Secondary 4 Elementary Mathematics Preliminary Examination Paper 5

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 4

Answer Key & Marking Scheme
Version 5 of 5

Section A: Short Answer Questions

1.
Using Cosine Rule: $AC^2 = 12^2 + 9^2 - 2(12)(9)\cos(110^\circ)$
$AC^2 = 144 + 81 - 216(-0.3420)$
$AC^2 = 225 + 73.87 = 298.87$
$AC = \sqrt{298.87} \approx 17.3$ cm
Answer: 17.3 cm [2]
(1 mark for correct substitution, 1 mark for answer)

2.
Reflex $\angle AOC = 360^\circ - 130^\circ = 230^\circ$.
Angle at circumference is half angle at centre.
$\angle ABC = \frac{1}{2} \times 230^\circ = 115^\circ$.
Answer: 115$^\circ$ [2]

3.
Reference angle: $\sin^{-1}(0.6) \approx 36.9^\circ$.
Sine is negative in 3rd and 4th quadrants.
$x = 180^\circ + 36.9^\circ = 216.9^\circ$.
$x = 360^\circ - 36.9^\circ = 323.1^\circ$.
Answer: $216.9^\circ, 323.1^\circ$ [2]

4.
Area $= \frac{1}{2}r^2\theta = \frac{1}{2}(8^2)(1.5) = \frac{1}{2}(64)(1.5) = 32(1.5) = 48$.
Answer: 48 cm$^2$ [2]

5.
Area $= ab \sin \theta = 10 \times 6 \times \sin(60^\circ) = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \approx 52.0$.
Answer: 52.0 cm$^2$ [2]

6.
Distance $= \sqrt{(8-2)^2 + (1-5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52} \approx 7.21$.
Answer: 7.21 units [2]

7.
If $\tan \theta = \frac{3}{4}$, opposite=3, adjacent=4. Hypotenuse $= \sqrt{3^2+4^2}=5$.
$\cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{4}{5}$.
Answer: $\frac{4}{5}$ or 0.8 [2]

8.
Bearings: North at A, North at B, North at C.
Angle $N_B BA = 180^\circ - 55^\circ = 125^\circ$ (co-interior? No, alternate interior with North lines).
Actually, simpler: Angle inside triangle at B.
Bearing $A \to B = 055^\circ$. Back bearing $B \to A = 235^\circ$.
Bearing $B \to C = 140^\circ$.
$\angle ABC = 235^\circ - 140^\circ = 95^\circ$.
Triangle $ABC$ is isosceles ($AB=BC$).
$\angle BCA = \angle BAC = (180^\circ - 95^\circ)/2 = 42.5^\circ$.
Bearing $C \to B = 140^\circ + 180^\circ = 320^\circ$.
Bearing $C \to A = 320^\circ - 42.5^\circ = 277.5^\circ$.
Answer: 277.5$^\circ$ [3]

9.
Sine Rule: $\frac{9}{\sin 45^\circ} = \frac{7}{\sin R}$.
$\sin R = \frac{7 \sin 45^\circ}{9} \approx 0.550$.
$R_1 = \sin^{-1}(0.550) \approx 33.4^\circ$.
$R_2 = 180^\circ - 33.4^\circ = 146.6^\circ$.
Check validity: $45^\circ + 146.6^\circ < 180^\circ$ (Valid).
Answer: $33.4^\circ$ and $146.6^\circ$ [3]

10.
Perpendicular bisects chord. Half-chord = 5 cm. Radius = 13 cm.
$d^2 + 5^2 = 13^2 \Rightarrow d^2 = 169 - 25 = 144$.
$d = 12$ cm.
Answer: 12 cm [2]

Section B: Structured Questions

11.
(a) Area $= \frac{1}{2}(15)(12)\sin(75^\circ) = 90 \sin(75^\circ) \approx 86.9$ cm$^2$. [2]
(b) $BC^2 = 15^2 + 12^2 - 2(15)(12)\cos(75^\circ) = 225 + 144 - 360(0.2588) = 369 - 93.17 = 275.83$.
$BC = \sqrt{275.83} \approx 16.6$ cm. [3]
(c) Sine Rule: $\frac{16.6}{\sin 75^\circ} = \frac{15}{\sin C}$.
$\sin C = \frac{15 \sin 75^\circ}{16.6} \approx 0.869$.
$C = \sin^{-1}(0.869) \approx 60.4^\circ$. [2]

12.
(a) $AC = \sqrt{10^2 + 8^2} = \sqrt{164} \approx 12.8$ cm. [2]
(b) $AM = \frac{1}{2} AC = \frac{\sqrt{164}}{2} \approx 6.40$ cm.
$\tan(\angle VAM) = \frac{VM}{AM} = \frac{12}{6.40}$.
$\angle VAM = \tan^{-1}(1.875) \approx 61.9^\circ$. [3]
(c) Slant height of face $VAB$: Midpoint of $AB$ is $M_{AB}$. $M_{AB}M = 4$ cm (half BC).
$VM_{AB} = \sqrt{12^2 + 4^2} = \sqrt{160} \approx 12.65$ cm.
Area $VAB = \frac{1}{2}(10)(12.65) = 63.25$ cm$^2$.
Slant height of face $VBC$: Midpoint of $BC$ is $M_{BC}$. $M_{BC}M = 5$ cm.
$VM_{BC} = \sqrt{12^2 + 5^2} = 13$ cm.
Area $VBC = \frac{1}{2}(8)(13) = 52$ cm$^2$.
Base Area $= 80$ cm$^2$.
Total Surface Area $= 80 + 2(63.25) + 2(52) = 80 + 126.5 + 104 = 310.5$ cm$^2$. [4]

13.
(a) Angles in same segment: $\angle ACD = \angle ABD = 38^\circ$. [1]
(b) In $\triangle ABX$: $\angle AXB = 180^\circ - (24^\circ + 38^\circ) = 118^\circ$.
$\angle AXD = 180^\circ - 118^\circ = 62^\circ$ (angles on straight line).
Alternatively, exterior angle of $\triangle ABX$: $\angle AXD = \angle BAX + \angle ABX$? No.
In $\triangle ADX$: Need $\angle DAC$. $\angle DAC = \angle DBC$ (same segment).
Let's use $\triangle ABX$: $\angle AXD$ is exterior to $\triangle ABX$? No, $\angle AXD$ and $\angle AXB$ are supplementary.
Wait, $\angle AXD$ is angle at intersection.
$\angle ABD = 38^\circ$, $\angle BAC = 24^\circ$.
In $\triangle ABX$, $\angle AXB = 180 - 38 - 24 = 118^\circ$.
$\angle AXD = 180 - 118 = 62^\circ$. [2]
(c) $AD=DC \Rightarrow$ chords equal $\Rightarrow$ arcs equal $\Rightarrow$ angles at circumference equal.
$\angle DAC = \angle DCA$.
In $\triangle ADC$: $\angle ADC = \angle ABC$? No.
$\angle ACD = 38^\circ$ (from a). Since $AD=DC$, $\triangle ADC$ is isosceles with base $AC$? No, $AD=DC$ means vertex is D.
So $\angle DAC = \angle DCA$.
We found $\angle ACD = 38^\circ$. So $\angle DAC = 38^\circ$.
Check: $\angle ADC = 180 - 38 - 38 = 104^\circ$.
Answer: $\angle DAC = 38^\circ$. [3]

14.
(a) Angle $PQR$: Bearing $Q \to P = 250^\circ$. Bearing $Q \to R = 160^\circ$.
$\angle PQR = 250^\circ - 160^\circ = 90^\circ$.
Right-angled triangle. $PR = \sqrt{40^2 + 30^2} = 50$ km. [3]
(b) $\tan(\angle QPR) = \frac{30}{40} = 0.75 \Rightarrow \angle QPR = 36.9^\circ$.
Bearing $P \to Q = 070^\circ$.
Bearing $P \to R = 070^\circ + 36.9^\circ = 106.9^\circ$.
Bearing $R \to P = 106.9^\circ + 180^\circ = 286.9^\circ$. [3]

15.
(a) Arc length $s = r\theta = 10(1.2) = 12$ cm. [2]
(b) Area Sector $= \frac{1}{2}(10^2)(1.2) = 60$ cm$^2$.
Area Triangle $= \frac{1}{2}(10)(10)\sin(1.2 \text{ rad}) = 50 \sin(1.2) \approx 50(0.932) = 46.6$ cm$^2$.
Area Segment $= 60 - 46.6 = 13.4$ cm$^2$. [3]

16.
(a) $XZ^2 = 14^2 + 10^2 - 2(14)(10)\cos(120^\circ) = 196 + 100 - 280(-0.5) = 296 + 140 = 436$.
$XZ = \sqrt{436} \approx 20.9$ cm. [3]
(b) Area $\triangle XYZ = \frac{1}{2}(14)(10)\sin(120^\circ) = 70(0.866) = 60.62$ cm$^2$.
Also Area $= \frac{1}{2}(XZ)(YW) = \frac{1}{2}(20.88)(YW)$.
$60.62 = 10.44 YW \Rightarrow YW = 5.81$ cm. [3]

17.
(a) Complete square: $(x-3)^2 - 9 + (y+4)^2 - 16 - 11 = 0$.
$(x-3)^2 + (y+4)^2 = 36$.
Centre $(3, -4)$. [2]
(b) $r^2 = 36 \Rightarrow r = 6$. [2]

18.
(a) $\angle AOT = \frac{1}{2} \angle AOB = 50^\circ$.
$\tan(50^\circ) = \frac{TA}{5} \Rightarrow TA = 5 \tan(50^\circ) \approx 5.96$ cm. [3]
(b) Area $OATB = 2 \times$ Area $\triangle OAT = 2 \times (\frac{1}{2} \times 5 \times 5.96) = 29.8$ cm$^2$. [2]

19.
(a) $\cos \theta = \frac{1.5}{5} = 0.3 \Rightarrow \theta = \cos^{-1}(0.3) \approx 72.5^\circ$. [2]
(b) New base $= 1.5 + 0.5 = 2.0$ m.
New height $h = \sqrt{5^2 - 2^2} = \sqrt{21} \approx 4.58$ m.
Old height $h_0 = \sqrt{5^2 - 1.5^2} = \sqrt{22.75} \approx 4.77$ m.
Slide $= 4.77 - 4.58 = 0.19$ m. [3]

20.
(a) $BD^2 = 13^2 - x^2 = 169 - x^2$. [1]
(b) $CD = 14 - x$. $BD^2 = 15^2 - (14-x)^2 = 225 - (14-x)^2$. [1]
(c) $169 - x^2 = 225 - (196 - 28x + x^2)$.
$169 - x^2 = 225 - 196 + 28x - x^2$.
$169 = 29 + 28x$.
$140 = 28x \Rightarrow x = 5$.
$BD = \sqrt{169 - 5^2} = \sqrt{144} = 12$ cm. [3]