# Mathematics Grade 12 Paper 2 MP Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 2 question paper and memo from the MP Mock Exam 2023, 32 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-mp-mock-exam-question-paper-and-memorandum-paper-2/59c096e1-f7de-4b3a-b565-056ebc032e2a
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: MP Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 32

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## About this document

Mathematics · Grade 12 · MP Mock Exam · 2023. Question paper and memorandum, 32 pages. Read online or download the PDF.

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### Page 1

education
MPUMALANGA PROVINCE

NATIONAL
SENIOR CERTIFICATE

MARKS: 150
TIME: 3 hours

This question paper consists of 13 pages and 1 information sheet

and an answer book is provided.

### Page 2

Mathelaaevanm@adiatr ‘om Stanmor ephysics.com Mpumalanga/Preparatory 2023
CAPS/KABV ~ Grade/Graad 12

INSTRUCTIONS AND INFORMATION
Read the following-instructions carefully before answering the questions.

1. The question|paper consists of 10 questions.

2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

3. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining the answers.
4. Answers only will NOT necessarily be awarded full marks.

5. You may use an approved scientific calculator (non-programmable and non-graphical) unless stated
otherwise.

6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise.
7. Diagrams are NOT necessarily drawn to scale.

8. Write neatly and legibly.

### Page 3

MathernaticR OMIAAGED from Stanmor ep hysics.com Mpumalanga/Preparatory 2023

CAPS/KABV ~ Grade/Graad 12

QUESTION 1

The table below shows the masses (in kg) of 15 randomly chosen weight lifters of a certain

gymnasium. . ira >
danny
11 | 83 | 88 oa! 92 | 92 | 95 | 97 | 104 | 108 | 109] 110 | 11 | 115] 129
TaTey
~
L.1 Calculate the mean mass of the weight lifters. (2)
1.2 Calculate the standard deviation of the masses of the weight lifters. (1)
1.3 What percentage of weight lifters fall in the feather weight division, if the criteria is
that your mass must be below one standard deviation below the mean? (3)
1.4 If all the weight lifters loose 3kg, what will be the new
1.4.1 Mean mass of the weight lifters? (1)
1.4.2 Standard deviation of their masses? (1)

[8]

### Page 4

maneheeMaleaded.from Stanmorephysics.com

CAPS/KABV ~ Grade/Graad 12

QUESTION 2

Mpumalanga/Preparatory 2023

2.1 The box-and-whisker plots show the wages of workers (in Rand) at two companies for
the same-type of work. Both companies have 20 workers.

Company A

Company B

200

300

400

(Wages in Rand)

500

600

2.1.1 State whether the following statement is TRUE or FALSE:
All the workers at Company A earn more than 25% of the workers at

700

800

Company B. qa)
2.1.2 Comment on the skewness of the data for company B. ()
2.1.3 Which company has the biggest range? Motivate your answer with the necessary (2)
calculations.
2.1.4 How many workers at Company B earn more than R200? (2)
2.2 A table of data, showing the price of crude oil at the end of each year in US Dollars ($)
(to the nearest dollar) per barrel, is given.
Year | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 | 2019 | 2020 | 2021
Price 90 92 98 54 37 54 60 61 49 75
2.2.1 Determine the equation of the least squares regression linesfor the price of crude
oil per year (3)
2.2.2 Calculate the value of the correlation coefficient. ()
2.2.3 Use calculations to predict the price of crude oil (in US $) per barrel at the end
of 2018. Discuss the validity of the answer that you obtained by using your
answer in 2.2.2. (2)

[12]

### Page 5

ManemaidoeMnlegded from Stanmorephysics.com

CAPS/KABV ~ Grade/Graad 12

Mpumalanga/Preparatory 2023

QUESTION 3

In the diagram A(2 ; 6), B(11; 1) and C(-1; —3) are the vertices of A ABC.
Point D is shown in thediagram such that BD | BC. N is the x-intercept of BC. AON =86.

D
ty
A(2; 6)
B11; 1)
8 >
N T «x
C(-1; -3)
3.1 If the gradient of BC is ; and the gradient of AC is 3, calculate
3.1.1 the x-coordinate of N, (2)
3.12 the size of ACB. (5)
3.2 Determine the equation of AC. (2)
If it is further given that point D lies on AC produced such that BC _L BDI calculate
3.3 (5)
the coordinates of D. A000]

[14]

### Page 6

M sine WORMED from Stanmorephysics.com Mpumalanga/Preparatory 2023
CAPS/KABV - Grade/Graad 12

QUESTION 4

In the diagram A(-3; 11) and C(1; 3) are points on the circumference of a circle with diameter AB and
centre T. The equation of AB is given by y= 3x+20.

ay

A(-3; 11

Ne (1; 3)

wa Oo *

B

41 Determine the equation of the perpendicular bisector of AC. 4
42 Show that the coordinates of the centre of the circle are (—5 ; 5). (6)
43 Calculate the length of diameter AB. (3)
44 Write down the equation of the circle. (2)
45 The tangent to the circle at A cuts the y-axis at (0 ; p). Calculate the numerical value of

p. (4)
4.6 If the circle through A, B and C is moved 3 units to the right and 2 units upwards, and

the radius is halved, write down the equation of the new circle. (3)
4.7 A new circle with equation (x—2)?+ (y-3 =4 and centre P is given. Will this circle

intersect the original circle or not? Motivate your answer_with the necessary (4)

calculations.

[23]

### Page 7

Mathemadgownieaded fr ‘om Stanmorephysics -com Mpumalanga/Preparatory 2023

CAPS/KABV — Grade/Graad 12

QUESTION 5

3
5.1 If @isa reflex: angle, and tan@ = — 4 , determine without the use of a calculator and

with the aidjofa sketch, the value of:

5.1.1 (2)
5.1.2 cos26 (3)
5.1.3 cos (8 + 30°) (3)
5.2 If x=4 sino and y=4 cosa, calculate the value of x? + y2. (2)
5.3 Simplify the expression to a single trigonometric ratio:
sin(900°~ x) . cos(—x) — sin(x—-180°).. sin (90° +x) (6)
Given the following the identity: intl Sia sin4x
5.4 2cos3x
5.4.1 Prove the identity. (4)
5.4.2 For which values is the identity above, undefined. Determine the general
solution of x for which the identity is undefined. QB)
5.5 Calculate the general solution of xif 2sin (3x+20°) = 2cosx (6)

[29]

### Page 8

Mather eueHAade dh from Stanmorephysics.com Mpumalanga/Preparatory 2023
CAPS/KABV ~ Grade/Graad 12

QUESTION 6

In the diagram below, the graphs of f(x) = 2 cos x and g(x) = cos 2x are drawn for the interval
x €[-45°;180°].

6.1 Write down the period of g. qd)
6.2 Write down the values of x for which the graph of f is increasing in the given interval. (2)
63 Write down the range of y=3g(x)-1. (2)
6.4 Determine the values of x for which f (x) 2 ; in the given interval. (4)
6.5 Determine the minimum value of cos" te 7 in the interval x ¢[0°; 180°] (4)

(13)

### Page 9

MathemaicdROWweHAn ded fr ‘om Stanmarephysics.com Mpumalanga/Preparatory 2023

CAPS/KABV ~ Grade/Graad 12

QUESTION 7

In the diagram AD is a3m vertical pole with support cables attached at T and A to a point R
which is x meters fro uf foot of the pole, D. D and R are in the same horizontal plane. T is 1m
above the foot of the ‘The angle of elevation of A from R is 6.

TRA=a iim

A,
2
T
1
6
D x R
Show that TR = > —
7 cos(O- @) @)
72 Provethat x= 2008@cos(0 - a) “
sin @&
73 If it is further given that 9 = 68,33° and @ = 28° calculate the
Area of AATR by using the formulas in 7.1 and 7.2. (4)

[10]

### Page 10

Math aMrlwadederr ‘om Stanmorephysics .com Mpumalanga/Preparatory 2023

CAPS/KABV — Grade/Graad 12

Give reasons for your statements and calculations in QUESTIONS 8, 9 and 10.

QUESTION 8

onny
8.1 In the diagram below, O is the centre of the circle ABC. AB and AC are chords. OB
and OGare joined.

Prove the theorem that states that BOC =2x BAC . (5)

### Page 11

Mathematicsowealeaded from Stanmor ephysics - COM Mpumalanga/Preparatory 2023

CAPS/KABV = Grade/Graad 12

8.2 O is the centre of circle CBE. DB is a tangent to the circle at B. EC produced meets
BD in D and intersects OB at F. CD = CB. OE and BE are joined. B, = x.

D

8.2.1 Name, with reasons, TWO other angles each equal to x. (4)
8.2.2 Express EOB in terms of x. 3)
8.2.3. Determine, with reasons, the size of Bo in terms of x. (2)
8.2.4 Show that DC =CF. QB)

{17]

### Page 12

manereRaadecns rom Stanmor ephysics -com Mpumalanga/Preparatory 2023

CAPS/KABV — Grade/Graad 12

QUESTION 9
In quadrilateral ABDC, diagonals AD and BC intersect at E.
AB =8cm Simm
BD=I5cm fp

40nn
DC=18cm fp
AC=10cm aw
BC=12cm J

A 8 B
15
10
Cc 18 D

9.1 Prove that ABCA ||| ACDB. (4)
9.2 Hence prove that AB || CD. (2)
9.3 Calculate the length of CE. (4)

[10]

### Page 13

Mathemaicdpowinloaded from Stanmorephysics.com Mpumalanga/Preparatory 2023

CAPS/KABV ~ Grade/Graad 12

QUESTION 10

In the diagram O is the centre of circle PSRT. TR produced intersects SV in V. ST bisects PTR
and TS = SV. TOP, OSvand SR are joined.
sonar

pe

10.1 Determine with reasons the size of PST . (2)
10.2 Determine the size of Sa. (5)
10.3 Prove that ATSO||| ATVS. (3)
10.4. Show that 2VS? = PT.TV (4)
[14]

/ TOTAL: 150

### Page 14

manirawmnaadedvfrom Stanmorephysics.com

Mpumalanga/Preparatory 2023

CAPS/KABV ~ Grade/Graad 12

INFORMATION SHEET
1 —b+ Vb’ -4ac
2a
Ina? 7
A= P(L+ni) Jappp A= P(1~ni) A=P(1-i) A= P(i+i)

aay

es

T,=a+(n

in n

$,= Ska +(n-Dd]

ia |
T, =ar"'! S eae) sre S$, = 3-1e<r<l

psi i] pail (1+i)"
F ,

ny — tim LEENA FO

F@)= fn, h

d=y(x,-x%,) +0, -yy

y=mxte

(x-a) +(y-8 =?
ab
sinA sinB

a’ =b? +c? —2be.cosA

__¢

In AABC: ——
sinC

area AABC = 5 absin (e

sin(a + #)=sin a.cos 8 + cosa.sin 8

cos(a + £)=cosa.cos 8 - sin a.sin B
cos’ a—sin’? @

cos2a =41-2sin’? a

2cos? a-1
yx
n

n(A)

n{S)

ypoathx

x

P(A) =

yy; = m(x- x)

A tX. Vit y2

2 2

y27Y
mat m=tan@
xy %

sin(a — 8) =sin a.cos B —cosa.sin B

cos(a — £)=cosa.cos + sin a.sin B

sin 2a = 2sina.cosa

n

(x, - XY —>
1

n th
On]

P(A or B) = P(A) + Peay P(Aand B)

p — t= FY W)
Gx

—

### Page 15

morephysics.com

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS PAPER 2

SEPTEMBER 2023

MARKING GUIDELINES

MARKS: 150

This marking guidelines consist of 18 pages:

NOTE:

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai aseblief om

### Page 16

mattbawntoaded from Stanmorephysics.com MPU/September 2023

CAPS/KABV — Grade/Graad 12 - Marking Guideline/Nasienriglyn

e  Ifacandidate answered a question TWICE, mark only the FIRST attempt.

e Ifacandidate crossed out an answer and did not redo it, mark the crossed-out answer.
« Consistent accuracy applies to ALL aspects of the marking guidelines.

e Assuming values/answers in order to solve a problem is unacceptable.

LET WEL:

As ‘'n kandidaat’n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.
As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde
antwoord na.

° = Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyn van toepassing.

e — Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los.

QUESTION/VRAAG 1

71 | 83 | 88 | 9L | 92 | 92 | 95 | 97 | 104 | 108 | 109 | 110) 111 } 115 | 129

1.1 1495
5) = 99,67 V¥ 99,67
(2)
1.2 o =14,06 ¥SD
(1)
1.3 99,67 —14,06 = 85,61 ¥ boundary
2
= 4100 =13,33% v=
15 15
v answer (3)
1.4.1 | 96,67 v answer
(1)
1.4.2 | 14,06 v answer
(1)
[8]

### Page 17

matbawntoaded from Stanmorephysics.com MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION/VRAAG 2
2.1.1 | True Y answer
()
2.1.2 | Positively-skewed / Skewed to the right Y answer
)
2.1.3 | Range for company A = 800-200 = 600
Range for company B = 600-100 = 500 Y both ranges
Biggest range : Company A ¥ Company A
(2)
2.1.4 | 20x 75% =15 workers V75%
¥ 15 workers
(2)
2.2.1 | y=5965,51-—2,93x ¥ 5329,84
v -2,61
Y equation
(3)
2.2.2 | r=-0,49 vy
)
2.2.3 | y = 5965,51—2,93(2018) v prediction
=52,77
The correlation coefficient is moderate thus the value is fairly reliably | Y answer + reason
predicted. (2)
[12]

### Page 18

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION/VRAAG 3
D
y
ar
ae A(2; 6)
B(11: 1)
0 .
T x
| —
C(-1;-3)
3.1.1) N:(x;0) Y subst into gradient
0-(-3) 1 formula
Mnc = ae
x-(-l) 3
x+1=9
x=8 Y answer (2)
OR

y-(3)= Fo 1)

¥ subst into equation

1,8
“3 3
For y=0:x=8 v answer
ria

5

l

i

### Page 19

matbowntoaded from Stanmorephysics.com MPU/September 2023
CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn
3.1.2 | tand =3 ¥ substitution
0=711,565... v0
tan BN ¥ substitution
B 8,434...° v BNT
C=71,565...°-18,434...°
= 53,13° v answer
Accept C= 53,14° (5)
3.2 | y-6=3(x-2) v substitution
y=3x Y equation (2)
3.30 | m, =-3 Ym, =-3
y-1=-3(x-11) y :
y=-3x434 equation
3x =-3x +34 . .
Y setting equations
6x = 34
equal
x= 7 5,67
“3° ¥ value of x

17
=3/—]=17
* (5

¥ value of y (5)

[14]

### Page 20

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION/VRAAG 4
y
A(-3:11)
D
C(1;3)
+
fe
B
4.1 m= 11-3 -8__, ¥ gradient of AC.
aC 3-1-4
¥ Midpt AC
mdpt yc =(-137) ep
y-T= 3(*-(-0) ¥ Lgradient and subst
ye l x+ IS Y equation
2 2? (4)
42 i. + 15 3x42 Y set equations equal
2 2
— 2,5x =12,5
x=—5 v solve x
y=3(-5)+20=5 Vsolve y 6)
a centre(—5;5)

43 radius = |(-5-(-3))" +(5-11)
= 40 = 210

diameter = 4/10 units

= 12,65

Y¥ substitution in formula
¥tadius

¥ diameter

(3)

44 | (x45) +(v-5) =40

Y (x +5), (7-5)
v 40 (2)

### Page 21

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

as |m -_lic5 _3

rad ~ (5)

1

Y gradient

¥ subst in form

Y¥ equation
Y value of p
(4)
2 v
46 | (x+2) +(y-7S= (Jo) ae
(x+2/ +(y-7) =10 Y10 (3)
47 Centres: (2; 3) and (—5; 5)
Distance between centres = ./(2 —(-5)? + (3-5) v subst of centres
¥ distance
= /53 =7,28
Sum of radii =2 + 2 J10 =8,32 ¥ Sum of radii
Distance between centres < Sum of radii
.. They will intesect. Y¥ conclusion (4)
[23]

### Page 22

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION 5

5.1

aN
y,

S11 | (3+ =r
2
= 25 v value of r
r=5
. - v answer
sin@ = = Q)
5.1.2 | cos20 =cos?6 — sin’0 ¥ expansion
= (=) - (=) ¥ substitution
5 5
_lo 9 v answer
25 25 (3)
ae
25
5.1.3 | cos(@+30°) = cos0cos30° — sinOsin30° ¥ expansion
-(3) 3 -(2\3) ¥ substitution
S5)\ 2 5 )\2
= 43 +3 v answer
10 (3)
5.2 (4sina)” +(4cosa) ae
, ¥ simplification
=l6sin’a+16cos’@ trararg
=16(sin? a +cos’ @) Janay
=16(1) =16 as (2)

### Page 23

matbewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

53 sin(900° - x) : cos(-x) - sin(x - 180°) : sin(90° + x)
=sin(180°— x) . cos(360° — x) — sin(180° + x) .sin(90°+ x) | ¥ sinx
= sin xcosx —(-sinx)cosx Vcosx
¢ > —smnx
¥ cosx
Y simplification
v double angle identity
‘ (6)
5.4.1 | sin7x+sinx .
————_ = sin4x
2cos3x Y both brackets
sin(4x+3x)+sin(4x—3x
LHS= ( ) ( ) v sin(4x+3x)
2cos3x .
_ sin4xcos3x + cos 4xsin 3x + sin 4x cos 3x — cos 4x sin 3x expansion
2cos3x ¥ sin(4x—3x)
_ sin4xcos3x+sin 1x cos 3x expansion
2cos3x . : :
_ 2sin4xcos3x ¥ simplification
2cos3x (4)
=sin4x =RHS
54.2 | 9 c0s3x=0 ¥ 2cos3x=0
cos3x=0
90°+360% v3 -{% +300 oe kez
= or kez _909° 0
90° + 360°k 90° + 360°
re 30° +120°h _ [30°+120°
30° +120°% ~)_39°4.120°%
OR oR (3)
2cos3x=0 ¥ 2cos3x=0
cos3x=0 90°+360°%
‘0 0 = or kez
_ {90° #360% 270° +360°k
270° + 360°%
30° + 120°% 30° +120°%
x= Vx=
90° + 120° 90+ 120°K

### Page 24

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

5.5

sin (3x + 20°) =cosx
sin (3x + 20°) = sin(90°— x)
3x+20°=90°-x+360° kez
4x = 7054360°K
x =17,524/ 90°
OR
3x + 20° = 180° —(90°- x) +360°%% kez
3x + 20° = 180° —90° + x + 360°
2x = 70° + 360°k
x =35°+180°%
OR

sin(3x + 20°) =cosx
cos[90° — (3x + 20°)] =cosx
—3x+70°=x+360°h kez

—4x =-70° + 360°

x =17,5°+90°%
OR

—3x+ 70° =-x+ 360° kez

—2x =—70° + 360°

x =35°+180°K

¥ sin(90-x)
¥ 3x+ 20° =90°-x+360°% kez

¥ x =17,5°+90°K

¥ 3x+ 20° =180°—(90°- x) +360°K
¥ 2x = 70° + 360°K
¥ x=35°+180°%
(6)
¥ cos[90° — (3x + 20°)]
¥ Ax =-70° + 360°
¥ x =17,5°+90°K
¥-3x + 70° =-x+300°% kez

v¥—2x =—70° +360°%%
vv x=35°+180°%

[29]

### Page 25

matbowntoaded vfrom Stanmorephysics.com

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

MPU/September 2023

QUESTION 6

6.1 180° ¥ answer
(1)
6.2 Y critical values
¥ notation
(2)
6.3 Range of g: y e[-I; 1]
Range of 3g: y €[-3; 3] V critical values
Range of y=3g(x)-l: 4S y<2 or ye|-+4;2] ¥ notation Q)
6.4 2 1 ¥ equation
cosx=—
2
1
cosx= 7 ¥ x= 75,5°
x =75,5°
° ° ¥ critical values
~45° <x S75,5 ¥ notation (4)
65 1, 1
=cos* x — —
2
= * (2cos? x-1)
4 ut
= dco 2x) lv double angle
4 KE
~ ri ) ¥ min of cos2x
1
“4 v answer (4)

### Page 26

matbowntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

| | [13] |
QUESTION 7
1 TRD=0-a ;
% ¥TRD
cos(0-a) =—
TR ¥ cos definition
x
TR = ————_
cos(8-@) (2)
72 A
A=90°-0
TR _ AT
sin(90°-6) sina Vx
TR 2
cos@ sing ¥ TR
TR = 20088
sm a@
x _ 2cos8 ¥ Subst in area-rule
cos(0-a) sina
_ 2cosAcos(0-a) ¥ answer
* sing (4)

### Page 27

mattDewntoaded vfrom Stanmorephysics.com MPU/September 2023

CAPS/KABV — Grade/Graad 12 - Marking Guideline/Nasienriglyn

73 a 2.cos68,33cos(68,33 — 28°)
7 sin 28°
vx
Vv TR
ATR = 130,33°
Area AATR = 320145730) sin130,33° v subst in area formula
=1,1992... ¥ answer (4)
= 1, 20units?
[10]

### Page 28

mattDewntoaded vfrom Stanmorephysics.com

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION 8

MPU/September 2023

8.1 Draw line from A through O to T
let Ai =x and A> =y

Ai =B =x (2's opp = sides)
Oo: =2x (ext Z of A)

similarly O2 =2y

BOC =2x+2y=2(x+y)

“BOC =2xBAC

¥ construction

vS/R

v SIR

vs

¥ deduction

(6)

### Page 29

matbewntoaded vfrom Stanmorephysics.com MPU/September 2023
CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn
8.2
a
fo
/
/
/
|
\
\
\
\
\
NN
A vS v
8.2.1 D=x (Z opp = sides) Ss vR
A VS v
Ez =x (tan-chord theorem) Ss vR (4)
2.2 A vSIR
8 C, =2x (extZ of A)
A A . vS
2C, =EOB=4x (Zatcentre=2x Z atcircumf ) VR @)
823 | OBD=90° (tan L rad) Vs
A VR
Bz =90°-x
(2)
824 | OBD=90° (tan 1 rad)
Aa A Vv
F2 =180°-x-OBD (sum Zs A) ve
F, =90°-
2 =90°-x SiR
”. BC = FC (opp 2's =) (3)
“. DC = BC = FC

[ [17]

### Page 30

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

QUESTION 9
A 8 B
E
10 12
Cc D
9.1 In ABCA and ACDB
Be _ 2 2 VS
cD 18 3
AB ft Sem2 vs
BC 12 3
AC _10_2 vs
DB 15 3
v
ABCA ||| ACDB (sides in proportion) sR 4)
A A v
9.2 ABC=BCD (ABCA ||| ACDB) s
-AB||CD. (alt Zs=) VR (2)
9.3 In AABE and ADCE:

ABC=BCD (proven)

AEB=CED (vertically opp Z)
AABE ||| ADCE (222)

AB BE ((lia's)
DC CE

8 x

18 12-x
8(12—x) = 18x

96 —8x =18x
96 = 26x

x = 3,69

CE = 8,31

V AABE ||| ADCE
vy AB _ BE
DC CE

wis __*
18 12-x

¥ CE=8,31

(4)

### Page 31

matbewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 — Marking Guideline/Nasienriglyn

| | [10) |
QUESTION 10
mM Vv
i
A ¥vS VR
10.1 PST =90° (Zinsemi-circle) Q)
A x v
10.2 T, =T2 (STis abisector) s
T> =V (Zopp=sides) vs
T=V
R =P (ext Z of cyclic quad) YS VR
Ss =PST =90°(sum Z of A) VS+R
(5)
10.3 ATSO ||| ATVS

Tet (STisa bisector)

S:=V= [om=F:

O1 =TSV (3rd ZA)
ATSO ||| ATVS (777)

(3)

### Page 32

mattDewntoaded vfrom Stanmorephysics.com

MPU/September 2023

CAPS/KABV — Grade/Graad 12 - Marking Guideline/Nasienriglyn

10.4 TS _OS

— =— _ ATSO||ATVS (2422

TV VS ll ( ) vs

TS.VS=OS.TV YS

but VS given

and SPT both radii ¥S+R

1 VSVS= SPIT Y substitution

©. 2VS? = PT.TV (4)

[14]
TOTAL: 150

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