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

> Grade 12 Mathematics Paper 2 question paper and memo from the KZN Mock Exam 2023, 29 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-kzn-mock-exam-question-paper-and-memorandum-paper-2/54a8b465-de5a-448d-8cc3-e1187e470fb1
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 29

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## About this document

Mathematics · Grade 12 · KZN Mock Exam · 2023. Question paper and memorandum, 29 pages. Read online or download the PDF.

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## Document text

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

f
£000]

N.B. This question paper consists of f2%pfages”and 1 information sheet.
This paper/hascan;Answer Booklet.

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.
1. This questiom/paper consists of 11 questions.

2. Answer ALL the questions.

3. Number the answers correctly according to the numbering system used in this question
paper.

4. Clearly show ALL calculations, diagrams, graphs, etc. that you have used in
determining your answers.

5. Answers only will NOT necessarily be awarded full marks.

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

7.  Ifnecessary, round off answers to TWO decimal places, unless stated otherwise.
8. Diagrams are NOT necessarily drawn to scale.
9. An information sheet with formulae is included at the end of the question paper.

10. Write neatly and legibly.

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QUESTION 1

Mr Siphokazi supplements his pension by mowing lawns for customers. He measures the areas
(x) (in m?) of 12 of his.customers' lawns and the time (y) in minutes, it takes him to mow these
lawns. He works 8 a day. He recorded the data.

Area (x) & 120 | 845 | 602 | 1 190 | 530 | 245 | 486 | 350 | 1.005 | 320 | 250

(m”) .
Time (y) 50 | 28 | 130} 75 | 120 | 95 | 55 |70 | 48 | 110 | 55 | 60
(minutes)
Lt Determine the equation of the least squares regression line. (3)
12 Calculate the value of r, the correlation coefficient for the data. (2)

‘i Dal

1.3 okazi charges a flat call out fee of R150, as well as R50 per half

f), estimate the charge for mowing a customer's lawn that has an

(For example: 100 minutes would be taken as 2 hours). (3)

1.4 The local high school wants Mr Siphokazi to mow their rugby field which is
rectangular, 100 meters long by 70 meters wide.

1.4.1 Use the regression equation found in 1.1 to calculate the time it would take
to mow this area. (1)

1.4.2 Is it possible for him to complete this job in a day?
Give a reason for your answer. qd)

[10]

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QUESTION 2

The following table gives the frequency distribution of the daily travelling time (in minutes)

from home to work for the employees of a certain company.

Daily travelling
time x (in minutes)

Number of employees

0)

Midpoint of Interval

O0<x<10

20

10<x<20

35

20<x<30
30<x<40

30
10

40 <x <50

5

2.1 Calculate the estimated mean travelling time.

2.2. Write the modal class of the data.

2.3. Anogive was drawn for the given data.

Construct a box-whisker plot for the-data in the ANSWER BOOK.

1

Number of employees

00

50

y

G3)

(2)

(3)

me

(ea

10

20

30

40

Daily travelling time (in minutes)

2.4 State whether the following statements are TRUE or FALSE.

x
50 minutes

2.4.1 The distribution of these travelling times is positively skewed. ()

2.4.2 The inter-quartile range for the data is 25.

2.4.3 35 employees take less than 20 minutes.

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()
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QUESTION 3

Trapezium ABCD is drawn below with AD || BC is drawn. The coordinates of the vertices are
A(1 ; 7), Bp ; gq), C6 = -8) and D(-4 ; -3). BC intersects the x — axis at F. DCB =a.
AD intersects the y =axis at E.

yt adn
E
Bp sq)
eT 0 > x
F
D(-4; -3)
N
a
C(-2 ; -8)
v
3.1 Calculate the gradient of AD. (2)
3.2. Determine the equation of BC in the form y = mx + c. (3)
3.3. Determine the coordinates of F. (2)
3.4 AMCD isa parallelogram, with M on BC. Determine the coordinates of M. (2)
3.5. Show that a = 48,37". (4)
3.6 Calculate the area of ADCF. (4)
{17]

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QUESTION 4

4.1 The cquation of a circle is x? + y? — 8x + 6y = 15.

4.1.1 Show-that P(2 ; -9) lies on the circle. (2)
4.1.2 Deterhtine the equation of the tangent to the circle at point P(2 ; -9). (6)
4.1.3 A tangent is drawn from Q(-10 ; 12) to the circle. Calculate the length of (4)

the tangent.

4.2 The circle, with centre T, and equation (x — 3)? + (y + 2)? = 25 is given below.
B is the y — intercept of the circle.

a’
B Cc
‘ [e) > x
T
v
4.2.1 Determine the coordinates of B. (4)
4.2.2 Write down the coordinates of C, if C is the reflection of B in the line x = 3. (2)

4.2.3 Another circle with centre M and equation (x — 12)? + (vy— 10)? = 100 is given.

(a) Calculate the distance, TM, between the centres. (2)
(b) Do these circles touch or intersect each other? Justify your/answer. (2)
[22]

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QUESTION 5

5.1 If sin 38° = p, determine the valuc of the following, without using a calculator:

5.1.1 cos 268° (3)
5.1.2 cos 4, (3)
5.1.3 sin26 cos 26° (2)

5.2 Evaluate the following trigonometric expression without using a calculator:

2sin 165° cos 195° (5)
cos 45° sin 15° — cos 15° sin 45°

5.3. Given: K = ¥3cosx + sinx.

5.3.1 Write K in the form of ¢sin («+6). (3)
5.3.2 Hence, calculate the value of t and 0. (1)
5.3.3. Write down the maximum value of K. (1)

5.4 Prove the identity:

5.4.1 2tan @—sin 26 = tan 8 (6)
2 sin? @ = tan
5.4.2 Hence, determine the values of 8, Qe [l 80°; 360°] which will make the
above identity undefined. (2)
[26]
QUESTION 6

6.1 Sketch the graphs of f(x) = 2sinx and g(x) = cos(x — 30°) forx €
[—180°; 180°] on the grid in the ANSWER BOOK. Indicate the intercepts with the
axes and also the turning points. (6)

6.2 Use your graphs to answer the following questions:

6.2.1 Write down the period of g. (1)

6.2.2. Netermine the values of x for which f(x) > g(x). (4)

6.2.3. Write down the values of x for which f(x) = 1,5 + g(x). (2)
[13]

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QUESTION 7

In the diagram below, PQ is a vertical mast. R and S are two points in the same horizontal

plane as the foot of tl inast, Q. QRS = a, QSR = 6 , SR=8 — 2x and SQ =x.
The angle of clevati OF P, the top of the mast from R, is 6.

foaar

P ee

Ss
7A Express PQ in terms QR and a trigonometric ratio of 0. (1)
7.2 Show that: PQ = 7222s"? (4)
sina
7.3 If B = 60°, show that the area of AQSR = 2y3x — 2 x?, GB)
74 Determine the value of x for which the area of AQSR will be at a maximum. (3)
faa? (yy

lana?

donot

don

i

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QUESTION 8
8.1 In the diagram O is the centre of the circle and M is a point on the circumference of
the circle. B subtends AOB at the centre of the circle and M at the
circumfere: f the circle.
M
A B
Use the diagram to prove the theorem that states that AOB=2M. (5)

8.2 Froma point A outside the circle, center O, two tangents AD and AV are drawn.AO
and VD intersect at M. BOD is a diameter of the circle. BV and VO are drawn.

VAD = 40°

8.2.1 Prove that quadrilateral VODA is cyclic. (2)

8.2.2 Calculate the magnitude of 0. (2)

8.2.3. Prove that BV || OA. (5)
[14]

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QUESTION 9

9.1 Complete the following statement: A line drawn parallel to one side of a triangle ... (2)

9.2 In the figufeSICL || QR. M and N are points on QR such that KN || PR and
LM || PQ. /-PK= 3 units, PL = 4 units, LR = 6 units and MN = 1,8 units.

P
K L
Q M oN R
9.2.1 Calculate the length of KQ. (2)
9.2.2 Prove that QM = NR. (2)

[6]

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QUESTION 10

In the figure, two circles intersect at A and B. AB produced to M bisects QAR. Tangents

MQ and MR meet the-circles at Q and R respectively. QBR is a straight line. AQ and AR
are drawn.

Prove:
10.1 AMQA ||| AMBQ. (3)
10..2 MR? = AM.MB (3)
10.3 MQ?
MR2 1 (4)
[12]

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QUESTION 11

11 AABC is right angled at C. ED 1 AB with E on CB and D on AB.
AC = 4,8 cm-and AB = 8 cm. AD = DB.

Cc
E
A D B
11.1 Calculate BC, correct to 1 decimal digit. (2)
11.2 Complete: ABAC ||| .... (1)
11.3. Hence, or otherwise calculate the area of ADEC. (5)

[8]
GRAND TOTAL: 150

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September 2023 Preparatory Examination.

INFORMATION SHEET: MATHEMATICS

INLIGTING BLADSY

_—b+Vb7 Ade

2a
A=P(l+ni) 4 = P(l—ni) A= P(1-i)" A=P(1+i)"
T, =a+(n—Nd S, = 5 2a+(n-Nd)
T =a"! a(r" -1) -— 7%.
| =ar s -* r#l S,,.=——3-l<r<l

" r-l l-r

pase = pasa")

i —_—_——

 S(Xth)- f(x)
"(x)= lim ——7-—
LG) h>0 h

xy tx yyty
d= (x, 4)? +02 -y,)? eed
yy — Vy.
y=mxte y-y, =m(x-x) m=22 1 m=tan0
xX -X
(s-a) +2) =P
In AABC: 4 = = £ a? =b? 407 —2be.cos A area ABC = Lab.sin C
sn A snB sinC 2

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

cos(a + f)= cos@.cos #—sin a.sin 8
cos’ asin? @
cos2a ={1—-2sin* a

2cos* a@-1

M
2

Rl
Il

_n(A)

n(S)

p=ath

P(A)

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sin(a - B) = sin a.cos # —cosa.sin B

cos(a - B) =cosa@.cos #+sina@.sin 2

sin 2a =2sina@.cosa

P(A or B) = P(A) + P(B) - P(A and B)

dXO-*o-y~)

65-0’
yo-

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< am OINAL,

; KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

MARKS: 150

This marking guideline consists of 14 pages.

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NB: CA APPLIES TO ALL SUB-QUESTIONS IN THIS MARKING GUIDELINE.

QUESTION 1
Ll VA Value of a
YA Value of b
as ;

y = 34,90 + 0,08x ¥CA_ Equation (3)

Answer only: Full Marks
1.2 | r=0,88 vAYA Answer (2)
1.3. |y = 34,90 + 0,08x

y = 34,90 + 0,08(560) ¥CA_ Substitution

y = 79.7 minutes = 1.33 hours YCA_ 79,7 minutes

Total cost = R150 + R150 = R300 ¥CA_ Answer (3)
1.4.1 | Area = 100 x 70 = 7000 square meters

y = 34,90 + 0,08(7000)

y = 594.9 minutes = 9.92 hours ¥CA Calculation (dl)
1.4.2 | No. The time taken will exceed his daily 8 hour working YCA Justification

hours. (1)

[10]

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QUESTION 2

Daily travelling Number of employees Midpoint of Interval fx
time x(in minutes) a) (x) :
0<x<10 20 5 100
10<x<20 35 15 §25
20<x<30 30 25 750
30<x< 40 10 35 350
40< x <50 5 45 225
Total 100 1950
y
100
rat J
an
8 Va
3 ai
E V4
2 50
ge
3
[=
=]
z HH
x
10 20 30 40 50 minutes
Daily travelling time (in minutes)
2.1 Estimated Mean = 2 = 19,5 YA 1950
i190 vA 100
Answer only: Full Marks “CA, Answer @)
2.2 | 10<x<20 VAWA Answer (2)
2.3 See Diagram YA — Minimum and Maximum value
YA 1 and 3™ Quartiles
YA 24 Quartile (3)
I ‘Tn BE 4
Five number summary: 0; 12; 18; 26; 50
Accept: +1 deviation on quartiles

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2.4.1 | True ¥A___Answer dd)
2.4.2 | False vA Answer da)
2.4.3 | True YA Answer qd)
[11]
QUESTION 3 hag
yt agen
E
Bip:q)
> x
F
|
D(-4;-3)
C(-2;-8)
3.1 _7+3 10 3 YA Substitution of points A and D
Ma = T4475 7 YCA Answer (2)
3.2. | map = Mgc = 2 .....(AD || BC) YCA_ Gradient of BC
y=mxeic
-8 =2(-2)+c YCA_ Substitution of point C and
gradient Lear
—4=c ———)
i :
y=2x-4 ¥CA_ Answer Ima @)

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3.3 Let y=0:
0=2x-4 YCA 0=2x-4
x=2 YCA x-—value
F(2;0) Q)
M@G3 ; 2) YA x-—value (2)
YA y-value
3.4
3.5 mp = —34+8 3 YA Gradient of CD
P4420 -2
Inclination of CD:
tan 0, = 180° — 68,2° = 111,8° YA Inclination of CD
Inclination of CF:
tan 6, =2 YA Inclination of CF
62 = 63,43"
Therefore
YA subtraction of the angles
a = 111,8° — 63,43°
= 48,37"
OR
DF - /2@+ 2+ 0+ 3)2 YA Distance of DF
= V45
YA Distance of DC
Dc = 29
CF = 80 VA Distance of CF
a 2 2 YA subst. into cosine rule
(V45) = (V29) + (¥80)? — 2V29 V80cos a (4)
a = 48,37°
3.6 | pc= (3+ 8)? + (—4 + 2) = ¥29 YA Length of CD
FC =/(0 + 8)? + (2 + 2)? = V80 YCA Length of FE
Area of ADCF = ; (V29)(V80) sin 48,37° ¥CA_ Substitution into area formula
= 18 square units. ¥CA_ Answer (4)
[17]

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QUESTION 4

4.1.1 | x? + y? —8x + 6y = 15

LHS = (2)? + (—9)? — 8(2) + 6(-9) YA Subst. of point
= 44 81> 16-54 YA Simplification
=15 (2)
= RHS

41.2 | oct 4 [OP

4 + 43" =40

Point of contact (x:y)

vA writing the equation as
(x — 4)? + (y +3)? = 40 (x — 4)? + (vy + 3)? = 40

(6)
YCA_ Centre of circle

Centre: C(4;-3) P(2;-9

entre: C( ) PC ) ¥YCA_ Gradient of radius

_ —34+9 _ 6 _
MRadius ~ ~q 2 27 YCA_ Gradient of tangent
Mrangent = “3
Equation of Tangent:
y=mx+e
YCA_ Substitution

-9==5(2) +c

25
38 ¥CA_ Answer

Pes
Y= Tata

4.1.3 | r?: = 4)2-+4(y + 3)? = 40 YCA Calculation of r?
r? = 40
(distance Q to the centre)? = (—10 — 4)? + (12 + 3)? Y¥CA Bilnce calculation
= 421 ¥CA JTangent calculation

(Length of tangent)? = 421 — 40 = 381 (4)
Length of tangent = ¥381 ¥CA_ Answer

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4.2.1 | (x—3)? +(y +2)? = 25

Let x = 0:

(0-3)? + (y +2)? =25 vA Letting x=0

(y +2)? = 16

y+2=44 YA — Simplification

y=-6 or jar2 YCA y-—values

B(O; 2) y ¥CA_ Answer (4)
42.2|C(6;2) aia YCA x—value (2)

7 YCA_y—value

4.2.3
(a) T(3 ;—2) and M(12; 10) YA Coordinates of M

TM? = (12-3) + (10+2) = 225

T™ = 15 units ¥CA_ Answer (2)
(b) Radius, center T = 5 units and Radius, center M = 10 units ¥CA_ Sum of radii

Sum of radii = 15 units

Circles touch . ¥CA_ Justification (2)

TM= Sum of radii

[22]
QUESTION 5
5.1
52°
1
P
38°
1-p?

5.1.1 | cos 218° YA Calculation of 1 — p?

= —cos 38° YA Reduction

1—p?
rl ¥CA_ Answer (3)

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5.1.2 | cos 14°
= cos(52° — 38°) YA Writing as difference
= cos 52’ cos 38° + sin52° sin38" vA Expansion
= ¥CA_ Answer
3)
cos 24° = sin 76° ¥ sin 76°
= sin2(38°)
= 2sin 38° cos 38°
v
=2p Jie double angle
v answer
5.1.3 | sin26° cos 26
1. YA Double angle
==sin52
2
= fi-p ¥YCA_ Answer
2 (2)
5.2 2sin165 cos 195"
cos 45° sin 15° — cos 15° sin45°
_ 2 sin 15°.(—cos 15°) ;
~ cos 45° sin 15° — cos 15° sin 45° vA —cos15
—2sin 30° y ;
= ————— A  —2sin30
sin(15° — 45 ° °
( ) YA sin(15 —45)
_ —2sin 30°
~ sin(—30°)
= T2sin 30° YA —sin 30°
—sin 30
=2 ¥CA_ Answer (5)
5.3.1] K =V3cosx+sinx
v3 1 }
K=2 (Fess +zsinx
2 2 vA 2(Scosx +4sinx)
K = 2(sin60 cosx + cos60 sinx) YA sin60° and cos 60°
K = 2sin (60° +x) YA 2sin(60°+ x) (3)

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°

5.3.2|}t =2 and @ = 60° YCA t-—value and 6 = 60 qd)
5.3.3 | 2 ¥CA_ Answer qd)
S41 LHs = 2 tan 0 —sin 20
nO
(] in cosd va sind
A Suny
_ _\cosé x cos0 cos8
pee cos8 VA 2sin 0 cos 6
_ 2sin0 ~2sin @cos* @
7 2sin20cos0 YA __ simplification
. re
_ 2sin0(I-cos’0)
2sin’Ocos0
_ 2sin® . sin’6 sin20
2sinO . cosd fain iysics.co
_ sin@ YA __ simplified to sind
cos co
= tan 0
= LHS (6)
5.4.2 |2sin? 6 = 0
sind = 0
=. 0 = 180° and 360° Y 180° and 360° ()
0 = 270° ¥ 270°
[26]

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QUESTION 6

6.1
2
15
1
ai
-150— 120 -90 “60 30 O30 60 90 12 150
“05
4
“1S
St
2
Graph of f: LA mark for x — intercepts
1A marks for minimum and maximum points
1A mark for shape
Graph of g: 1A mark for end points
1A mark for x — intercepts
1A mark for y— intercept (6)
6.2.1 | 360° YA Answer (1)
6.2.2 [-180° ;-150°)
x € [-180° ;—150°) U (30°; 180°] YA: Notation YCA : values (4)
(30°; 180°]
YA: Notation “CA: values
6.2.3 | f(x) = 1.5 + g(x)
fF) —g@)=15 YA f@)—g@) =15
x = 90° orx = 150° YA Answer (2)
[13]

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QUESTION 7
P
[1 0
Q a? R
x
8-—2x
B
Ss
71 PQ
aR = tand
PQ = QRtan@ vA Answer qd)
7.2 | QR_ QS VA _— Sine rule formula

sinS sinQRS
. YA _ Subs. Sine rule

QR x

sinB sina

VA Making QR a subj. of the

i formula
vA Subst. of QR
x sin B tan 6
PQ =
sina :
73 | Area of AQSR = 5 (x)(8 — 2x) sin 60° YA _— Subst. into Area rule
“1 ic
= 5 (x)(8 - 2x) (F) YA
=4(x)(6 - 2x)
= 1 VA
= V3(x) (2-34)
v3
= vax @)

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74
For Max Area: x = — 2a YA Formula
___(2v3)
7 v3
X= 2 Y A Substitution into formula | (3)
v CA Answer
[11]
QUESTION 8
8.1
M
A B
Constr: Join MO and produce to D. vA Construction
AOD = OAM+ AMO... (Ext. 0 ZfA) YA S/R
BOD = OBM+ BMO... (Ext. Z of A)
But OAM = AMO and OBM = BMO ...(Radii =) vA
.. AOD + BOD = 2AMO + 2BMO vA
AOB = 2(AMO + BMO) vA S
AOB = 2M (5)
NOTE No construction : No marks
8.2.1 | OVA = ODA = 90 .... (Radius L Tangent) YA S/R
VODA is a cyclic quad. (Converse of opposite angles of VA LR (2)
quad. Supplementary)
8.2.2 | 0, = 40" ...(Exterior angle of cyclic quad = int. opp. Angle) | ¥S&=“4R (2)

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823) 9 = WO = = 70" ....(sumof Zs of A; radii) Y oSR

180-40"

ADM = = 70° ....(sumof Zs of A; Tangents drawn Vv
from a common point A)

S/R

loa?

ADM = 0, bu ... (Angles subtended by common chord VS YR

AV)

BV ||OA pg weeeeee (Converse of Alt Zs or Alt Zs are=) | v R (5)
OR

In AOVA and AODA

OV=OD......... (radii)

vA
AOVA = AODA.......(SSS)
Al= Adve (= As)

=20°
A2=V).......(Zs in the same segment ) vA SIR
=20°
V=90° — 20°=70° VA V=10°
0.=180° -[ovA+hr]
=180° -[90°+20° |
= 70°
VA O2=70°
wy, t 62.1... (both - 70°)
.. BY||OA...... (altZs are = or conv.altZs )
YA R
[14]
40.00

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### Page 27

Mathempsiss Vhloaded from Stanmer hysics.com 2023 Preparatory Examination

QUESTION 9
91 Divides the other two sides, proportionally. YS divides the other two sides
YS ___ proportionally (2)
9.2
P
4
L
6
Q M oN R
1,8
9.21 | eae (Prop. Thm; KL || QR) vy SIR
KQ = 4,5 units v Answer (2)
9.2.2 | KL = QM ....(Opposite sides of |" QKLM are v SIR
equal) v¥ S/R
KL=NR .... (Opposite sides of ||" NKLR are (2)
equal)
Therefore QM = NR.
ox - os R a - “ Prop Thm, KN||PR
Q tar Q
45_3 OR
3? MR RL
MR_ RL —_—=— Prop Thm, ML||QP
QM =oL Prop Thm, ML||QP QM PL
-6_3
4 2
QN _MR
NR QM
QM+1,8 _ NR+1,8
NR QM
“QM (QM+1,8)=MR (NR+1,8) / OMELS _ NRH8
“. QM=NR NR QM

[16]

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### Page 28

Mathenpsiss nloaded from Stanmer livsics.com 2023 Preparatory Examination

QUESTION 10

10.1 | InA’s MQA and MBQ

M,= My... (Common) v S/R
Ay =Qy wees (Tan-Chord Theorem) v JR
AQM = By ee (Remaining 4s of A’s)
AMQA ||| AMBQ ...... (AAA) v R GB)
10.2 | In A’s MAR and MRB
My = Mg oes (Common) s
Ag = Ry weeeeee eee (Tan-Chord Thm) S/R
ARM = RBM .......... (Remaining Zs of A’s)
AMAR ||| AMRB ...... (AAA) v R
MA MR
—_— = *s simi YS AYR
MR MBO (A’s similar)
MR? = AM.MB ©)
10.3 |MQ_ MA MA
—-= — v —
MB MQ (from 10.1) MQ
MQ? = MB.MA =.......( 1)
Also MR? = AM.MB v = MB. MA
r>
Now Ma: = MBMA yy MQ _ MB.MA
MR2 AM.MB MR2 ~~ AM.MB
_ MQ?
” MR2 (4)
{12}

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### Page 29

Mathempsiss Vhloaded from Stanmer if (sigsgom 2023 Preparatory Examination

QUESTION 11

Cc
enn
40)
lniny 4.8
A D B
8
11.1 | BC? = 8? — 4,8? ........ ( TOP) v S
BC =6,4cm v S (2)
11.2 | ABED ys ()
11.3
A AC Cc
Ba _ AC LBC ABAC ||| ABED vA S
BE ED BD
8 48 64
BE ED 4
48x4 YCA_ Value of ED
a7 3cm
Area of AEDB
1
=5 (4)(3) = 6 cm? YCA_ Area of AEDB
Area of AABC
1
= 5 4,8)(64) = 15,6 cm? YCA_ Area of AABC
Therefore, Area of ADEC
= 15,6-6=9,6cm?
(5)
ti

GRAND TOTAL: 150

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