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WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019
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WYNBERG BOYS HIGH SCHOOL
SUBJECT: MATHEMATICS
PAPER: 2
GRADE: 12
DATE: 6 SEPTEMBER 2019
TIME: 3 HOURS
MARKS: 150
EXAMINER: G VAN DER WESTHUIZEN
MODERATOR: C. HULL
NAME: MEMO
TEACHER: BG ED HU LR VZ
QUESTION 1 2 3 4 5 6 7 8 9 10 11 12 TOTAL
MAX.
MARKS 11 5 6 20 20 22 12 8 7 8 14 17 150
MARKS
OBTAINED
This answer book consists of 20 pages
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WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019
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10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70
QUESTION 1
1.1 𝑚𝑒𝑎𝑛 =1155
20 ✓ = 57,75𝑘𝑔 ✓ a div by 20 and answer
𝑠𝑡𝑑 𝑑𝑒𝑣. = 6,74✓✓ a correct answer to 2 dec. places.
(4)
1.2.1 22 𝑏𝑜𝑦𝑠✓ a correct answer
(1)
1.2.2 𝑚𝑒𝑎𝑛 =1320
22 ✓ = 60𝑘𝑔✓ a div by 22 and correct answer
(2)
1.3 1155+5𝑥
25✓ = 60✓ a LHS and RHS
1155 + 5𝑥 = 1500✓ a simplification
5𝑥 = 345
𝑥 = 69𝑘𝑔✓ a answer
𝐸𝑎𝑐ℎ 𝑏𝑜𝑦𝑠 𝑚𝑢𝑠𝑡 𝑤𝑒𝑖𝑔ℎ 69𝑘𝑔
(4)
QUESTION 2
2.1 𝑚𝑖𝑛 = 12✓ a min and max
𝑄1 = 17𝑄2 = 30𝑄3 = 38
} a quartiles
𝑚𝑎𝑥 = 65✓
(2)
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2.2 𝐼𝑄𝑅 = 38 − 17✓ ca subtraction
= 21✓ ca answer
(2)
2.3 Data negatively skewed✓ a
(1)
QUESTION 3
Time (𝑡) 0,5 1,0 1,5 2,0 3,0 4,0 5,0
Devon’s pulse rate (𝑃) 125 113 104 95 82 85 72
3.1 �̂� = 122,67✓ − 10,74𝑥✓ a A and B
(2)
3.2 �̂� = 122,67 − 10,74(2,5)✓ ca substitution
�̂� ≈ 96✓ ca answer
(2)
3.3 Devon is fitter✓ since his pulse rate returns to normal far quicker because the
gradient of his regression line is greater✓ (negative). a answer and reason
(2)
EXTRA WRITING SPACE
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QUESTION 4
4.1 𝑚𝑃𝑅 =
5 − 1
−4 − 4
= −1
2✓ a answer for gradient
𝑦 − 1 = −1
2(𝑥 − 4)✓ a subst of point
𝑦 = −1
2𝑥 + 2 + 1
∴ 𝑦 = −1
2𝑥 + 3✓ ca answer
(3)
4.2 𝑚𝑄𝑆 = 2 (𝑃𝑅 ⊥ 𝑄𝑆)✓ ca from 4.1
𝑦 + 4 = 2(𝑥 + 1)✓ ca subst of point
𝑦 = 2𝑥 + 2 − 4
𝑦 = 2𝑥 − 2✓ ca answer
(3)
4.3 −1
2𝑥 + 3 = 2𝑥 − 2✓ ca equating
−𝑥 + 6 = 4𝑥 − 4
10 = 5𝑥
2 = 𝑥✓ ca value of x
⇒ 𝑦 = 2(2) − 2✓ ca subst of x
𝑦 = 2✓ ca value of y
∴ 𝑇(2; 2)
(4)
R(4;1)
F
S
P(-4;5)
Q(-1;-4)
T
𝜃
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4.4 Extend PR to cut x-axis
𝐼𝑛𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑅 = tan−1(−1
2)✓ ca from 4.1
= 180° − 26,57°
= 153,43°✓ ca but must be obtuse
𝜃 = tan−1 (1−(−4)
4−(−1))✓ a subst of points
𝜃 = 45°✓ a answer (must be acute)
∴ 𝑃�̂�𝐹 = 153,43° − 45°
∴ 𝑃�̂�𝐹 = 108,43°✓ ca (5)
4.5 𝑃�̂�𝑄 = 71,57° (angles on a str. line)✓ ca from 4.4
𝑃𝑅 = √(−4 − 4)2 + (5 − 1)2
𝑃𝑅 = √80✓ a distance
𝑅𝑄 = √(−1 − 4)2 + (−4 − 1)2
𝑅𝑄 = √50✓ a distance
𝐴𝑟𝑒𝑎 ∆𝑃𝑄𝑅 = 1
2(√80)(√50)𝑠𝑖𝑛71,57° ✓ ca
∴ 𝐴𝑟𝑒𝑎 ∆𝑃𝑄𝑅 = 30 𝑢𝑛𝑖𝑡𝑠2✓ ca
(5)
EXTRA WRITING SPACE
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QUESTION 5
5.1.1 𝐷 (
(−1 + 5)
2;8 + 0
2)
𝐷(2; 4)✓ a midpoint
(𝑥 − 2)2 + (𝑦 − 4)2 = 𝑟2
(5 − 2)2 + (0 − 4)2 = 𝑟2 ✓ a subst
25 = 𝑟2✓ a value of 𝑟2
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛: (𝑥 − 2)2 + (𝑦 − 4)2 = 25✓ ca equation of circle
(4)
5.1.2 𝐴𝐶 ⊥ 𝐶𝐵 (𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 𝑏𝑦 𝑑𝑖𝑎𝑚. 𝐴𝐵)
⇒ 𝐶(−1; 0)✓, 𝑠𝑖𝑛𝑐𝑒 𝑥𝐴 = 𝑥𝐶 a answer (must be co-ordinate)
OR
Using centre of circle x-co-ordinate
(1)
5.1.3 𝑚𝐷𝐶 =
4 − 0
2 − (−1)
=4
3✓ ca gradient
∴ 𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = −3
4 (𝑟𝑎𝑑 ⊥ 𝑡𝑎𝑛)✓ ca
𝑦 − 0 = −3
4(𝑥 + 1)✓ ca subst
𝑦 = −3
4𝑥 −
3
4✓ ca equation of tangent
(4)
A(-1;8)
D
C B(5;0) O
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5.1.4 𝑥2 − 4𝑥 + 𝑦2 − 8𝑦 − 29 = 0
(4)
𝑥2 − 4𝑥 + 4 + 𝑦2 − 8𝑦 + 16 = 29 + 4 + 16 ✓ a com sq
(𝑥 − 2)2 + (𝑦 − 4)2 = 49✓ (𝑐𝑖𝑟𝑐𝑙𝑒 2) a simplify
𝑐𝑒𝑛𝑡𝑟𝑒 (2; 4)✓ a co-ord
∴ 𝑐𝑖𝑟𝑐𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑖𝑐 (same centre)✓ a conc
5.1.5 𝑐𝑖𝑟𝑐𝑙𝑒 2: 𝑟𝑎𝑑𝑖𝑢𝑠 = 7✓ a value
(2)
𝑐𝑖𝑟𝑐𝑙𝑒 1: 𝑟𝑎𝑑𝑖𝑢𝑠 = 5
∴ 𝑐𝑖𝑟𝑐𝑙𝑒 𝑙𝑖𝑒𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝐴𝐶𝐵✓ a concl.
5.2 𝑥2 − 2𝑥 . cos 𝜃 + 𝑦2 − 4𝑦 . cos 𝜃 = −2
𝑥2 − 2𝑥. cos 𝜃 + (cos𝜃)2 + 𝑦2 − 4𝑦 cos 𝜃 + (2cos𝜃)2 = −2 + cos2 𝜃 + 4 cos2 𝜃✓ a
∴ 𝑟2 = 5 cos2 𝜃 − 2✓a
max 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑜𝑠2𝜃 = 1✓a
∴ max 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑟2 = 3✓a
⇒ max 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑟 = √3✓a
(5)
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QUESTION 6
6.1
✓a correct quad
✓a correct adj. side
(5)
sin(A + 30°) = sinAcos30° + cosAsin30°✓ a compound formula
= (−3
7) (
√3
2) + (
−2√10
7) (
1
2)✓ ca subst
=−3√3−2√10
14✓ ca answer
6.2.1 1−sin (180°+θ)∙cos (90°+θ)
cos (−θ)
=1−(−sinθ)✓∙(−sinθ)✓
cosθ✓ a reductions
=1−sin2θ
cosθ
=cos2θ✓
cosθ a
= cosθ ✓ a answer
(5)
6.2.2 Undefined when cos 𝜃 = 0
Undefined when θ = 90°✓or θ = 270°✓ a
(2)
-3 7
−2√10 A
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6.3 LHS = 4cosθ −cos2θ+1
cosθ
=4cos2θ−(2cos2θ−1)✓−1
cosθ✓ a double angle formula
=4cos2θ−2𝑐𝑜𝑠2θ+1−1
cosθ a common denominator
=2cos2θ
cosθ ✓ a simplification
= 2cosθ
RHS =sin2θ
sinθ
=2sinθcosθ✓
sinθ a double angle formula
= 2cosθ ✓ a simplification
∴ LHS = RHS penalty of 1 if no conclusion
(5)
6.4 LHS = cos32°√1 − sin228° − sin28°√1 − cos232° ✓
= cos32° ∙ cos28° − sin28° ∙ sin32° ✓ a simplification
= cos (32° + 28°)✓ a compound formula
= cos60°✓ a simplification
=1
2 ✓ a answer
= RHS penalty of 1 if no conc.
(5)
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QUESTION 7
7.1 𝑎 = 2 ✓ a (1)
7.2
(3)
7.3 2𝑠𝑖𝑛𝑥 = 2cos (𝑥 − 30°)
𝑠𝑖𝑛𝑥 = cos (𝑥 − 30°)✓ a division by 2
𝑠𝑖𝑛𝑥 = sin (90° − (𝑥 − 30°)
𝑠𝑖𝑛𝑥 = sin (120° − 𝑥)✓ a reduction
𝑥 = 120° − 𝑥 + 𝑘. 360°, 𝑘𝜖𝑍 ✓ a equation
∴ 2𝑥 = 120° + 𝑘. 360°
𝑥 = 60° + 𝑘. 180°✓ a general solution
∴ 𝑥 = 60° 𝑜𝑟 𝑥 = −120° ✓ a answers
(5)
7.4 𝑘 = 3✓ a
(1)
7.5 𝑔 is shifted 60°✓to the right✓ a both
(2)
−180 180
−3
−2
−1
1
2
3
x
y
T(−90°; −2)
−60° 120°
(30°; 2)
√3
✓a x-int (both) ✓a y-int ✓a TP
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QUESTION 8
8.1 𝑠𝑖𝑛2𝑥 =𝐴𝐵
𝑟
∴ 𝐴𝐵 = 𝑟𝑠𝑖𝑛2𝑥 ✓a
(1)
8.2 𝐴�̂�𝐶 = 90° + 𝑥 (𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 ∆𝐶𝐵𝐾)✓ a
(1)
8.3 𝐴𝐾
𝑠𝑖𝑛𝑥=
𝑟
sin (90°+𝑥) ✓ ca sine rule
𝐴𝐾 =𝑟𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥✓ a answer
(2)
8.4 𝐴𝐾
𝐴𝐵=
2
3
𝑟𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
𝑟𝑠𝑖𝑛2𝑥=
2
3 ✓ a subst
𝑟𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥×
1
𝑟2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥✓=
2
3 a double angle formula
1
2𝑐𝑜𝑠2𝑥=
2
3
4𝑐𝑜𝑠2𝑥 = 3
𝑐𝑜𝑠2𝑥 =3
4 ✓ a simplification
𝑐𝑜𝑠𝑥 =√3
2
∴ 𝑥 = 30°✓ a answer
C
A K B
x x
r
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(4)
EXTRA WRITING SPACE
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QUESTION 9
9.1 𝐵�̂�𝐷 = 110° (∠′𝑠 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎 𝑝𝑜𝑖𝑛𝑡)✓ a statement and reason
∴ �̂� = 55° ✓ (∠ 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒 = 2∠ 𝑎𝑟𝑐)✓ a statement and reason each
(3)
9.2 𝐵�̂�𝐷 = 125° (𝑂𝑝𝑝 ∠′𝑠 𝐶𝑄 𝐴𝐵𝐶𝐷)✓ ca statement and reason
(1)
9.3 𝐵𝐸 = 𝐸𝐶 (𝑔𝑖𝑣𝑒𝑛)
∴ 𝐶1 = 65° ✓ (∠′𝑠𝑜𝑓 ∆𝐸𝐵𝐶, ∠′𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠)✓ a statement and reason each
∴ 𝐶2 = 60°✓ a answer
(3)
EXTRA WRITING SPACE
A
B
E
O
1 D
C
2
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QUESTION 10
10.1 𝐿𝑒𝑡 𝐴3 = 𝑥
∴ �̂� = 𝑥 (∠′𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠)✓ a statement and reason
∴ �̂� = 𝑥 (∠′𝑠 𝑖𝑛 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔)✓ a statement and reason
∴ 𝐴𝐺 = 𝐺𝐷 (𝑠𝑖𝑑𝑒𝑠 𝑜𝑝𝑝 = ∠′𝑠)✓ a statement and reason
(3)
10.2 𝐸2 = 𝑥 ✓ (𝑡𝑎𝑛 − 𝑐ℎ𝑜𝑟𝑑 𝑡ℎ𝑚)✓ a statement and reason each
∴ 𝐴2 = 𝑥 (∠′𝑠 𝑜𝑛 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔)✓ a statement and reason
∴ 𝐴𝐻 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 𝐵�̂�𝐸
(3)
10.3 𝐸2 = 𝑥
𝐴3 = 𝑥 ✓ a statement
∴ 𝐺𝐸 𝑖𝑠 𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝐴𝐸𝐻
(𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑒 𝑡𝑎𝑛 − 𝑐ℎ𝑜𝑟𝑑 𝑡ℎ𝑚)✓ a reason - must be converse
(2)
A
B G
C
E
D
H
1 2 3
3 2
1 2 3
4
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R
T
P
S
Q
QUESTION 11
11.1 𝑅𝑇𝑃:
𝑃𝑆
𝑆𝑄=
𝑃𝑇
𝑇𝑅
𝐶𝑜𝑛𝑠𝑡𝑟: 𝐷𝑟𝑎𝑤 ℎ ⊥ 𝑃𝑆 𝑎𝑛𝑑 𝑘 ⊥ 𝑃𝑇. 𝐽𝑜𝑖𝑛 𝑄𝑇 𝑎𝑛𝑑 𝑆𝑅 ✓a
𝑃𝑟𝑜𝑜𝑓: 𝐴𝑟𝑒𝑎 ∆𝑃𝑆𝑇
𝐴𝑟𝑒𝑎 ∆𝑆𝑇𝑄=
1
2𝑃𝑆.ℎ
1
2𝑆𝑄.ℎ
=𝑃𝑆
𝑆𝑄✓a
𝐴𝑟𝑒𝑎 ∆𝑃𝑆𝑇
𝐴𝑟𝑒𝑎 ∆𝑆𝑇𝑅=
1
2𝑃𝑇.𝑘
1
2𝑇𝑅.𝑘
=𝑃𝑇
𝑇𝑅✓a
𝑏𝑢𝑡 𝑎𝑟𝑒𝑎 ∆𝑆𝑇𝑄 = 𝑎𝑟𝑒𝑎 ∆𝑆𝑇𝑅 (𝑠𝑎𝑚𝑒 𝑏𝑎𝑠𝑒 𝑆𝑇 𝑎𝑛𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑎𝑚𝑒 ∥ 𝑙𝑖𝑛𝑒𝑠)✓a
∴𝐴𝑟𝑒𝑎 ∆𝑃𝑆𝑇
𝐴𝑟𝑒𝑎 ∆𝑆𝑇𝑄=
𝐴𝑟𝑒𝑎 ∆𝑃𝑆𝑇
𝐴𝑟𝑒𝑎 ∆𝑆𝑇𝑅✓a
𝑖. 𝑒 𝑃𝑆
𝑆𝑄=
𝑃𝑇
𝑇𝑅✓a
(6)
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11.2
11.2.1 𝑃𝐴
𝑃𝑄=
4
9
𝐿𝑒𝑡 𝑃𝐴 = 4𝑘
∴ 𝐴𝑄 = 5𝑘 (𝑃𝑄 = 9𝑘)✓ a
𝐿𝑒𝑡 𝐵𝑃 = 𝑝 ∴ 𝑅𝐵 = 2𝑝 (2𝑃𝐵 = 𝐵𝑅) 𝐶𝐴
𝑃𝐴=
2𝑝
3𝑝 (𝑝𝑟𝑜𝑝 𝑡ℎ𝑚, 𝐵𝐶 ∥ 𝐷𝐴)✓ a statement and reason
∴ 𝐶𝐴 =8𝑘𝑝
3𝑝
=8𝑘
3✓ a
∴𝐵𝐷
𝐷𝑄=
𝐴𝐶
𝐴𝑄 (𝑝𝑟𝑜𝑝 𝑡ℎ𝑚, 𝐵𝐶 ∥ 𝐷𝐴)✓ a
∴ =
8𝑘3
5𝑘
=8
15✓ a answer
(5)
R
B
P C Q A
D
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11.2.2 𝐷𝑟𝑎𝑤 ⊥ ℎ𝑡 𝑓𝑟𝑜𝑚 𝑅 𝑡𝑜 𝑃𝑄✓ a constr.
𝑇ℎ𝑖𝑠 ℎ𝑒𝑖𝑔ℎ𝑡 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛 𝑡𝑜 ∆𝑃𝑅𝐴 = ∆𝑄𝑅𝐴
∴𝐴𝑟𝑒𝑎 ∆𝑃𝑅𝐴
𝐴𝑟𝑒𝑎 ∆𝑄𝑅𝐴=
𝑃𝐴
𝐴𝑄 (𝑠𝑎𝑚𝑒 ℎ𝑒𝑖𝑔ℎ𝑡𝑠)✓ a statement and reason
=4𝑘
5𝑘
=4
5✓ a answer
(3)
EXTRA WRITING SPACE
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WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019
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QUESTION 12
12.1 𝐵2 = 𝑥 ✓ (𝑂𝐵 = 𝑂𝐸 𝑟𝑎𝑑𝑖𝑖, ∠′𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠, ∠′𝑠 𝑖𝑛 ∆𝑂𝐵𝐸)✓ a statement and reason
𝐵4 = 𝑥✓ (𝑡𝑎𝑛 − 𝑐ℎ𝑜𝑟𝑑 𝑡ℎ𝑚)✓ a statement and reason
�̂� = 𝑥 ✓ (𝑐𝑜𝑟𝑟𝑒𝑠𝑝. ∠′𝑠, 𝐵𝐷 ∥ 𝐴𝑂)✓ a statement and reason
(6)
12.2 𝐵1 = 90 − 𝑥 (∠′𝑠 𝑜𝑛 𝑠𝑡𝑟 𝑙𝑖𝑛𝑒)✓ a statement and reason
∴ 𝐹2 = 90° (∠′𝑠 𝑖𝑛 ∆𝐵𝐴𝐹)✓ a statement and reason
∴ 𝑂𝐹 ⊥ 𝐵𝐸
∴ 𝐹 𝑖𝑠 𝑎 𝑚𝑖𝑑𝑝𝑡 (𝑙𝑖𝑛𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑒 ⊥ 𝑡𝑜 𝑐ℎ𝑜𝑟𝑑 𝑏𝑖𝑠𝑒𝑐𝑡𝑠 𝑐ℎ𝑜𝑟𝑑)✓ a reason
(3)
12.3 𝐼𝑛 ∆𝐶𝐵𝐷 𝑎𝑛𝑑 ∆𝐶𝐸𝐵:
1) 𝐵4 = 𝐸 = 𝑥 ✓ (𝑝𝑟𝑜𝑣𝑒𝑛 𝑖𝑛 1.2.1) a statement and reason
2)�̂� 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛✓ a statement and reason
3)𝐵�̂�𝐶 = 𝐸�̂�𝐶 (𝑠𝑢𝑚 𝑜𝑓 ∠′𝑠 𝑖𝑛 ∆)
∴ ∆𝐶𝐵𝐷‖|∆𝐶𝐸𝐵 (𝐴𝐴𝐴) ✓ a reason
Penalty of 1 if no 3rd ∠
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WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019
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(3)
12.4 𝐼𝑛 ∆𝐵𝐹𝐴 𝑎𝑛𝑑 ∆𝑂𝐹𝐵:
1)�̂�2 = �̂�1 (𝑏𝑜𝑡ℎ 90°, ∠′𝑠 𝑜𝑛 𝑠𝑡𝑟 𝑙𝑖𝑛𝑒)✓ a statement and reason
2)�̂� = �̂�2 (𝑝𝑟𝑜𝑣𝑒𝑛 𝑖𝑛 1.2.1)✓ a statement and reason
3)𝐵1 = 𝑂2 (𝑠𝑢𝑚 𝑜𝑓 ∠′𝑠 𝑖𝑛 ∆)
∴ ∆𝐵𝐹𝐴‖|∆𝑂𝐹𝐵 (𝐴𝐴𝐴)✓ a reason 𝐴𝐹
𝐵𝐹=
𝐵𝐹
𝑂𝐹 (𝑐𝑜𝑟 𝑠𝑖𝑑𝑒𝑠 𝑖𝑛 𝑝𝑟𝑜𝑝)✓ a statement and reason
𝐵𝐹2 = 𝐴𝐹. 𝑂𝐹
(1
2𝐵𝐸)
2
= 𝐴𝐹. 𝑂𝐹✓ a subst.
𝐵𝐸2 = 4𝐴𝐹. 𝑂𝐹
(5)
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WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019
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EXTRA WRITING SPACE
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