle OAC = (180-130)/2 = 25^\circ$. So $\angle ACB = 25^\circ$. Alternative: Angle at centre $\angle AOB$? No. Angle $\angle AOP = 50^\circ$. This is exterior to $\triangle AOC$? No. $\angle AOB$? $B$ is on the segment $PO$. So $\angle AOB = 50^\circ$? No, $P-B-O-C$. So $\angle AOB$ is not defined as a central angle for arc $AB$ in the standard sense if $B$ is just a point on the secant. Actually, $B$ is on the circumference. So $\angle AOB$ is the angle at centre subtending arc $AB$. $\angle AOB = 180 - 50 = 130$? No. $P, B, O, C$ are collinear. $\angle AOP = 50^\circ$. Since $B$ is between $P$ and $O$, $\angle AOB = 180 - 50$? No, $A, O, B$ form triangle? Angle $\angle AOB$ is supplementary to $\angle AOP$ only if $A, O, P$ is a line? No. $P-B-O-C$ is a line. $\angle AOP = 50^\circ$. Therefore $\angle AOC = 180 - 50 = 130^\circ$. $\angle ABC$ is angle at circumference subtending arc $AC$? No. Question asks for $\angle ACB$. $\angle ACB$ subtends arc $AB$. Angle at centre for arc $AB$ is $\angle AOB$. Since $P-B-O$ is a line, $\angle AOB = 180 - \angle AOP$? No. $\angle AOP$ is the angle between $OA$ and $OP$. $B$ lies on $OP$. So $\angle AOB = \angle AOP = 50^\circ$. Angle at circumference $\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2}(50) = 25^\circ$.
• Answer: 25
• Marks: [2]

14. (a)

• Concept: Volume of composite solid.
• Working: Volume of Cylinder $= \pi r^2 h = \pi(4^2)(10) = 160\pi$. Volume of Hemisphere $= \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(4^3) = \frac{128\pi}{3}$. Remaining Volume $= 160\pi - \frac{128\pi}{3} = \frac{480\pi - 128\pi}{3} = \frac{352\pi}{3}$. $\frac{352\pi}{3} \approx 368.58...$
• Answer: 369
• Marks: [4]

(b)

• Concept: Surface Area of composite solid.
• Working: Curved Surface Area of Cylinder $= 2\pi r h = 2\pi(4)(10) = 80\pi$. Base Area of Cylinder $= \pi r^2 = 16\pi$. Curved Surface Area of Hemisphere $= 2\pi r^2 = 2\pi(16) = 32\pi$. (Note: The circular top of the cylinder is removed, replaced by the hemisphere's curved surface). Total SA $= 80\pi + 16\pi + 32\pi = 128\pi$. $128\pi \approx 402.12...$
• Answer: 402
• Marks: [4]

15. (a)

• Concept: Cosine Rule.
• Working: $BC^2 = 12^2 + 10^2 - 2(12)(10)\cos(40^\circ)$ $BC^2 = 144 + 100 - 240(0.7660...)$ $BC^2 = 244 - 183.85... = 60.14...$ $BC = \sqrt{60.14...} \approx 7.755$
• Answer: 7.76
• Marks: [3]

(b)

• Concept: Area of triangle.
• Working: Area $= \frac{1}{2} ab \sin C = \frac{1}{2}(12)(10)\sin(40^\circ)$. Area $= 60 \sin(40^\circ) \approx 38.567...$
• Answer: 38.6
• Marks: [2]

(c)

• Concept: Median length formula or Cosine Rule on sub-triangles.
• Working: Using Apollonius theorem: $AB^2 + AC^2 = 2(AM^2 + BM^2)$. $BM = \frac{BC}{2} \approx 3.877$. $12^2 + 10^2 = 2(AM^2 + 3.877^2)$. $244 = 2(AM^2 + 15.03)$. $122 = AM^2 + 15.03$. $AM^2 = 106.97$. $AM = \sqrt{106.97} \approx 10.34$.
• Answer: 10.3
• Marks: [3]

16. (a)

• Concept: Volume of prism.
• Working: Area of Trapezium $= \frac{1}{2}(8+14)(6) = \frac{1}{2}(22)(6) = 66 \text{ cm}^2$. Volume $= \text{Area} \times \text{Length} = 66 \times 20 = 1320$.
• Answer: 1320
• Marks: [3]

(b)

• Concept: Surface Area of prism.
• Working: Two trapezium faces: $2 \times 66 = 132$. Rectangular faces: Bottom: $14 \times 20 = 280$. Top: $8 \times 20 = 160$. Vertical side: $6 \times 20 = 120$. Slanted side: Need length. Horizontal projection of slant $= \frac{14-8}{2} = 3$ (assuming isosceles trapezium as implied by "non-parallel sides are equal"). Slant length $= \sqrt{6^2 + 3^2} = \sqrt{36+9} = \sqrt{45} \approx 6.708$. Area of slanted face $= \sqrt{45} \times 20 \approx 134.16$. Wait, the question says "non-parallel sides... are equal". So there are two slanted sides? No, "perpendicular height... is 6". Usually implies one side is perpendicular if not specified isosceles? "The non-parallel sides of the trapezium are equal in length." -> Isosceles Trapezium. So there are TWO slanted sides in the cross-section? No, a trapezium has 4 sides. 2 parallel, 2 non-parallel. If it is isosceles, both non-parallel sides are slanted. So the prism has: 2 Trapezium ends. 4 Rectangular lateral faces:
  1. Base $14 \times 20$.
  2. Top $8 \times 20$.
  3. Slant 1 $\times 20$.
  4. Slant 2 $\times 20$. Slant length calculation: Drop perpendiculars from top vertices to base. Base segments: $(14-8)/2 = 3$ on each side. Slant $= \sqrt{6^2 + 3^2} = \sqrt{45}$. Lateral Area $= 20(14 + 8 + \sqrt{45} + \sqrt{45}) = 20(22 + 2\sqrt{45})$. $2\sqrt{45} = 2(6.708) = 13.416$. Lateral Area $= 20(35.416) = 708.32$. Total SA $= 132 + 708.32 = 840.32$.
• Answer: 840
• Marks: [5]

17. (a)

• Concept: Tangent ratio.
• Working: In $\triangle ADE$, $\tan(30^\circ) = \frac{DE}{AD} = \frac{3}{AD}$. $AD = \frac{3}{\tan(30^\circ)} = \frac{3}{1/\sqrt{3}} = 3\sqrt{3} \approx 5.196$.
• Answer: 5.20
• Marks: [2]

(b)

• Concept: Trigonometry in $\triangle BCE$.
• Working: $BC = AD = 3\sqrt{3}$. $EC = 5$. $\triangle BCE$ is right-angled at $C$. $\tan(\angle EBC) = \frac{EC}{BC} = \frac{5}{3\sqrt{3}}$. $\angle EBC = \tan^{-1}\left(\frac{5}{3\sqrt{3}}

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Answer Key and Marking Scheme

TuitionGoWhere Practice Paper - Elementary Mathematics O-Level Topic: Geometry & Trigonometry (Version 3)

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Section A

1.

• Concept: Pythagoras' Theorem.
• Working: $AC^2 = AB^2 + BC^2$ $AC^2 = 12^2 + 5^2 = 144 + 25 = 169$ $AC = \sqrt{169} = 13$
• Answer: 13 cm
• Marks: [2] (1 for substitution, 1 for answer)

2.

• Concept: Inverse trigonometric ratios.
• Working: $x = \tan^{-1}(0.5)$ $x \approx 26.565...$
• Answer: 26.6
• Marks: [2] (1 for correct inverse operation, 1 for correct rounding)

3.

• Concept: Angle at centre is twice angle at circumference.
• Working: The angle at the centre $\angle AOC = 130^\circ$. The angle at the circumference $\angle ABC$ subtends the same arc $AC$. $\angle ABC = \frac{1}{2} \times \angle AOC$ $\angle ABC = \frac{1}{2} \times 130^\circ = 65^\circ$
• Answer: 65
• Marks: [2] (1 for stating relationship, 1 for answer)

4.

• Concept: Cosine ratio in right-angled triangle.
• Working: Let $\theta$ be the angle with the ground. Adjacent side = 2.5 m, Hypotenuse = 6 m. $\cos \theta = \frac{2.5}{6}$ $\theta = \cos^{-1}\left(\frac{2.5}{6}\right)$ $\theta \approx 65.375...^\circ$
• Answer: 65.4
• Marks: [2] (1 for correct ratio, 1 for answer)

5.

• Concept: Parallel lines and angles (Zig-zag theorem).
• Working: Draw a line through $E$ parallel to $AB$ and $CD$. The angle at $E$ is split into two parts: alternate interior to $\angle BAE$ and alternate interior to $\angle CDE$. $\angle AED = \angle BAE + \angle CDE$ $\angle AED = 40^\circ + 35^\circ = 75^\circ$
• Answer: 75
• Marks: [2] (1 for method/reasoning, 1 for answer)

6.

• Concept: Volume of a cylinder.
• Working: $V = \pi r^2 h$ $500 = \pi r^2 (10)$ $r^2 = \frac{500}{10\pi} = \frac{50}{\pi}$ $r = \sqrt{\frac{50}{\pi}} \approx 3.989...$
• Answer: 3.99
• Marks: [3] (1 for formula, 1 for substitution/rearrangement, 1 for answer)

7.

• Concept: Cosine Rule.
• Working: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)$ $PR^2 = 8^2 + 10^2 - 2(8)(10)\cos(60^\circ)$ $PR^2 = 64 + 100 - 160(0.5)$ $PR^2 = 164 - 80 = 84$ $PR = \sqrt{84} \approx 9.165...$
• Answer: 9.17
• Marks: [3] (1 for formula, 1 for substitution, 1 for answer)

8.

• Concept: Interior angles of regular polygons.
• Working: Sum of interior angles $= (n-2) \times 180^\circ$. For hexagon, $n=6$: $(6-2) \times 180 = 720^\circ$. One interior angle $= \frac{720}{6} = 120^\circ$. Alternatively: Exterior angle $= \frac{360}{6} = 60^\circ$. Interior $= 180 - 60 = 120^\circ$.
• Answer: 120
• Marks: [2] (1 for method, 1 for answer)

9.

• Concept: Trigonometric identities / Pythagorean theorem in triangles.
• Working: Consider a right-angled triangle with opposite side 3 and hypotenuse 5. Adjacent side $= \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}$
• Answer: $\frac{4}{5}$ (or 0.8)
• Marks: [2] (1 for finding adjacent side, 1 for ratio)

10.

• Concept: Total Surface Area of a Cone.
• Working: $TSA = \pi r^2 + \pi r l$ $TSA = \pi (3)^2 + \pi (3)(5)$ $TSA = 9\pi + 15\pi = 24\pi$
• Answer: $24\pi$
• Marks: [3] (1 for curved surface area, 1 for base area, 1 for total)

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Section B

11. (a)

• Concept: Cosine Rule in $\triangle ABD$.
• Working: $BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(60^\circ)$ $BD^2 = 7^2 + 8^2 - 2(7)(8)(0.5)$ $BD^2 = 49 + 64 - 56 = 57$ $BD = \sqrt{57} \approx 7.55$
• Answer: 7.55 cm
• Marks: [3]

(b)

• Concept: Cosine Rule in $\triangle BCD$.
• Working: $BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD)$ $57 = 5^2 + 6^2 - 2(5)(6)\cos(\angle BCD)$ $57 = 25 + 36 - 60\cos(\angle BCD)$ $57 = 61 - 60\cos(\angle BCD)$ $60\cos(\angle BCD) = 61 - 57 = 4$ $\cos(\angle BCD) = \frac{4}{60} = \frac{1}{15}$ $\angle BCD = \cos^{-1}\left(\frac{1}{15}\right) \approx 86.18^\circ$
• Answer: 86.2$^\circ$
• Marks: [3]

12. (a)

• Concept: Trigonometry in right-angled $\triangle STB$.
• Working: $\tan(45^\circ) = \frac{h}{TB} \implies 1 = \frac{h}{TB} \implies TB = h$
• Answer: $h$
• Marks: [1]

(b)

• Concept: Trigonometry in right-angled $\triangle STA$.
• Working: $\tan(30^\circ) = \frac{h}{TA} \implies TA = \frac{h}{\tan(30^\circ)} = h\sqrt{3}$
• Answer: $h\sqrt{3}$ (or $\frac{h}{\tan 30^\circ}$)
• Marks: [1]

(c)

• Concept: Forming and solving equations.
• Working: $TA - TB = AB$ $h\sqrt{3} - h = 20$ $h(\sqrt{3} - 1) = 20$ $h = \frac{20}{\sqrt{3} - 1}$ $h \approx \frac{20}{1.732 - 1} = \frac{20}{0.732} \approx 27.32$
• Answer: 27.3 m
• Marks: [4]

13. (a)

• Concept: Radius is perpendicular to tangent.
• Answer: 90
• Marks: [1]

(b)

• Concept: Angles in $\triangle OAP$.
• Working: In $\triangle OAP$, sum of angles is $180^\circ$. $\angle AOP = 180^\circ - \angle OAP - \angle APO$ Note: $\angle APO$ is the same as $\angle APT = 40^\circ$. $\angle AOP = 180^\circ - 90^\circ - 40^\circ = 50^\circ$
• Answer: 50
• Marks: [2]

(c)

• Concept: Angle at centre vs circumference / Isosceles triangle.
• Working: $\triangle OAC$ is isosceles ($OA=OC=radius$). Angle at centre $\angle AOC = 180^\circ - \angle AOP = 180^\circ - 50^\circ = 130^\circ$ (Angles on a straight line P-O-C). Wait, P-B-O-C is a line. So $\angle AOC$ and $\angle AOP$ are supplementary. $\angle AOC = 180 - 50 = 130^\circ$. In $\triangle OAC$, base angles are equal: $\angle OAC = \angle OCA$. $2(\angle OCA) + 130^\circ = 180^\circ \implies 2(\angle OCA) = 50^\circ \implies \angle OCA = 25^\circ$ $\angle ACB$ is the same as $\angle OCA$ (since O, B, C are collinear? No, B is on the segment OC? No, P-B-O-C. So C is on the circle. B is between P and O? "PBC is a secant line passing through the centre O". Usually implies order P-B-O-C or P-O-B-C. Given $\angle APT=40$, A is "above". Let's assume standard diagram where B is the near intersection and C is the far intersection. So P-B-O-C. Then $\angle ACB$ subtends arc AB? No. Let's use the property: Angle between tangent and chord equals angle in alternate segment. Chord is AB? No, chord is AC? Let's stick to centre angle. $\angle AOP = 50^\circ$. This is the exterior angle to $\triangle AOC$? No. $\angle AOC = 180 - 50 = 130^\circ$. $\triangle AOC$ is isosceles. $\angle OCA = (180-130)/2 = 25^\circ$. Since B lies on the line segment OC (or extension), $\angle ACB$ is the same angle as $\angle ACO$.
• Answer: 25
• Marks: [2]

14. (a)

• Concept: Volume subtraction.
• Working: Volume of Cylinder $V_{cyl} = \pi r^2 h = \pi (4^2)(10) = 160\pi$. Volume of Hemisphere $V_{hem} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (4^3) = \frac{128\pi}{3}$. Remaining Volume $= 160\pi - \frac{128\pi}{3} = \frac{480\pi - 128\pi}{3} = \frac{352\pi}{3}$. $\frac{352\pi}{3} \approx 368.61...$
• Answer: 369 $\text{cm}^3$
• Marks: [4]

(b)

• Concept: Surface Area.
• Working: Curved Surface Area of Cylinder $= 2\pi r h = 2\pi(4)(10) = 80\pi$. Base Area of Cylinder (bottom only) $= \pi r^2 = 16\pi$. Curved Surface Area of Hemisphere (inner) $= 2\pi r^2 = 2\pi(16) = 32\pi$. Top rim area is removed (it's a hole). Total SA $= 80\pi + 16\pi + 32\pi = 128\pi$. $128\pi \approx 402.12...$
• Answer: 402 $\text{cm}^2$
• Marks: [4]

15. (a)

• Concept: Cosine Rule.
• Working: $BC^2 = 12^2 + 10^2 - 2(12)(10)\cos(40^\circ)$ $BC^2 = 144 + 100 - 240(0.7660...)$ $BC^2 = 244 - 183.85... = 60.14...$ $BC = \sqrt{60.14...} \approx 7.755...$
• Answer: 7.76 cm
• Marks: [3]

(b)

• Concept: Area of triangle.
• Working: $Area = \frac{1}{2} ab \sin C = \frac{1}{2}(12)(10)\sin(40^\circ)$ $Area = 60 \sin(40^\circ) \approx 38.567...$
• Answer: 38.6 $\text{cm}^2$
• Marks: [2]

(c)

• Concept: Median length formula or Cosine Rule on sub-triangles.
• Working: Using Apollonius Theorem: $AB^2 + AC^2 = 2(AM^2 + BM^2)$. $BM = \frac{BC}{2} \approx 3.877$. $12^2 + 10^2 = 2(AM^2 + 3.877^2)$. $244 = 2(AM^2 + 15.03)$. $122 = AM^2 + 15.03$. $AM^2 = 106.97$. $AM = \sqrt{106.97} \approx 10.34$.
• Answer: 10.3 cm
• Marks: [3]

16. (a)

• Concept: Volume of prism.
• Working: Area of Trapezium $= \frac{1}{2}(8 + 14) \times 6 = \frac{1}{2}(22)(6) = 66 \text{ cm}^2$. Volume $= \text{Area} \times \text{Length} = 66 \times 20 = 1320$.
• Answer: 1320 $\text{cm}^3$
• Marks: [3]

(b)

• Concept: Surface Area.
• Working: Two trapezium faces: $2 \times 66 = 132$. Rectangular faces: Bottom: $14 \times 20 = 280$. Top: $8 \times 20 = 160$. Slanted sides: Need slant height. Horizontal projection of slant $= \frac{14-8}{2} = 3$. Height $= 6$. Slant length $= \sqrt{3^2 + 6^2} = \sqrt{9+36} = \sqrt{45} \approx 6.708$. Two slanted faces: $2 \times (6.708 \times 20) = 2 \times 134.16 = 268.32$. Total SA $= 132 + 280 + 160 + 268.32 = 840.32$.
• Answer: 840 $\text{cm}^2$
• Marks: [5]

17. (a)

• Concept: Trigonometry in $\triangle ADE$.
• Working: $\tan(30^\circ) = \frac{DE}{AD} = \frac{3}{AD}$. $AD = \frac{3}{\tan(30^\circ)} = 3\sqrt{3} \approx 5.196$.
• Answer: 5.20 cm
• Marks: [2]

(b)

• Concept: Trigonometry in $\triangle BCE$.
• Working: $BC = AD = 3\sqrt{3}$. $EC = 5$. $\tan(\angle EBC) = \frac{EC}{BC} = \frac{5}{3\sqrt{3}}$. $\angle EBC = \tan^{-1}\left(\frac{5}{3\sqrt{3}}\right) \approx \tan^{-1}(0.962) \approx 43.9^\circ$.
• Answer: 43.9$^\circ$
• Marks: [3]

18. (a)

• Concept: Bearings and geometry.
• Working: Bearing of B from A is $050^\circ$. Bearing of A from B is $050 + 180 = 230^\circ$. Bearing of C from B is $140^\circ$. $\angle ABC = 230^\circ - 140^\circ = 90^\circ$.
• Answer: 90
• Marks: [2]

(b)

• Concept: Pythagoras' Theorem (since $\angle B = 90^\circ$).
• Working: $AC^2 = AB^2 + BC^2 = 100^2 + 80^2 = 10000 + 6400 = 16400$. $AC = \sqrt{16400} \approx 128.06$.
• Answer: 128 km
• Marks: [3]

(c)

• Concept: Bearings.
• Working: In right $\triangle ABC$, $\tan(\angle BAC) = \frac{80}{100} = 0.8$. $\angle BAC = \tan^{-1}(0.8) \approx 38.66^\circ$. Bearing of B from A is $050^\circ$. Bearing of C from A is $050 + 38.66 = 088.66^\circ$. We need bearing of A from C. Bearing of A from C = Bearing of C from A + $180^\circ$ (if $<180$) or $-180^\circ$. $88.66 + 180 = 268.66^\circ$.
• Answer: 269$^\circ$
• Marks: [3]

19. (a)

• Concept: Angle in a semicircle.
• Answer: 90
• Marks: [1]

(b)

• Concept: Angle in a semicircle.
• Answer: 90
• Marks: [1]

(c)

• Concept: Angles in same segment / Cyclic quad.
• Working: In $\triangle ABD$, $\angle ADB = 90^\circ$, $\angle ABD = 35^\circ$. $\angle BAD = 180 - 90 - 35 = 55^\circ$. $\angle BAC = 25^\circ$. $\angle CAD = \angle BAD - \angle BAC = 55 - 25 = 30^\circ$. $\angle BDC$ and $\angle BAC$ subtend the same arc BC? No. $\angle BDC$ subtends arc BC. $\angle BAC$ subtends arc BC. Therefore $\angle BDC = \angle BAC = 25^\circ$.
• Answer: 25
• Marks: [3]

20. (a)

• Answer: $\frac{1}{3}\pi r^2 h$
• Marks: [1]

(b)

• Concept: Scaling.
• Working: $V_2 = \frac{1}{3}\pi (2r)^2 (3h) = \frac{1}{3}\pi (4r^2)(3h) = 12 (\frac{1}{3}\pi r^2 h) = 12V$.
• Answer: $12V$
• Marks: [2]

(c)

• Concept: Area scaling.
• Working: Linear scale factor $k=2$. Area scale factor $k^2 = 2^2 = 4$.
• Answer: $4A$
• Marks: [2]