# Mathematics Grade 12 Paper 2 KwaZulu Natal Mock 2022 Memo

> Grade 12 Mathematics Paper 2 memorandum from the KwaZulu Natal Mock Exam 2022, 30 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-grade-12-sept-2022-p2-and-memo/5b8825db-c21d-4819-91c6-7a011a04d713
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: KwaZulu Natal Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 30

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### Page 1

Naumlac ‘ ‘ fram ‘ , > ww

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P2

PREPARATORY EXAMINATION

SEPTEMBER 2022

3 hours

This question paper consists of 13 pages, 1 information sheet and
an answer book with 22 pages.

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Mathematics/P2

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NSC —- GRADE 12

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1.

2.

10.

11.

This question paper consists of 11 questions.

Read the questions carefully.

Answer ALL the questions.

Number your answers exactly as the questions are numbered.

Clearly show ALL calculations, diagrams, graphs, etc. which you have used in detente

your answers. (@)
G

Answers only will NOT necessarily be awarded full marks. . 2”

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

If necessary, round off answers correct to TWO decimal cess stated otherwise.
Diagrams are NOT necessarily drawn to scale. &

An information sheet with formulae is included and of the question paper.
Write neatly and legibly. e?

K\
fe)
&
ww
ee

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Mathematics/p2 . “com
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QUESTION 1

The weight (in kg) of 20 boys in the soccer squad of school A are given below:

ll

1.2

1.3

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40 47 48 31 53 57 58 458 59 59

60 60 60 60 61 62 63 64 66 69

Calculate:
Ld. the mean Weight of the boys in this soccer squad. &
1.1.2 the standard deviation of this data, 9

Determine the number of boys that have a Weight within one sti deviation of

the mean. Y)
The following information was obtained from the coh i soccer squad of

school B:
2 ‘
> x, = 1320?
n=l Ss

1.3.1 How many boys are in the 1B squad?

1.3.2 Calculate the mean yee of a boy in the soccer Squad of school B.

Assume that the Mean weight Of the boys in the soccer Squad at school B is 60 kg.
Five boys of equal wi a € added to the school A Squad so that the means of both
school squads are thes e. Calculate the weight of each of these five boys.

Or
Oo
x

ss

KZN Preparatory Examination 2022

Q)

(1)

(2)

()

(D

(4)
(11)

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D
Mathematics/P2 ownloaded from Stanmorephysics.com KZN Preparatory Examination 202?
NSC - GRA EI -

QUESTION 2
A survey was done on 250 people to determine the distances they travel to work daily.

The results are shown in the table below.

CUMULATIVE
DISTANCE, d (in km) FREQUENCY FREQUENCY,
O<dsi 8
p< dso} eo
10<dsi5 63
ee | oe el
15<d<20 52
ee rr ee
20<d<25 4!
25<d 530 38
25 <a SW ———=—
30<d $35 7
TOTAL
2.1 Complete the cumulative frequency column, on the attached ANSWER SHEET. (3)
2.2 Draw a cumulative frequency graph (ogive) for the given data on the grid provided on
the attached ANSWER SHEET. (4)
23 Use the graph to determine the median distance travelled. Indicate on your graph the
median distance. (2)

19]

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5

Mathematics/P2 KZN Preparatory Examination 2022

NSC — GRADE 12
QUESTION 3

In the diagram below, A(-5 ; 2), B(-3 ; 8) and C are vertices of AABC.
E(1 ; 0) is the midpoint of AC. D isa point on the x-axis such that AD is a line perpendicular

to the x-axis. a is the angle of inclination of AC.

B(3:8) y
M
A(5;2)4)8
Pw x
D [e} E(1 ; 0)
Cc

3.1 Determine the coordinates of M , the midpoint of AB. (2)
3.2 Write down the coordinates of point D. (1)
3.3 Show that the coordinates of C are (7; -2) (2)
3.4 Calculate the length of line AC. (Leave answer in simplest surd form) (2)
3.5 Determine the coordinates of F , if F lies in the first quadrant and CABF isa

parallelogram. (2)
3.6 Determine the equation of the perpendicular bisector of AB. (4)
3.7 Calculate the size of @ , the angle of inclination of line AC. (3)
3.8 Determine the equation of the line parallel to AB passing through E. (2)
3.9 Calculate the size of angle @ . (2)
3.10 Calculate the area of AABC. (4)

[24]

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Mathematics/P2 6 KZN Preparatory Examination 2022
NSC — GRADE 12

QUESTION 4

In the diagram, the circle TS centred at C(3;~1) has aradius CT of 10 units. PTR, where
R(k ; 21), is a tangent to the circle at T. PS is a tangent to the circle at S and PS || x-axis.
PC, TC and CR are drawn. TR = 20 units.

y R (&; 21)
20
T,
10
x
(6) DY
CB;-l
P S i
4.1 Give areason why CT LTR. qd)
42 Calculate the value of & , where R is in the first quadrant. (4)
43 Write down the equation of the given circle. (2)
44 Write down the equation of PS. qd)
4.5 The equation of tangent PTR is 3y=4x+35.
4.5.1 Calculate the coordinates of P. (2)
4.5.2 Calculate the length of PT. (3)
[13]

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NSC — GRADE 12

KZN Preparatory Examination 2022

QUESTION 5

5.1 If 5cosA= 26 where Ae [90°;360°] , calculate, without using a calculator and

with the aid of a diagram, the values in simplest form of:

5.1.1 V6. tanA (4)
5.1.2 sin2A (4)
5.2 Given: sin18°= p

Without using a calculator, determine each of the following in terms of p.

3.2.1 cos 18° (2)
5.2.3 cos 48° (5)
5.2.3 sin 9° (3)
[18]
QUESTION 6
6.1 Without using a calculator, simplify the following expression fully:

sin(180° — x).tan(x-180°).cos(360° +x)
sin?(180°+ x) + sin?(90°— x) (6)

6.2 Without using a calculator, determine the value of:

cos 330°. tan 150°.sin 12°

tan 675°. cos 258° (7)
6.3 Given the identity: cosa + 60820 = Ose +
sin2a —-sina sing
6.3.1 Prove the identity. (4)
6.3.2 For which other values of @ is the identity undefined? (5)
[22]

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Mathematics/P2 8 KZN Preparatory Examination 2022

NSC —- GRADE 12
QUESTION 7
Given: f(x) =sin2x and g(x) = cos(x+a) where x <[—180" ; 180°]

The graphs of f and g intersectat B and D. E is the y-intercept of g, and C isa turning
point of g. A isa turning point of both f and g.

| Ags)
E
r
x
-180° Hy? fe) ny 180°
D B (165°; ~ %)
g
c |
al Waite down the value of a. (1)
72 State the period of (1)
73 Determine the coordinates of C and E. GB)
74 Write down the amplitude of A if h(x) =3 f(x). ()
75 Determine for which value(s) of x , if xe[0°;180°], will:
7.5.1 g(x) > f(x) 2)
75.2 g(x). f'®) 20 (2)
7.6 Without solving the equation, use the above graphs to show how you would solve the

following equation:

(3)
[13]

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V2sin2x = cosx + sinx

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Mathematics/P2 KZN Preparatory Examination 2022

NSC — GRADE 12

QUESTION 8
8.1 Complete the following statement:
The line drawn from the centre of the circle perpendicular to the chord ......... (ld)
8.2 The circle below with centre O has chord AB = 8 cm.

OMT | AB with MT=2 cm. The radius of the circle is r cm.

O
r
cq
A M B
T
8.2.1 Write down, with a reason, the value of AM. (2)
8.2.2 Calculate the length of the radius of the circle. (4)

(7]

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Mathematics/P2 10 KZN Preparatory Examination 2022
NSC - GRADE 12

QUESTION 9

9.1 In the diagram below, O is the centre of the circle. P, A and T are points on the
circumference of the circle. PA, TA, PO and TO are drawn.

T

Prove the theorem which states that POT = 2PAT. (5)

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Mathematics/P2 KZN Preparatory Examination 2022
NSC — GRADE 12

9.2 AOB and COB are diameters of circle ACDBE with centre O.
Chords AC, CB, AE, AD and DE are drawn. D=50°.

D

9.21 Calculate, with reasons, the size of the following angles:
(a) 0, (2)
(b)  E, @)
9.2.2 Prove that AE || CB. (4)

[14]

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Mathematics/P2 12

KZN Preparatory Examination 2022
NSC - GRADE 12

QUESTION 10

In the diagram DRS is a tangent to the circle TMAR at R. AT bisects MTR.
AT intersects MR at P. AR is drawn. T, =x.

D
R
A
S
Li
T
M
10.1 Prove, giving reasons, that:
10.1.1 R,=R,. (4)
10.1.2 AAPR ||| AMPT . (3)
3 PT

10.2 If AR= 3MT , then calculate the value of PR” (3)
[10]

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Mathematics/P2 KZN Preparatory Examination 2022

NSC — GRADE 12
QUESTION 11

In the diagram below, AABC has D and E on BC. BD =6cmand DC=9cm.
AT:TC=2:1 and AD | TE.

A
T
B D E C
. . CE
11.1 Write down the numerical value of ED’ qd)
11.2 Show that D is the midpoint of BE. Q)
11.3 If FD =2 cm, calculate the length of TE. (2)
11.4 Calculate the numerical value of :
Wad Areaof AADC 7
_ Area of AABD (2)
142 Areaof ATEC 3
_ Area of AABC (3)

19]

TOTAL: 150

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Mathematics/P2 KZN Preparatory Examination 2022
NSC - GRADE 12

INFORMATION SHEET: MATHEMATICS

_ —b+\b" —4ac

2a
A=P(1+ni) A=P(1-ni) A=P(-i"

T, =a+(n—ld S, =5 (2a+(n—Na)}

alr” -1
T, = ar" s, <4 De S.2 os -l<rsl
r- -r

x{(1+i)’ -1] > x[l-a+i"]

i i

19) = jim LE =)

X, +X, VY+typ
d= V(x, -x,)? +0, -y)* ea m = tan0

227M yy

yemxte YY =mM(x-%)
Xy

mee

In AABC: -~___¢
sinA sinB sinC

a? =h? +e7 —2bc.cosA
area AA BC =Fab.sinc

sin(a+B)=sina.cosB +cosa.sinB sin(a—f) =sin a.cosB —cosa.sinB

cos(a+f)=cosa.cosB—sino.sinB cos(a—B) = cosa. cosB +sin a.sinB

cos’ a—sin? «

cos 2a =41—2sin’? a sin 20 = 2sina.coso
2cos? a—1

y= fi »_Elan3)’
=a o = ——___

n n
pcay = 2) P(A or B)=P(A)+P(B) —P(A and B

=) or B)= P(A)+P(B) —P(A and B)
pease p= X= 9)

Sons

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 150

TIME: 3 hours

These marking guideline consists of 15 pages.

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NOTE:

NSC — Marking Guidelines

If a candidate answered a QUESTION TWICE, mark only the FIRST attempt.

If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.
Consistent accuracy applies to ALL aspects of the marking guidelines.

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

GEOMETRY
A mark for a correct statement
Ss (A statement mark is independent of a reason.)
R A mark for a correct reason

(A reason mark may only be awarded if the statement is correct.)

S/R Award a mark if the statement AND reason are both correct.
QUESTION 1
40 47 48 51 53 57 58 58 59 59
60 60 60 60 61 62 63 64 66 69
1.1.1 — 1155 AY 1155
xX =—
20 Answer only: full marks
X =57,75 kg CAY answer (2)
Penalty 1 mark for rounding
here for the entire paper
1.1.2. | o =6,73702 ~ 6,74 kg AY answer (1)
1.2 (X-o;X+o)
(57,75-6,74 ; 57,75+6,74)
limit = (51,01 ; 64,49) CAV’ interval 2)
14 boys CAY answer
1.3.1 | 22 AY answer (1)
13.2 | _ 1320
xX =—
22
X = 60 kg CAY based on 1.3.1 (1)
1.4
ya ok t1I55 _ ey CAY 5x+1155
25 25
5x +1155 =1500 CAY equation
5x =345 CAV’ simplification
=69 k
x g CAY answer (4)
(11]

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NSC — Marking Guidelines

QUESTION 2
2.1
ic ‘ CUMULATIVE
DISTANCE, d(inkm) | FREQUENCY FREQUENCY
O<d<5 8 8
v
5<d<=10 41 49 Av Se®
l0<d<I5 63 112
CAY
i5<d=20 52 164 ae
20<d<25 41 205
25<d<30 38 243 cav 205,243 and
30<d <35 7 250 20
TOTAL 250
(3)
2.2
CUMULATIVE FREQUENCY GRAPH
(OGIVE)
250 =r
225
sa AY — grounded
175 at (0 ; 0)
CAY cumulative
f 150 frequencies
= as y, for
2 y-coords
z 100 CAY Sother —
50 AY
25 4
¥
0 $ 10 isA 30 2s 30 35
DISTANCE (in kms)
(4)
2.3 See ogive for dotted lines and point marked A. CAV indication
Median = 16 km on graph
CAY approx. (2)
OR median
Answer only: Full marks (accept 15,16,17)
[9]

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NSC — Marking Guidelines

QUESTION 3
B(-3:8) y
M
A(-5;2)Q)0
Py x
D oO E(1 ; 0)
Cc
3.1 M x +% . Yot YM
2° 2
Vv woe
5-3 248 A Substitution of A
M| —— ;—— and B into midpoint
2 2 Answer only: full marks
formula
M(-4;5) CAY answer (2)
3.2 D(-5;0) AY answer (1)
3.3 5+ Xo 4 AY — midpoint ITO x
2 AY midpoint ITO y
WX =7
C(7;-2) (2)
OR
v ~
C(1 +6; 0-2) Using transformations A x-value
C7 :-2) Av _ y-value (2)
3.4 _ 2 2
AC = (x) - x) +02 -Y)
as 5 : - AY - substitute A and C
AC = y(-5-7)" +(2+2) into distance formula
AC =4V10 CAY answer (2)

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NSC — Marking Guidelines

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3.5 Method: translation
AB: (x;y) > (x+2; y +6)
.. by symmetry: C > F: C(7;-2) > F(7+2;-2 +6) AY — x-coordinate
“. FQ; 4) Av _ y-coordinate (2)
OR OR
Midpoint of intersection of diagonals = T (2; 3)
Let coordinates of F be (a ; b)
a-5 b+2
7 = 2 and =3
a=9 andb=4
-. F934) AY — x-coordinate
AY _ y-coordinate (2)
_ Vv =
3.6 mAB= 23 A mAB =3
-34+5
1
.. gradient of perpendicular bisector: “3 1
1 CAY Myerp bisector =>
Ww ys-ixte
73
b(-4;5): 5=-2(-4)+
sub (-4;5): 5 = 3° )+e CAY substitution
Lu
3
. . dicular bi 1 1 CAY equation
“. equation of perpendicular bisector: y = ~3*t 3 (4)
v
3.7 mAC = 24+2 _1 A mAC
5-7 3
1
tana=—) CAY tana@=-=
afl
q@=tan “3 +180°
a =161,57° CAY answer
“a 162° (3)
OR
Accept method using Cosine Rule
3.8 MAB =3
. gradient of new line = 3 (/ limes =gradients ) CAY equal gradients
y =3x +c
sub E(1;0):0=3(1) +c
=-3
y=3x-3 CAY answer (2)

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NSC — Marking Guidelines

3.9 _ 1
mAB x mAC =3 x 3 AY mABxmAC=~<1
“MAB x mAC =-1
“ ABLAC
“0 =90° AY answer (2)
Accept methods using Cosine Rule & Trig ratios
3.10 | AC =4¥10 units
AB= [(-3+5) +(8—2) = 210 units AY © substitute A andB
into distance formula
1 CAY length of AB
“. Area of AABC = 3 AC.AB
. Area of AABC = 5 (4V10) (200) CAY substitution
. Area of AABC = 40 units’ CAY answer (4)
[24]

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NSC — Marking Guidelines

QUESTION 4
y R(k; 21)
20
T,
10
x
S/
C331)
P Ss
41 Radius is perpendicular to tangent AY — reason ()
42 | cr= io’ +20°=V500 —_ (pythag) Avy 500
CR = (k—3)° + (2141? = J(k-3)? +484
2.4 (k-3) + 484 = /500 Av equate
.. (k—3)? +484 =500
. 2
“(k=3y =16 CAv simplification
wk-3=44
“k=-1 or k=7 CAV both values and
but k #-1 (given that R is the first quadrant) rejection of k =—1
k =7 only
(4)
43° | (x-3)' +(y+1) =100 AY (x-3) +(y+1)’
2
AY 100 (2)
44 | S@;-1D
wy=-ll AY answer (1)
45.1 | y=-ll Eq1
3y=4x+35 Eq 2
Sub Eq 1 into Eq2: 3(-1) = 4x 435 CAv‘ substitution
4x = -68
xeol? CAY x-value
“.P(-17;-11) (2)

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NSC — Marking Guidelines

45.2) pc=f(3417)? +(-1411)? =10V5 cAY PC? =500
TC= 10 PC=10V5
PT? = PC? -TC? (pythag)
CAY Pythagoras
pT =4/(tovs) —(10)'
PT = 20 units CAY answer (3)
OR
PT=PS (tangents from common point) Av SIR
PS = 3—(-17)= 20units CAY Substitution
“PT = 20 units CAY answer
(3)
[13]
QUESTION 5
5.1.1
y
A IG
las 2/6
SN Lo AY — diagram
i]
I
5 il
I
|
y= Ir? -x
2
y=\(5) -(2V6) =1 AY y=-1
“ay=-l (quadrant 4)
-V6. tanA
-1
~~ Gx{ =e CAY tana = 206
=i CAY answer
(4)

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NSC — Marking Guidelines

5.1.2 | sin2A = 2sinA.cosA Av double angle identity
= (2 26 CAY sinA=—
5 5
4/6 CAY  cosA=—— ws
25 CAY answer
(4)
5.2
72°
1
P
CT 18°
1-p?
5.2.1 | oosig° =f, P AY — diagram
AY — answer (2)
Answer only full marks
5.2.2 | cos48° =cos(30° +18°) AY (30°+18°)
= cos 30° cos18°—sin 30°sin18° AY — expansion
AY special angle
3\( fi-p?) (1
(24210) substitution
CACAY “substitution of
VB./1- p? - p sin18° and cos18°
So
_\¥3-3p" -P
2 (5)
5.2.3 | cosl8° =1-2sin’9° Av double angle
—2sin? 9° = cos18°-1 expansion
sin? 9° _ cos18°-1 -1
Av making sin9° the
sin9° = ar =1 subject
>. |¥l-p?-1 1-,1- p°
sin9 _jee =) = | 2 CAY answer (3)
[18]

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NSC — Marking Guidelines

QUESTION 6

6.1

sin(180° — x). tan(x —180°).cos(360° + x)

sin’ (180° + x) + sin?(90°— x) AY - sinx
: AY tan x
sin x. tan x.COs X i
= AY — sin’x
sin’ x+ Cos x
. AY cos’x
=sinx. sink cos X sinx
cos x" AY tanx=——
a cos x
= sin’ x CAY answer (6)
6.2 cos 330°. tan150°.sin12°
tan 675°.cos 258°
a
_ (cos30°).(— tan 30°).(sin 12°) Ay cos 30
tan 45°).(—cos 78°) Ay ~tan30
30°) (tan 30°),(sin12° AY — coratio
__ (cos 30°).(1 an Dsin ) Av —cos78°
(tan 45°),(sin 12°) AY tands°
3/8
2 3
——1 AY - special angles
1 CAY answer
2 %)
6.3.1 LHS: cosa + cos 2a
sin2a-sina V doubl 1
A cos double angle
cosa@ +(2cos’ a -1
= Losas (2cos' aI) AY - sin double angle
(2sina@cosa)—sina
_ 2cos* @ + cosa@—1
2sina cosa —sina
(2cosa—1)(cosa +1) AY numerator factors
= sin a(2cos@—1) AY — denominator factors
_ (cos @ +1)
sing
= RHS (4)
6.3.2 | sin2a—sina=0 AY — equating to 0
2sinacosa—sina=0
sina (2cosa—1)=0 AY factors
- sina =0or 2cosa—1=0 CAY 0°+k.180
cosa == CAY +60°+k.360
a=0°+k.180;k EZ
a=+60°+k.360;k EZ AY kez
N.B. If sina = 0 is solved only — max. 3/5 marks (5)
[22]

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Mathenkakawwdoaded from Stanmore ephysics. CPi8paratory Examination September 2022
NSC — Marking Guidelines

QUESTION 7
: Yo A(45*1)
E
f
x
H 180)
B (165°; — 4)
71 a=-45° AY answer (1)
7.2 180° AY answer qd)
7.3 C(-135°;-1) E: — y=cos(0°+45°) AY C(-135°;-1)
v2
= AVY © solving for y
V2
roe av e(o;,%2
2 (3)
7.4 Amplitude = 3 AY answer (1)
7.5.1 x €[0°;165°);x # 45° AY — [0°;165")
OR AY an AS
x € [0°;45°) U (45% 165%) AY critical values
OR AY ~ notation —
O°<x< 165%x # 45° ]
Av “075 x <165°
AY x #45° (2)
7.5.2 O°<x < 135° AY critical values
AY __ notation (2)
76 V2 sin 2x = cos x + sinx
sin 2x = 008% 4 sin x
v2 V2
sin 2x = cosx a + sinx AY division of V2
v2 v2
sin 2x = cos x.cos 45° + sin x.sin 45° Av _ identifying
. _ are compound 4
sin 2x = cos(x— 45°) cos(x — 45°)

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Matheri2axwwvoaded from StanmoreprySsics. Cpsparatory Examination September 2022
NSC — Marking Guidelines

Equating the 2 functions gives the points of intersection of | AY —_ conclusion
the 2 graphs. (3)

[13]

QUESTION 8
8.1 bisects the chord. AY __answer dd)
8.2.1 | AM=4cm (line from centre | chord) AVY S AYR (2)
8.2.2 | OM = (r—2) AY OM = (r-2)
re=(r-2) +4 (pythag) CAY Pr =(r-27 +4
rsr'—4r+4+16 CAV simplification
4r = 20
r=Scm
CAV answer (4)
[7]
QUESTION 9
9.1 Constr:
Draw line AO and extend to D. AY construction
Proof: OP = OT (radii)
A, =P (Zs opp = sides) AY S/R
A AW v
but O, =A, +P (ext Z of A) A S/R
0, =2A, AY Ss
Similarly 0, =2A,
0,+0, =2(A,+A,) AY Ss
“POT = 2PAT (5)

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Matheniaawwvoaded from Stanmorepibysics. CeBparatory Examination September 2022
NSC — Marking Guidelines

9.2 Do not mark this question!!! (Maximum for Paper 2 — 141)
Cc D

A B
E
9.2.1a | O, =100° (Zat centre=2x Z at circ) AV Ss AYR (2)
9.2.1b | 6, =100°
A=E, (4s opp = radii) AY Ss AYR
. 199°
“EB, = 180° = 100° (sum 4s A)
2 CAY S/R
E, =40° Q)
9.2.2 | A =B Zs opp = radii Av S/R
SE (4s opp ) AY SIR
but A, =C, (Zsin same segment)
EB, =C, AY Ss
» AEUCB (alt Zs =) AY OR
(4)
OR
A, =E, (Zs opp = radii) AY SIR
~ a v
but E, =B (Zsin same segment) A S/R
“A, =B AY Ss
-. AEOCB alt Zs =
( ) AVY R (4)
OR
AY S/R
AVY SIR

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Matherinaxwwvoaded from Stanmorepiaysics. Cpsparatory Examination September 2022
NSC — Marking Guidelines

CAE = 90° (Zsinsemiil )
A AV Ss
ACB = 90° (4sinsemil] ) (4)
CAE+ ACB =180° Av oR
.. AEDCB (co - int Zs)
[14]
QUESTION 10
D
R
A
S
T
M
10.1.1 R, = T, (4s in same segment) Av Ss AYR
R,=T, (tan chord theorem) AY S AYR
but i, = T, (given AT bisects MTR)
~R,=R, (4)
10.1.2 | In AAPR and AMPT
1. A=M (Zsin same segment) AY SIR
2. B, =P, (vert opp Zs) AY S/R
3. R,=T, (remaining 7s A)
AAPR ||| AMPT (422) AY R (3)

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Matheri2axwwvoaded from Stanmorepisysics. Cpxsparatory Examination September 2022
NSC — Marking Guidelines

10.2. | AR_PR
wr PT ((\| As) AY S AYR
AR= 3 MT (given)
AR _3
MT 2
_PT_2 AY 5
PR 3 (3)
[10]
QUESTION 11
A
T
B D E C
1.1 | CE 1
ED 72 AY Ss @
11.2 2
DE =, @cm) = 6cm AY Ss (1)
11.3 TE BE 2 .
> =m" zu triangles) AV SIR
- TE =4cm
Accept solution using the Midpoint theorem AY § (2)
11.4.1 1 AY Areas
4, 5DC x
Area of AADC _2 _ 6cm =1 AY Answer (2)
Area of AABD BD x 6cm

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Mathentaawwvoaded from Stanmorepieysics. CeBparatory Examination September 2022
NSC — Marking Guidelines

11.4.2 | Areaof ATEC
Area of AABC
_ Areaof ATEC y Area of ATBC AY Ss
Area ATBC Area AABC
_ Ec aey av Ss
BC AC
11
=-—x=
4 3 QB)
i Av Answer
12
OR
Areaof ATEC= (Area of ATBC) (common vertex=altit.)
41 Av $s
== (; Area of aan (common vertex=altit.)
Areaof ATEC _ 1. AY Ss
Areaof AABC 12
OR AY Ss (3)
Accept area rule with the use of the angle of C.
[9]
TOTAL: 150

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