WYNBERG BOYS HIGH SCHOOL SUBJECT: MATHEMATICS … CAPS 2019 Prelim Papers... · 2020. 8. 29. · WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019 P a g e 1 | 20

WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019

P a g e 1 | 20

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


P a g e 2 | 20

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)


P a g e 3 | 20

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


P a g e 4 | 20

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


P a g e 5 | 20

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


P a g e 6 | 20

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


P a g e 7 | 20

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)


P a g e 8 | 20

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


P a g e 9 | 20

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)

EXTRA WRITING SPACE


P a g e 10 | 20

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


P a g e 11 | 20

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


P a g e 12 | 20

(4)

EXTRA WRITING SPACE


P a g e 13 | 20

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


P a g e 14 | 20

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


P a g e 15 | 20

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)


P a g e 16 | 20

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


P a g e 17 | 20

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


P a g e 18 | 20

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 ∠


P a g e 19 | 20

(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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WYNBERG BOYS HIGH SCHOOL SUBJECT: MATHEMATICS … CAPS 2019 Prelim Papers... · 2020. 8. 29. · WBHS Mathematics Exam Answer Booklet Grade 12 Paper 2 September 2019 P a g e 1 | 20