Secondary 4 Elementary Mathematics Practice Paper 5

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 4

Answer Key and Marking Scheme (Version 5)

Total Marks: 90

Section A: Short Answer Questions (45 marks)

1. (a) $\angle ACB = 65^\circ$ [1] (Angle at centre = 2 × angle at circumference: $\angle ACB = \frac{1}{2} \times 130^\circ = 65^\circ$)

(b) $\angle ABD = 40^\circ$ [2]

• $\angle ADB = \angle ACB = 65^\circ$ (angles in same segment) [1]
• In $\triangle ABD$, $\angle ABD = 180^\circ - 65^\circ - 25^\circ - 65^\circ = 25^\circ$? Wait, recalculate.
• $\angle ADB = 65^\circ$ (angles in same segment as $\angle ACB$)
• $\angle BAD = \frac{1}{2} \times (360^\circ - 130^\circ) = 115^\circ$? No.
• Alternative: $\angle ACD = \frac{1}{2} \times 130^\circ = 65^\circ$. $\angle ABD = \angle ACD = 65^\circ$ (angles in same segment). Wait, check.
• $\angle ABD$ subtends arc $AD$. $\angle ACD$ also subtends arc $AD$. $\angle ACD = \angle ACB + \angle BCD$? Not given.
• Better: $\angle AOB = 130^\circ$, so reflex $\angle AOB = 230^\circ$. $\angle ADB = \frac{1}{2} \times 230^\circ = 115^\circ$ (angle at centre).
• In $\triangle ABD$, $\angle ABD = 180^\circ - 115^\circ - 25^\circ = 40^\circ$. [1] for method, [1] for answer.

2. (a) $PR^2 = 8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos 72^\circ$ [1] $PR^2 = 70.56 + 125.44 - 188.16(0.3090) = 196 - 58.14 = 137.86$ $PR = \sqrt{137.86} = 11.7$ cm (3 s.f.) [1]

(b) Area $= \frac{1}{2} \times 8.4 \times 11.2 \times \sin 72^\circ$ [1] $= 47.04 \times 0.9511 = 44.7$ cm$^2$ (3 s.f.) [1]

3. (a) Arc length $= r\theta = 15 \times \frac{5\pi}{6} = \frac{75\pi}{6} = \frac{25\pi}{2} = 39.3$ cm (3 s.f.) [1]

(b) Sector area $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 15^2 \times \frac{5\pi}{6}$ [1] $= \frac{1}{2} \times 225 \times \frac{5\pi}{6} = \frac{1125\pi}{12} = \frac{375\pi}{4} = 295$ cm$^2$ (3 s.f.) [1]

4. Using cosine rule: $\cos B = \frac{7^2 + 9^2 - 12^2}{2 \times 7 \times 9}$ [1] $= \frac{49 + 81 - 144}{126} = \frac{-14}{126} = -0.1111$ [1] $\angle ABC = \cos^{-1}(-0.1111) = 96.4^\circ$ (1 d.p.) [1]

5. Let $d$ be the horizontal distance. $\tan 28^\circ = \frac{85}{d}$ [1] $d = \frac{85}{\tan 28^\circ}$ [1] $d = \frac{85}{0.5317} = 160$ m (3 s.f.) [1]

6. (a) $\angle ADC = 180^\circ - 55^\circ = 125^\circ$ (opposite angles of cyclic quadrilateral sum to $180^\circ$) [1]

(b) In $\triangle ABX$: $\angle AXB = 180^\circ - 55^\circ - 70^\circ = 55^\circ$ [1] $\angle AXC = 180^\circ - 55^\circ = 125^\circ$ (angles on a straight line) [1]

7. (a) Let height be $h$. $h^2 + 2.5^2 = 6.5^2$ [1] $h^2 = 42.25 - 6.25 = 36$ $h = 6$ m [1]

(b) Let $\theta$ be the angle. $\cos \theta = \frac{2.5}{6.5}$ or $\sin \theta = \frac{6}{6.5}$ or $\tan \theta = \frac{6}{2.5}$ [1] $\theta = \cos^{-1}(\frac{2.5}{6.5}) = 67.4^\circ$ (1 d.p.) [1]

8. Area $= \frac{1}{2} \times 10 \times 14 \times \sin 35^\circ$ [1] $= 70 \times 0.5736 = 40.2$ cm$^2$ (3 s.f.) [1]

9. (a) $\angle PQR = 145^\circ - 55^\circ = 90^\circ$ (or using bearings: $180^\circ - 55^\circ - (180^\circ - 145^\circ) = 90^\circ$) [1] $PR^2 = 12^2 + 9^2 = 144 + 81 = 225$ [1] $PR = 15$ km [1]

(b) $\tan(\angle QPR) = \frac{9}{12} = 0.75$ [1] $\angle QPR = 36.87^\circ$ [1] Bearing of $R$ from $P = 055^\circ + 36.87^\circ = 092^\circ$ (nearest degree) [1]

10. (a) $\angle OTP = 90^\circ$ (tangent perpendicular to radius) [1]

(b) In $\triangle OPT$: $\angle POT = 180^\circ - 90^\circ - 34^\circ = 56^\circ$ [1]

(c) $\angle PQT = \frac{1}{2} \times \angle POT = \frac{1}{2} \times 56^\circ = 28^\circ$ (angle at centre = 2 × angle at circumference) [1] Or: $\angle PQT = \angle OPT = 34^\circ$? No, alternate segment theorem: $\angle PQT = \angle OPT$? Check: tangent $PT$, chord $QT$. $\angle PQT$ is angle in alternate segment to $\angle OPT$? Actually $\angle PQT = \angle PTQ$? Let's use angle at centre. $\angle PQT = 28^\circ$. [1] for correct theorem, [1] for answer.

11. Largest angle is opposite longest side (14 cm). $\cos \theta = \frac{8^2 + 11^2 - 14^2}{2 \times 8 \times 11}$ [1] $= \frac{64 + 121 - 196}{176} = \frac{-11}{176} = -0.0625$ [1] $\theta = \cos^{-1}(-0.0625) = 93.6^\circ$ (1 d.p.) [1]

12. $BC^2 = 6^2 + 8^2 - 2(6)(8)\cos 120^\circ$ [1] $= 36 + 64 - 96(-0.5) = 100 + 48 = 148$ [1] $BC = \sqrt{148} = 12.2$ cm (3 s.f.) [1]

Section B: Structured Questions (45 marks)

13. (a) $\angle ABC = 90^\circ$ (angle in a semicircle) [1]

(b) In $\triangle ABC$: $\angle BCA = 180^\circ - 90^\circ - 28^\circ = 62^\circ$ [1] for method, [1] for answer.

(c) $\angle BDC = \angle BAC = 28^\circ$ (angles in same segment) [1] for theorem, [1] for answer.

(d) $\angle BOC = 2 \times \angle BAC = 2 \times 28^\circ = 56^\circ$ (angle at centre = 2 × angle at circumference) [1] for theorem, [1] for answer.

14. (a) $\tan 31^\circ = \frac{12}{BC}$ [1] $BC = \frac{12}{\tan 31^\circ} = \frac{12}{0.6009} = 20.0$ m (3 s.f.) [1]

(b) $\tan 19^\circ = \frac{12}{BD}$ [1] $BD = \frac{12}{\tan 19^\circ} = \frac{12}{0.3443} = 34.9$ m (3 s.f.) [1]

(c) $CD = BD - BC = 34.9 - 20.0 = 14.9$ m [1] for method, [1] for answer.

15. (a) $\cos(\angle PQR) = \frac{15^2 + 18^2 - 12^2}{2 \times 15 \times 18}$ [1] $= \frac{225 + 324 - 144}{540} = \frac{405}{540} = 0.75$ [1] $\angle PQR = \cos^{-1}(0.75) = 41.4^\circ$ (1 d.p.) [1]

(b) Area $= \frac{1}{2} \times 15 \times 18 \times \sin 41.4^\circ$ [1] $= 135 \times 0.6613 = 89.3$ cm$^2$ (3 s.f.) [1] for substitution, [1] for answer.

(c) Shortest distance $h$ from $P$ to $QR$: Area $= \frac{1}{2} \times QR \times h$ [1] $89.3 = \frac{1}{2} \times 18 \times h$ $h = \frac{89.3}{9} = 9.92$ cm (3 s.f.) [1]

16. (a) Arc length $= r\theta = 10 \times 1.2 = 12.0$ cm (3 s.f.) [1]

(b) Sector area $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 10^2 \times 1.2$ [1] $= 50 \times 1.2 = 60.0$ cm$^2$ [1]

(c) Triangle area $= \frac{1}{2}r^2\sin\theta = \frac{1}{2} \times 10^2 \times \sin 1.2$ [1] $= 50 \times 0.9320 = 46.6$ cm$^2$ [1] Segment area $= 60.0 - 46.6 = 13.4$ cm$^2$ (3 s.f.) [1]

17. (a) After 2 hours: $OX = 30$ km, $OY = 40$ km. [1] Angle between paths: Bearing $120^\circ$ and $210^\circ$. Difference $= 210^\circ - 120^\circ = 90^\circ$. [1] $XY^2 = 30^2 + 40^2 = 900 + 1600 = 2500$ [1] $XY = 50$ km [1]

(b) $\tan(\angle OXY) = \frac{40}{30} = \frac{4}{3}$ [1] $\angle OXY = 53.13^\circ$ [1] Bearing of $Y$ from $X$: $X$ is on bearing $120^\circ$ from $O$. The line $XY$ makes angle $53.13^\circ$ with $OX$. Bearing $= 120^\circ + 180^\circ + 53.13^\circ = 353^\circ$ (nearest degree). Or: $120^\circ + 180^\circ - 53.13^\circ = 247^\circ$? Need careful vector approach. Vector $\vec{OX} = (30\cos 120^\circ, 30\sin 120^\circ) = (-15, 25.98)$ Vector $\vec{OY} = (40\cos 210^\circ, 40\sin 210^\circ) = (-34.64, -20)$ $\vec{XY} = \vec{OY} - \vec{OX} = (-19.64, -45.98)$ Bearing $= 180^\circ + \tan^{-1}(\frac{45.98}{19.64}) = 180^\circ + 66.9^\circ = 247^\circ$ (nearest degree). [1]

18. (a) $\angle BCD = 180^\circ - 75^\circ = 105^\circ$ (opposite angles of cyclic quadrilateral) [1]

(b) $\angle ADB = \angle ABD = 40^\circ$? No. $\angle ADB$ and $\angle ACB$ subtend same arc. Not helpful. Since $AB \parallel DC$, $\angle BDC = \angle ABD = 40^\circ$ (alternate angles) [1] for reasoning, [1] for answer.

(c) $\angle ADC = \angle ADB + \angle BDC$. Need $\angle ADB$. In $\triangle ABD$: $\angle ADB = 180^\circ - 75^\circ - 40^\circ = 65^\circ$ [1] $\angle ADC = 65^\circ + 40^\circ = 105^\circ$ [1]

(d) $\angle DBC = \angle DAC$ (angles in same segment) $\angle DAC = \angle DAB - \angle CAB$? Not given. Alternative: $\angle DBC = 180^\circ - \angle BCD - \angle BDC = 180^\circ - 105^\circ - 40^\circ = 35^\circ$ [1] for method, [1] for answer.

19. (a) $AB = \sqrt{(8-2)^2 + (5-1)^2} = \sqrt{36 + 16} = \sqrt{52} = 7.21$ units (3 s.f.) [1]

(b) $BC = \sqrt{(4-8)^2 + (9-5)^2} = \sqrt{16 + 16} = \sqrt{32} = 5.66$ units (3 s.f.) [1]

(c) $AC = \sqrt{(4-2)^2 + (9-1)^2} = \sqrt{4 + 64} = \sqrt{68} = 8.25$ units (3 s.f.) [1]

(d) Area $= \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$ [1] $= \frac{1}{2}|2(5-9) + 8(9-1) + 4(1-5)|$ $= \frac{1}{2}|2(-4) + 8(8) + 4(-4)|$ [1] $= \frac{1}{2}|-8 + 64 - 16| = \frac{1}{2}|40| = 20$ square units [1]

20. (a) $EF^2 = 9^2 + 13^2 - 2(9)(13)\cos 50^\circ$ [1] $= 81 + 169 - 234(0.6428) = 250 - 150.4 = 99.6$ $EF = \sqrt{99.6} = 9.98$ cm (3 s.f.) [1]

(b) Area of $\triangle DEF = \frac{1}{2} \times 9 \times 13 \times \sin 50^\circ$ [1] $= 58.5 \times 0.7660 = 44.8$ cm$^2$ (3 s.f.) [1]

(c) Area of $\triangle DEG = \frac{DG}{DF} \times$ Area of $\triangle DEF$ (same height from $E$ to $DF$) [1] $= \frac{5}{13} \times 44.8 = 17.2$ cm$^2$ (3 s.f.) [1]

END OF ANSWER KEY