MEMO

CLUSTER PAPER 2016

MATHEMATICS: PAPER II Time: 3 hours 150 marks

PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1. This question paper consists of 28 pages and an Information Sheet of 2 pages(i-ii).

Please check that your question paper is complete. 2. Read the questions carefully. 3. Answer ALL the questions on the question paper and hand this in at the end

of the examination. 4. You may use an approved non-programmable and non-graphical calculator, unless

otherwise stated.

5. All necessary working details must be clearly shown. 6. Round off your answers to one decimal digit where necessary, unless otherwise

stated. 7. Ensure that your calculator is in DEGREE mode. 8. It is in your own interest to write legibly and to present your work neatly.

9. The last pages can be used for additional working, if necessary. If this space is

used, make sure that you indicate clearly which question is being answered.

CLUSTER: PAPER II – SEPTEMBER 2016 Page 2 of 28

SECTION A QUESTION 1 Refer to the sketch below:

AD//BC. A is the point (2; 4) and D is the point (1; -5).

Determine:

1.1 the coordinates of B. (2)

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1.2 the equation of BC in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0. (4)

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𝐵(1; 4)

𝑚𝐵𝐶 = 𝑚𝐴𝐷 =4 + 5

−2 − 1 𝐴𝐷//𝐵𝐶

𝑚𝐴𝐷 =9

−3

𝑚𝐴𝐷 = −3

𝑦 − 4 = −3(𝑥 − 1)

𝑦 − 4 = −3𝑥 + 3

𝑦 + 3𝑥 − 7 = 0 𝑓𝑜𝑟 𝑦 ∈ [0; 7]

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1.3 the coordinates of C. (2)

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[8]

QUESTION 2

Refer to the sketch below:

∆𝐴𝐵𝐶 has vertices of A(0; 3), B(-3;-2) and C(5; 0).

2.1 Calculate the length of: (4)

2.1.1 AB

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2.1.2 BC

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A(0;3)

B(-3;-2)

C(5;0)

y

x

𝑥 − 𝑖𝑛𝑡: 𝐿𝑒𝑡 𝑦 = 0

3𝑥 = 7

𝑥 =7

3

𝐶 (7

3; 0)

𝐴𝐵 = √(0 + 3)2 + (3 + 2)2

𝐴𝐵 = √9 + 25 𝐴𝐵 = √34 𝑢𝑛𝑖𝑡𝑠

𝐵𝐶 = √(5 + 3)2 + (0 + 2)2

𝐵𝐶 = √64 + 4

𝐵𝐶 = √68 𝑢𝑛𝑖𝑡𝑠

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2.1.3 AC

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2.2 Hence, show that ∆𝐴𝐵𝐶 is a right angled isosceles triangle. (4)

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2.3 Determine the area of ∆𝐴𝐵𝐶. (2)

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2.4 Given that BC is a diameter of the circumscribed circle of ∆𝐴𝐵𝐶, show that the

centre of this circle is M, the point (1; −1). (2)

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[12]

𝐴𝐶 = √(5 + 0)2 + (0 − 3)2

𝐴𝐶 = √25 + 9

𝐴𝐶 = √34 𝑢𝑛𝑖𝑡𝑠

𝐴𝐶 = 𝐴𝐵 = √34 𝑢𝑛𝑖𝑡𝑠

∆𝐴𝐵𝐶 𝑖𝑠 𝑎𝑛 𝑖𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

𝐴𝐶2 + 𝐴𝐵2 = 34 + 34 = 68

𝐵𝐶2 = 68

𝐴𝐶2 + 𝐴𝐵2 = 𝐵𝐶2 ∴ �̂� = 90°

∆𝐴𝐵𝐶 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 =1

2√34. √34

𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 = 17 𝑢𝑛𝑖𝑡2

𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐵𝐶: (−3 + 5

2;−2 + 0

2) = (1; −1)

𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 =1

2√34. √34 𝑠𝑖𝑛90°

𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 = 17 𝑢𝑛𝑖𝑡2

𝐴𝐶 = 𝐴𝐵 = √34 𝑢𝑛𝑖𝑡𝑠

∆𝐴𝐵𝐶 𝑖𝑠 𝑎𝑛 𝑖𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

𝑚𝐴𝐵 =3 + 2

0 + 3=

5

3 𝑎𝑛𝑑

𝑚𝐴𝐶 =3 − 0

0 − 5= −

3

5

𝑚𝐴𝐵. 𝑚𝐴𝐶 = (5

3) (−

3

5) = −1 ∴ 𝐴𝐵𝐴𝐶

∆𝐴𝐵𝐶 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

CLUSTER: PAPER II – SEPTEMBER 2016 Page 5 of 28

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QUESTION 3 Find the equation of the line that is a tangent to the circle

(𝑥 + 1)2 + (𝑦 + 2)2 = 20 at (3; −4). (6)

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[6]

𝐶𝑒𝑛𝑡𝑟𝑒: (−1; −2)

𝑚𝑟 =−4 + 2

3 + 1

𝑚𝑟 = −1

2

𝑚𝑇 = 2 (𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡)

𝑦 + 4 = 2(𝑥 − 3)

𝑦 = 2𝑥 − 10

CLUSTER: PAPER II – SEPTEMBER 2016 Page 6 of 28

QUESTION 4

If 37𝑐𝑜𝑠𝜗 + 10 = −2 and 𝜗 ∈ [180°; 360°], determine, with the use of a diagram, the

value of

1

4 𝑠𝑖𝑛 2𝜗

𝑐𝑜𝑠 𝜗. (4)

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[4]

𝑐𝑜𝑠𝜗 =−12

37

𝑦 = ±√(37)2 − (−12)2 𝑃𝑦𝑡ℎ

𝑦 = ±√1225

𝑦 = ±35 𝑏𝑢𝑡 𝑖𝑛 𝐼𝐼𝐼 𝑦 = −35

14

𝑠𝑖𝑛 2𝜗

𝑐𝑜𝑠 𝜗

=

14

(2𝑠𝑖𝑛 𝜗. 𝑐𝑜𝑠 𝜗)

𝑐𝑜𝑠 𝜗

=1

2𝑠𝑖𝑛 𝜗

=1

2(−

35

37)

=−35

74

𝝑 −𝟏𝟐

𝟑𝟕

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QUESTION 5

5.1 Prove: 1 − 𝑐𝑜𝑠 (90° − 2𝜗)𝑡𝑎𝑛 (180° + 𝜗) = 𝑐𝑜𝑠 2𝜗 (5)

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5.2 For which values of 𝜗 is this identity in 5.1 undefined? (2)

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[7]

𝐿𝐻𝑆 = 1 − 𝑠𝑖𝑛 2𝜗 𝑡𝑎𝑛 𝜗

𝐿𝐻𝑆 = 1 − (2 𝑠𝑖𝑛 𝜗 𝑐𝑜𝑠 𝜗)𝑠𝑖𝑛 𝜗

𝑐𝑜𝑠 𝜗

𝐿𝐻𝑆 = 1 − 2𝑠𝑖𝑛2 𝜗

𝐿𝐻𝑆 = 𝑐𝑜𝑠 2𝜗

𝐿𝐻𝑆 = 𝑅𝐻𝑆

𝜗 = 90° + 𝑘. 180°; 𝑘 ∈ 𝑍

CLUSTER: PAPER II – SEPTEMBER 2016 Page 8 of 28

QUESTION 6 Evaluate without using a calculator:

6.1 𝑐𝑜𝑠 3°. 𝑐𝑜𝑠 318° + 𝑐𝑜𝑠 267°. 𝑠𝑖𝑛 42° (5)

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6.2 𝑡𝑎𝑛2210° − (1 + 𝑐𝑜𝑠 120°). 𝑠𝑖𝑛2225° (5)

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[10]

= 𝑐𝑜𝑠 3°. 𝑐𝑜𝑠 42° − 𝑐𝑜𝑠 87°. 𝑠𝑖𝑛 42°

= 𝑐𝑜𝑠 3°. 𝑐𝑜𝑠 42° − 𝑠𝑖𝑛 37°. 𝑠𝑖𝑛 42°

= 𝑐𝑜𝑠 (42° + 3°)

= 𝑐𝑜𝑠 45°

=√2

2

= (𝑡𝑎𝑛 30°)2 − (1 − 𝑐𝑜𝑠 60°)(−𝑠𝑖𝑛 45°)2

= (√3

3)

2

− (1 −1

2) (−

√2

2)

2

=1

3− (

1

2) (

1

2)

=1

3−

1

4

=1

12

= 𝑐𝑜𝑠 3°. 𝑠𝑖𝑛 48° − 𝑠𝑖𝑛 3°. 𝑠𝑖𝑛 42°

= 𝑐𝑜𝑠 3°. 𝑠𝑖𝑛 48° − 𝑠𝑖𝑛 3°. 𝑐𝑜𝑠 48°

= 𝑠𝑖𝑛 (48° − 3°)

= 𝑠𝑖𝑛 45°

=√2

2

CLUSTER: PAPER II – SEPTEMBER 2016 Page 9 of 28

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QUESTION 7

The table gives the Olympic pole vault records for the past 12 games Olympic

games.

Year Height in metres

1968 5,40

1972 5,50

1976 5,50

1980 5,78

1984 5,75

1988 5,90

1992 5,80

1996 5,92

2000 5,90

2004 5,95

2008 5,96

2012 5,97

7.1 Find the equation of the line of best fit for the height (𝑦) against the year (𝑥)

in the form 𝑦 = 𝑎 + 𝑏𝑥.

Determine your values for 𝑎 and 𝑏 rounded to 2 decimal digits. (2)

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7.2 Complete the scatter plot and draw the line of best fit on the scatter plot. (3)

19

68

19

84

19

92

20

12

20

08

19

72

19

76

19

80

19

96

20

00

20

04

19

88

𝑎 = 5,45 ; 𝑏 = 0,05

x x x

x x

x

CLUSTER: PAPER II – SEPTEMBER 2016 Page 10 of 28

7.3 Determine the value of the correlation coefficient rounded to 2 decimal digits

and explain clearly what can be deduced from this result. (3)

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7.4 Use the model to predict the record pole vault height for the 2016 Olympic

Games. (1)

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7.5.1 Use the model to predict the record pole vault height for the 2028 Olympic

Games. (1)

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7.5.2 Do you think the actual record in 2028 will be higher or lower than this

prediction? Give a reason for your answer. (2)

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[12]

𝑟 = 0,91

𝑉𝑒𝑟𝑦 𝑆𝑡𝑟𝑜𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛.

𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑂𝑙𝑦𝑚𝑝𝑖𝑐 𝑝𝑜𝑙𝑒 𝑣𝑎𝑢𝑙𝑡 𝑟𝑒𝑐𝑜𝑟𝑑𝑠 𝑠ℎ𝑜𝑢𝑙𝑑 𝑖𝑚𝑝𝑟𝑜𝑣𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟𝑠.

𝑦 = 5,45 + 0,05(13)

𝑦 = 6,10 𝑚

From Graph: 𝑦 = 𝐻𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟 2016 = 6,10 𝑚

OR

𝑦 = 5,45 + 0,05(16)

𝑦 = 6,25 𝑚

𝐿𝑜𝑤𝑒𝑟

𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑒𝑥𝑡𝑟𝑎𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑖𝑡 𝑓𝑎𝑙𝑙𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟 𝑖𝑡 𝑖𝑠

𝑛𝑜𝑡 𝑎 𝑟𝑒𝑙𝑖𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒.

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QUESTION 8 In the diagram, O is the centre of the circle passing through A, B, C and D. AB//CD

and �̂� = 20°.

Complete the following statements and reasons to prove that AOEC is a cyclic

quadrilateral. No extra steps/calculations are allowed. (5)

STATEMENTS REASONS

�̂�1 = … … alt. angles ; AB//CD

�̂�1 = 40° …………………..

�̂� = 20° ……………………...

�̂�1 = … … ext. angle of ∆

𝐴𝑂𝐸𝐶 𝑖𝑠 𝑎 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑 ……………………...

[5]

20°

40°

< 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒 = 2 × < 𝑜𝑛 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒

<′ 𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

𝑙𝑖𝑛𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑 =<′ 𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑖𝑑𝑒

CLUSTER: PAPER II – SEPTEMBER 2016 Page 12 of 28

QUESTION 9 9.1 Complete the following statement: The line from the centre of the circle perpendicular to the chord _____________

_______________________________ . (1)

9.2 Refer to the figure below:

In the circle with centre O. OTQP, OSPR, OT = 8 units, PQ = 30 units and

PR = 23 units.

Determine OS = 𝑥. (5)

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Q

O P

T

S

R

𝑥

𝑏𝑖𝑠𝑒𝑐𝑡𝑠 𝑡ℎ𝑒 𝑐ℎ𝑜𝑟𝑑

𝑄𝑇 = 𝑇𝑃 = 15 𝑢𝑛𝑖𝑡𝑠 𝑙𝑖𝑛𝑒 𝑓𝑜𝑟𝑚 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 𝑡𝑜 𝑐ℎ𝑜𝑟𝑑

𝑂𝑄 = √152 + 82 𝑃𝑦𝑡ℎ

𝑂𝑄 = 17 𝑢𝑛𝑖𝑡𝑠

𝑂𝑅 = 17 𝑢𝑛𝑖𝑡𝑠 𝑅𝑎𝑑𝑖𝑖

𝑅𝑆 = 𝑆𝑃 = 11,5 𝑢𝑛𝑖𝑡𝑠 𝑙𝑖𝑛𝑒 𝑓𝑜𝑟𝑚 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 𝑡𝑜 𝑐ℎ𝑜𝑟𝑑

𝑂𝑆 = 𝑥 = √172 − (11,5)2 𝑃𝑦𝑡ℎ

𝑥 = 12,5 𝑢𝑛𝑖𝑡𝑠

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[6] QUESTION 10

Refer to the figure below:

PQ and PS are tangents to circle QRST. ST, RS and RQ are chords and QS is joined.

PQ//ST. Let �̂�2 = 𝑥.

10.1 Calculate �̂�1 in terms of 𝑥. (1)

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�̂�1 = 𝑥 𝑡𝑎𝑛 − 𝑐ℎ𝑜𝑟𝑑 𝑇ℎ𝑒𝑜𝑟𝑒𝑚

𝑥

CLUSTER: PAPER II – SEPTEMBER 2016 Page 14 of 28

10.2 Prove: ∆𝑅𝑆𝑄///∆𝑅𝑄𝑃 (4)

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10.3 Prove: 𝑄𝑅2 = 𝑅𝑆. 𝑅𝑃 (2)

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[7]

TOTAL FOR SECTION A = 77 MARKS

�̂�2 = �̂�1 = 𝑥 𝑃𝑟𝑜𝑣𝑒𝑛

�̂�2 = �̂� <′ 𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡;

𝑏𝑢𝑡 �̂� = �̂�1 𝑎𝑙𝑡 <′ 𝑠; 𝑃𝑄//𝑆𝑇

∴ �̂�2 = �̂�1

�̂�2 + �̂�3 = �̂�1 ∑ 𝑜𝑓 <′ 𝑠 𝑜𝑓∆= 180°

∴ ∆𝑅𝑆𝑄///∆𝑅𝑄𝑃 <<<

𝐼𝑛 ∆𝑅𝑆𝑄 𝑎𝑛𝑑 ∆𝑅𝑄𝑃:

𝑄𝑅2 = 𝑅𝑆. 𝑅𝑃

𝑅𝑆

𝑄𝑅=

𝑄𝑅

𝑃𝑅 ∆𝑅𝑆𝑄///∆𝑅𝑄𝑃

CLUSTER: PAPER II – SEPTEMBER 2016 Page 15 of 28

SECTION B QUESTION 11

The circle, with equation 𝑥2 − 10𝑥 + 𝑦2 + 8𝑦 − 40 = 0, is given. A point 𝑃(𝑥; 𝑦)

on the circumference of a NEW circle, is such that it is always 3 units from the

circumference of the original circle, and outside the original circle.

Determine the equation of this new circle. (6)

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[6]

(𝑥 − 5)2 + (𝑥 + 4)2 = 81

(𝑥 − 5)2 + (𝑥 + 4)2 = 122

(𝑥 − 5)2 + (𝑥 + 4)2 = 144

𝑥2 − 10𝑥 + 25 + 𝑦2 + 8𝑦 + 16 = 25 + 16 + 40

Initial circle: Centre (5; −4) 𝑎𝑛𝑑 𝑟 = 9

New circle: Centre (5; −4) 𝑎𝑛𝑑 𝑟 = 12

CLUSTER: PAPER II – SEPTEMBER 2016 Page 16 of 28

QUESTION 12

Solve for 𝑥 correct to one decimal digit if 𝜗 ∈ [−180°; 180°]

𝑠𝑖𝑛(2𝜗 − 10°) = −𝑐𝑜𝑠 (𝜗 + 50°) (8)

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[8]

𝑠𝑖𝑛(2𝜗 − 10°) = −sin [90° − (𝜗 + 50°)]

𝑅𝐴 (2𝜗 − 10°) = 40° − 𝜗

𝜗 = 76,7° + 𝑘. 120°

𝜗 = {−163,3°; −43,3°; −30°; 76,7°}

𝑠𝑖𝑛(2𝜗 − 10°) = −𝑐𝑜𝑠 (𝜗 + 50°)

3rd 𝑘 ∈ 𝑍 4th

2𝜗 − 10° = 180° + 40° − 𝜗 + 𝑘. 360° 2𝜗 − 10° = 360° − (40° − 𝜗) + 𝑘. 360°

3𝜗 = 230° + 𝑘. 360° 𝜗 = 330° + 𝑘. 360°

CLUSTER: PAPER II – SEPTEMBER 2016 Page 17 of 28

PLEASE TURN OVER

QUESTION 13 The company, Mega Ski, has sales of ski equipment as given by

𝑆(𝑡) = 10[1 − 𝑐𝑜𝑠 (30°. 𝑡)]

where 𝑡 is the time, in months (𝑡 = 0 corresponds to 1 August 2015), and 𝑆(𝑡) is in

thousands of Rands.

13.1 Use the given graphpaper below to sketch the function of the sales on a 12-month

interval [0; 12]. (4)

CLUSTER: PAPER II – SEPTEMBER 2016 Page 18 of 28

13.2 What is the period of the function? (2)

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13.3 What is the minimum amount of sales? (1)

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13.4 What is the maximum and when does the maximum amount of sales occur? (3)

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[10]

12 𝑚𝑜𝑛𝑡ℎ𝑠

𝑅 0,00

Max: R20 000

When: after 6 months

1 𝐹𝑒𝑏𝑟𝑢𝑎𝑟𝑦 2016

CLUSTER: PAPER II – SEPTEMBER 2016 Page 19 of 28

PLEASE TURN OVER

QUESTION 14 At a particular time during the day, a tower of height 19,2 meters casts a shadow. At

the same time, a person who is 1,65 meters tall, casts a shadow 5 metres long.

What is the length of the shadow cast by the tower at that time? (6)

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[6]

𝑠ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑡𝑜𝑤𝑒𝑟

19,2=

5

1,65 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 ∆ // 𝑡𝑜 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒

∴ 𝑠ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑡𝑜𝑤𝑒𝑟 = 58,2 𝑚

𝑇𝑜𝑤𝑒𝑟//𝑃𝑒𝑟𝑠𝑜𝑛 𝑇𝑜𝑤𝑒𝑟 𝑎𝑛𝑑 𝑃𝑒𝑟𝑠𝑜𝑛 𝑖𝑠 𝑡𝑜 𝑔𝑟𝑜𝑢𝑛𝑑; 𝑐𝑜𝑟𝑟 < 𝑠 𝑒𝑞𝑢𝑎𝑙

1,65 m

Shadow of tower

19,2 m

5 m

5

1,65 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 ∆ // 𝑡𝑜 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒

∴ 𝑠ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑡𝑜𝑤𝑒𝑟 = 58,2 𝑚

OR

In small triangle: 𝑡𝑎𝑛𝜃 =1,65

5

𝜃 = 18,2628. . °

In big triangle: 𝑡𝑎𝑛𝜃 =19,2

𝑆ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑇𝑜𝑤𝑒𝑟

𝑆ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑇𝑜𝑤𝑒𝑟 =19,2

𝑡𝑎𝑛18,2628..°

∴ 𝑠ℎ𝑎𝑑𝑜𝑤 𝑜𝑓 𝑡𝑜𝑤𝑒𝑟 = 58,2 𝑚

CLUSTER: PAPER II – SEPTEMBER 2016 Page 20 of 28

QUESTION 15

The new fitness centre offers two different fitness classes. The attendance for each class

for 12 sessions is represented in the box-and-and whisker plot.

15.1 Determine the interquartile range of Class A. (1)

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15.2 Compare the data from Class A and Class B. How are they alike? (1)

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15.3 Compare the data from Class A and Class B. How are they different? (1)

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15.4 Which class has more variability? Show calculations to support your answer. (3)

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15.5 Which class has a higher mean deviation? Explain your answer. (2)

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[8]

𝐼𝑄𝑅 = 14 − 4 = 10

𝐵𝑜𝑡ℎ 𝑠𝑘𝑒𝑤𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡/𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒.

𝐵𝑜𝑡ℎ ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑡𝑡𝑒𝑛𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 22 OR

𝐶𝑙𝑎𝑠𝑠 𝐴: 𝑚𝑖𝑛 = 2 and 𝐶𝑙𝑎𝑠𝑠 𝐵: 𝑚𝑖𝑛 = 10 OR 𝐶𝑙𝑎𝑠𝑠 𝐴: 𝑟𝑎𝑛𝑔𝑒 =

20 and 𝐶𝑙𝑎𝑠𝑠 𝐵: 𝑟𝑎𝑛𝑔𝑒 = 12 OR any other reason.

𝐶𝑙𝑎𝑠𝑠 𝐴: 𝑅𝑎𝑛𝑔𝑒 = 20 𝑎𝑛𝑑 𝐼𝑄𝑅 = 10

𝐶𝑙𝑎𝑠𝑠 𝐵: 𝑅𝑎𝑛𝑔𝑒 = 12 𝑎𝑛𝑑 𝐼𝑄𝑅 = 4

𝐵𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 𝑎𝑛𝑑 𝐼𝑄𝑅 𝑜𝑓 𝐶𝑙𝑎𝑠𝑠 𝐴 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑡ℎ𝑜𝑠𝑒 𝑜𝑓 𝐶𝑙𝑎𝑠𝑠 𝐵,

∴ 𝐶𝑙𝑎𝑠𝑠 𝐴 ℎ𝑎𝑠 𝑎 𝑚𝑜𝑟𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦

𝐶𝑙𝑎𝑠𝑠 𝐴. 𝑆𝑘𝑒𝑤𝑒𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒𝑙𝑦: �̅� − 𝑀 > 0

𝐶𝑙𝑎𝑠𝑠 𝐴 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑎 ℎ𝑖𝑔ℎ𝑒𝑟 𝑚𝑒𝑎𝑛 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛,

𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 𝑖𝑡𝑒𝑚𝑠 𝑎𝑟𝑒 𝑓𝑢𝑟𝑡ℎ𝑒𝑟 𝑓𝑟𝑜𝑚 �̅�.

CLUSTER: PAPER II – SEPTEMBER 2016 Page 21 of 28

PLEASE TURN OVER

QUESTION 16 Refer to the sketch below: In the diagram O is the centre of the circle with AE and CE tangents to the circle.

�̂�1 = 45° and �̂�1 = 127°

Calculate the size of �̂�. (5)

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[5]

A B

C

D

E

O 127°

45°

1 2

1 2 3

1 2

1

�̂�1 + �̂�2 = 90° 𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡

∴ �̂�2 = 45°

�̂� = 360° − 2(90°) − 127° 𝐼𝑛𝑡 <′ 𝑠 𝑜𝑓 𝑞𝑢𝑎𝑑𝑟𝑖𝑙𝑎𝑡𝑒𝑟𝑎𝑙 = 360°

∴ �̂� = 53°

�̂�1 = 90° 𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡

�̂�1 + �̂�2 = 90° 𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡

∴ 𝐴𝐸𝐶𝑂 𝑖𝑠 𝑎 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑 𝑜𝑝𝑝 𝑖𝑛𝑡 <′ 𝑠 𝑜𝑓 𝑞𝑢𝑎𝑑 = 180°

�̂�1 + �̂� = 180° 𝑜𝑝𝑝 𝑖𝑛𝑡 <′ 𝑠 𝑜𝑓 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑

∴ �̂� = 53° �̂�1 = 127°

OR

�̂�1 = 90° 𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡

CLUSTER: PAPER II – SEPTEMBER 2016 Page 22 of 28

QUESTION 17

In the diagram below, points R, P, A, Q and T lie on a circle. RA bisects �̂� and

𝐴𝐵 = 𝐴𝑄.

RA and TQ produced meet at B.

Prove that:

17.1 AQ bisects 𝑃�̂�𝐵. (3)

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17.2 TR = TB. (2)

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�̂�3 = �̂�1 𝐸𝑥𝑡 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑐𝑦𝑐𝑙𝑖𝑐 𝑄𝑢𝑎𝑑

�̂�2 = �̂�2 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

∴ �̂�2 = �̂�3

∴ 𝐴𝑄 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 𝑃�̂�𝐵

�̂�1 = �̂�2 𝑅𝐴 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 �̂�

�̂�3 = �̂�1 𝑃𝑟𝑜𝑣𝑒𝑛 𝑖𝑛 17.1

∴ �̂�1 = �̂�

∴ 𝑇𝑅 = 𝑇𝐵 𝑠𝑖𝑑𝑒𝑠 𝑜𝑝𝑝 = 𝑎𝑛𝑔𝑙𝑒𝑠

�̂� = �̂�3 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠

CLUSTER: PAPER II – SEPTEMBER 2016 Page 23 of 28

PLEASE TURN OVER

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17.3 �̂� = 𝑇�̂�𝑃 (3) ________________________________________________________________________

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[8]

�̂� = �̂�1 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

�̂�3 = �̂�1 𝑝𝑟𝑜𝑣𝑒𝑛

∴ �̂� = 2�̂�1 𝑓𝑟𝑜𝑚 17.1

∴ �̂� = 𝑇�̂�𝑃

�̂�1 = 2�̂�3 𝑒𝑥𝑡 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 ∆; �̂� = �̂�3

∴ 𝑅𝑃//𝑇𝐵 = 𝑎𝑙𝑡 <′ 𝑠

�̂�2 + �̂�3 = �̂� 𝑎𝑙𝑡 <′ 𝑠 ; 𝑅𝑃//𝑇𝐵

𝑏𝑢𝑡 �̂�2 + �̂�3 = �̂�1 + �̂�2 𝑒𝑥𝑡 < 𝑜𝑓 𝑐𝑦𝑐𝑙𝑖𝑐 𝑞𝑢𝑎𝑑 𝑃𝑅𝑇𝑄

∴ �̂� = 𝑇�̂�𝑃

OR

�̂�2 = �̂�1 = �̂�2 = �̂�3 = �̂� 𝑝𝑟𝑜𝑣𝑒𝑛 𝑖𝑛 17.1 𝑎𝑛𝑑 17.2

CLUSTER: PAPER II – SEPTEMBER 2016 Page 24 of 28

QUESTION 18 Refer to the sketch:

In the circle below, AB and CD are chords intersecting at E.

If 𝐴𝐸 = 5, 𝐵𝐸 = 12 and 𝐶𝐸 = 6, what is the length of DE? (6)

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[6]

A

B

E

C

D

𝐼𝑛 ∆𝐴𝐶𝐸 𝑎𝑛𝑑 ∆𝐵𝐷𝐸:

�̂� = �̂� <′ 𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

�̂� = �̂� <′ 𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

𝐴�̂�𝐶 = 𝐷�̂�𝐵 ∑ 𝑜𝑓 <′ 𝑠 𝑜𝑓∆= 180°

∴ ∆𝐴𝐶𝐸///∆𝐷𝐵𝐸 <<<

𝐴𝐶

𝐷𝐵=

𝐶𝐸

𝐵𝐸=

𝐴𝐸

𝐷𝐸 ∆𝐴𝐶𝐸///∆𝐷𝐵𝐸

6

12=

5

𝐷𝐸

𝐷𝐸 =5 × 12

6

𝐷𝐸 = 10 𝑢𝑛𝑖𝑡𝑠

𝐶𝑜𝑛𝑠𝑡𝑟: 𝐴𝐶 𝑎𝑛𝑑 𝐷𝐵

OR

Similarly

𝐶𝑜𝑛𝑠𝑡𝑟: 𝐴𝐷 𝑎𝑛𝑑 𝐵𝐶

Prove: ∴ ∆𝐴𝐸𝐷///∆𝐶𝐸𝐵 <<<

CLUSTER: PAPER II – SEPTEMBER 2016 Page 25 of 28

PLEASE TURN OVER

QUESTION 19

Daniel decided to participate in this year’s Tour de France bicycle race. One of the

routes are around the Annecy Lake which is a circular lake that has a diameter of 4 km,

as shown in the diagram below. Points B and D are the opposite sides of Annecy Lake

and lie on a straight line through the centre of the lake, with each point 5 km from the

centre. The course of the race is ABCDE, where AB and DE are tangents to the lake.

Determine the length of the route. (6) ________________________________________________________________________

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B

A

C D

𝟒 𝒌𝒎

𝟓 𝒌𝒎

𝑨𝒏𝒏𝒆𝒄𝒚 𝑳𝒂𝒌𝒆

𝟓 𝒌𝒎

E

𝑟𝑎𝑑𝑖𝑢𝑠 = 𝐴𝐶 = 𝐶𝐸 = 2 𝑘𝑚 𝑟𝑎𝑑𝑖𝑢𝑠 =1

2 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

𝐴𝐵 = 𝐷𝐸 = √52 − 22 𝑃𝑦𝑡ℎ

𝐴𝐵 = 𝐷𝐸 = √21 𝑘𝑚

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑟𝑜𝑢𝑡𝑒 = 2√21 + 2(5)

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑟𝑜𝑢𝑡𝑒 ≈ 19,2 𝑘𝑚

𝐶𝑜𝑛𝑠𝑡𝑟: 𝐴𝐶 𝑎𝑛𝑑 𝐶𝐸 𝑟𝑎𝑑𝑖𝑖

�̂� = �̂� = 90° 𝑅𝑎𝑑𝑖𝑢𝑠𝑇𝑎𝑛𝑔𝑒𝑛𝑡

CLUSTER: PAPER II – SEPTEMBER 2016 Page 26 of 28

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[6]

CLUSTER: PAPER II – SEPTEMBER 2016 Page 27 of 28

PLEASE TURN OVER

QUESTION 20 The diagram shows triangles ABC and ABD with AD parallel to BC. The sides AC and

BD intersect at Y. The point X lies on AB such that XY is parallel to AD and BC.

20.1 Prove ∆ABC /// ∆AXY. (4)

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20.2 Hence, or otherwise, prove that 1

𝑋𝑌=

1

𝐴𝐷+

1

𝐵𝐷 (6)

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D

Y

C

A X B

𝐼𝑛 ∆𝐴𝐵𝐶 𝑎𝑛𝑑 ∆𝐴𝑋𝑌:

𝐶�̂�𝐵 = 𝐶�̂�𝐵 𝑐𝑜𝑚𝑚𝑜𝑛 𝑎𝑛𝑔𝑙𝑒

�̂� = 𝐴�̂�𝑋 𝑐𝑜𝑟𝑟𝑒𝑠𝑝 <′ 𝑠 ; 𝐶𝐵//𝑌𝑋

�̂� = 𝐴�̂�𝑌 𝑐𝑜𝑟𝑟𝑒𝑠𝑝 <′ 𝑠 ; 𝐶𝐵//𝑌𝑋 𝑜𝑟 ∑ 𝑜𝑓 <′ 𝑠 𝑜𝑓∆= 180°

∴ ∆𝐴𝐵𝐶///∆𝐴𝑋𝑌 <<<

CLUSTER: PAPER II – SEPTEMBER 2016 Page 28 of 28

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[10]

TOTAL FOR SECTION B = 73 MARKS

TOTAL: 150 marks

𝐴𝑋

𝐴𝐵=

𝑋𝑌

𝐵𝐶 ∆𝐴𝐵𝐶///∆𝐴𝑋𝑌

𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑖𝑠 ∆𝐴𝐵𝐷///∆𝑋𝐵𝑌.

𝐵𝑋

𝐵𝐴=

𝑋𝑌

𝐴𝐷 ∆𝐴𝐵𝐶///∆𝐴𝑋𝑌

∴𝐴𝑋

𝐴𝐵+

𝐵𝑋

𝐵𝐴=

𝑋𝑌

𝐵𝐶+

𝑋𝑌

𝐴𝐷

𝐵𝑢𝑡 𝐴𝑋 + 𝐵𝑋 = 𝐴𝐵

𝐴𝐵

𝐴𝐵= 𝑋𝑌 (

1

𝐵𝐶+

1

𝐴𝐷)

1

𝑋𝑌=

1

𝐵𝐶+

1

𝐴𝐷