# Mathematics Grade 11 P1 National Exemplar November 2019 QP

> Grade 11 Mathematics Paper 1 question paper from the National Exemplar November Exam 2019, 28 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/g11-math-p1-afr-eng-november-2019-question-paper/c8199d89-6e06-4357-8afc-e2e1f8507329
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper
- Year: 2019
- Exam period: National Exemplar November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 28

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## About this document

Mathematics · Grade 11 · National Exemplar November Exam · 2019. Question paper, 28 pages. Read online or download the PDF.

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                   testpapers.co.za

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Mathematics/P2/Wiskunde V2 2 DBE/November 2018
CAPS/KABV — Grade/Graad 11 - Memorandum

NOTE:
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.
e Consistent accuracy applies to ALL aspects of the marking memorandum.
¢ 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 memorandum van toepassing.
Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los.
QUESTION/VRAAG 1
4 | 12 | 13 | 16 | 17 | 18 | 20 | 22 | 22 | 25
1.1 | 4 minutes/ minute Y answer/ antwoord
Q)
. v
ba Mean/ gemiddel: =169 169 169
10 ¥ answer/ antwoord
_ (2)
1.3 | Standard deviation/ Standardafwyking = 5,79 ¥ answer/ antwoord
Q)
1.4 | (16,9-2x5,79;16,9 +2x 5,79) VY X-26
(5,32 ; 28,48) V¥X+20
.. | member of the team completed the obstacle race outside of 2 ¥ answer/ antwoord
standard deviations of the mean./
1 lid van die span het die hundernisbaan buite twee standardafwykings (3)
van die gemiddeld voltooi.
1.5 169+x+5 _J¢ ¥169+x45
20
x =18x 20-174 dividing by 20/
deel deur 20
x= 186 ¥ answer/ antwoord
@)
[10]

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Mathematics/P2/Wiskunde V2

QUESTION/VRAAG 2

3

CAPS/KABV — Grade/Graad 11 — Memorandum

DBE/November 2018

Cumulative frequency graph of the ages of soccer players/
Kumutlatiewe frekwensie -grafiek van sokkerspelers se ouderdomme

| 50 EERE Se er 4 | TI]

2 Ea = —
= === =a SEER
2 40 SSSSSe Soccee ===
2 Ss =t REE EEEEFEEH
yy == ==
| s == == =
—  & 30 Rot
3 t = aS =o t
| > T
Ba : =F SSSSenee

& f = :
| £ 10 ao 4

3 t Arn 4 J

| S|

| | a 1

9 I oo I ==
0 5 10 15 20 25 30 35 40
Ages of soccer players/ Ouderdomme van sokkerspelers
2.1.1 | 50 players/ spelers ¥ answer/ antwoord
1)
2.1.2 | 40-10 =30 players/ spelers v'40 and/ en 10
¥ answer/ antwoord
2)

2.1.3
Class interval/ Frequency/ Cumulative
Klas interval Frekwensie frequency
Kumulatiewe
Srekwensie
15sx<20 4 4
20<x<25 9 13
255% <30 24 37
30<x<35 10 47
35<x<40 3 50

¥ two correct
values/ twee
korrekte waardes

¥ three correct values/
drie korrekte waardes

¥ all correct valucs/
al die waardes korrek

(3)

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DBE/November 2018

2.1.4

Mathematics/P2/Wiskunde V2 4
CAPS/KABV — Grade/Graad 11 — Memorandum

Class interval/ Class midpoint/ Frequency/
Klas- interval Klas- middelpunt Frekwensie
15S<x<20 17,5 4

[ 20<x<25 22,5 9
25<x<30 27,5 24
30<x<35 32,5 10
35s x<40 37,5 3

Frequency polygon of the ages of soccer players/
Frekwensie- veelhoek van sokkerspelers se oudermme

Y using midpoints

At least 25% of class A have lower marks than any learner in class B./
ten minste 25% van klas A het laer punte as enige leerder in klas B.

25 + / gebruik
[ middelpunte
20 Y plotting the
& iY points correctly/
< korrekte punte
S15 P
a geplot
S we
x 10 Y points joined by
— v4 straight line/ punte
2 4 : 4
5 ! a verbind met ‘n
B55 4 reguitlyn
2 A :
iy Y grounding at/
0- TT AVL | = LON geanker by
0 5 10 15 20 25 30 35 40 (12,5;0) and/ en
Ages of soccer players/ Ouderdomme van sokkerspelers (42,5 ; 0)
(4)
2.2 | The claim is not valid. / Die bewering is nie geldig nie Y claim not valid/
Range of class/ Omvang van Mas A= 60 ewering me
Range of class/ Omvang van Has B = 40 getalg nie
The range of class A is bigger than the range of class B. Therefore the “ comment on the
marks of class A are more spread out than the class B./ overall spread/
Die omvang van klas A is groter as die omvang van klas B. Dus is die punte | kommentaar oor
in klas A meer verspreid as klas B die algehele
verspreiding

Y comparison of
the lower marks/

Class A performed worse at the bottom end. / vergelyk laer punte
Klas A het slegter gevorder aan die onderste groep GB)
[13]

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Mathematics/P2/Wiskunde V2 5 DBE/November 2018
CAPS/KABV — Grade/Graad 11 - Memorandum

QUESTION/VRAAG 3
y
B(6; 5)
A(-2; 2)
E N (50) x
M
C(O; -3)
D(x)
3.1 B(6;5) C(0;-3)
OR/OF mtg, = 222!
mm, <2 x, —X,
‘BC
oH 5-3)
-3-5 ~e.0 ¥ subst into correct
7 0-6 8 grad.form / verv in
_ -8 = 6 grad form.
= ; ¥ answer/ antwoord
“3 (2)

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Mathematics/P2/Wiskunde V2 6 DBE/November 2018
CAPS/KABV - Grade/Graad 11 — Memorandum
3.2 4 4
Man = Mac = 5 (AD || BC) Mp =
= 4. +e
3
2-42) +e Y subst of mand
3 point (-2;2) /
14 very. m en punt
—=e
3 (-2;2)
a4 414
Yass ¥ answer/ antwoord
OR/OF G)
4
my = 5 (ADI BC) ¥ my =z
y-2=4(¢-(-2)) Y subst of mand
3 point (-2;2) /
yoda verv. m en punt
—2;2)/
er (-2:2)
“y 30 3 Y answer/ antwoord
(3)
3.3 _4 xt 14
rhe)
0 =4,,14 Y subst/verv. y=0
3. 3
zl4_4
3. 3
azl4_-7 ¥ answer/ antwoord
4 2 (2)
3.4 2 2
AN= (x. -%) +(9,-y,)
ny ; ¥ subst. in distance
= () - (-2)] +(2- 0) formula/ very. in
: i afstand formule
4
=>
2 ¥ answer/ antwoord
Q)

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Mathematics/P2/Wiskunde V2 7

CAPS/KABV — Grade/Graad 11 —- Memorandum

DBE/November 2018

35 3y -3= 4. + 14 Y equating/ vergelyk
8 3 3
23 _ 23 ¥ simplifcation/
24 "= 3 vereeny.
x=-8 v x- value/ waarde
4 14
=—(—8)4—
agar
=-6 v y- value/ waarde
D(-8;-6) (4)
® 6-(-2) 8 B= 8
Mag = Myc
~ ABDC Y ABIIDC
but/maar AD||BC
.. ABCD isaparallelogram — [oppsidesare||/teenoorst sye is||] Vteason/ rede
OR/OF @B)
2 2
AD= (x. -%) +(9%)-¥)
2 2
= \((-2)-(-8)) +(2- 6)
= °° v length of AD/
7 lengte van AD
2 2
BC= (x. -x,) +(%,-»)
=, (6-0) +(5-(- 3) y Y length of BC/
= 100 lengte van BC
=10
-. AD =BC
but/maar AD||BC
A _ ¥ reason/ rede
«. ABCD isa parallelogram 2 oppsides are =and|i
teenoorst syeis =en || 6)

OR/OF

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Mathematics/P2/Wiskunde V2 8

CAPS/KABV — Grade/Graad 11 — Memorandum

DBE/November 2018

Mis the midpoint of AC
M is die middelpunt van AC
(-2)+0 2+(3)

Mh see)
2 2

a!
2

is the midpoint of BD
is die middelpunt van BD

XR RE

>

2 2

M -1-4)
2

di Is bisect each oth
-. ABCD isa parallelogram iagonals bisect each other

hoeklyne halveer mekaar

Y midpoint of AC/
middelpunt van AC

¥ midpoint of BD/
middelpunt van AC

Vv reason/ rede

(3)

3.7 | Mis the midpoint of AC [diagonals bisect]
M is die middelpunt van AC [hoelyne hat veer|

a C22, 229)

2 2

Y Substitution into

the correct formula/
Verv. in korrekte form.

¥x- value / waarde
vy- value / waarde

@G)
[19]

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Mathematics/P2/Wiskunde V2 9 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum
QUESTION/VRAAG 4
* G(5;2)
< B 04,04 .
x
F(2;-2)
C;y)
4.1 _ 2-(-2) Y subst. into correct
Meg = 5-2 gradient form./ vervang
4 in gradiént formule
=5 Y answer
(2)
42 tye = = @ciF6) Y mg= a
yt2  -3 ¥ equating gradients/
6-2 4 stel gradiénte gelyk
ys ¥answer/ antwoord
(3)
OR/OF

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Mathematics/P2/Wiskunde V2

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DBE/November 2018

10

CAPS/KABV ~ Grade/Graad 11— Memorandum

Mp X Mpg =—-1 (FC-LFG)

yr2 4
6-2 3
A(y+2)=-12 Yo Myo X Meg =I
yt+2=-3
¥ substitution/ verv.
yas
vanswer/ antwoord
(3)
43 | tanp=4 Y tanp=4
3 3
B=53,13° ¥ B=53,13°
0=104,04°—53,13° [ext Zof A/ buite Zvan A]
0 =50,91° ¥ answer/ antwoord
GB)
¥ subst. into distance
44 | FC=\(x,.-x,F +(y.-») su
( : J ez ») formula/ very. in afst.
= \(6-2) +52) form
Y length of FC /

=v16+9
=5

sin? = FC
NC
sin 50,91° = —
NC
ca 3
sin 50,91°
= 6,44 unit

lengte van FC

Y sin 50,91° = —
NC

Y answer! antwoord
(4)
[12]

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Mathematics/P2/Wiskunde V2 1 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum
QUESTION/VRAAG 5
ty
T(x;15)
P C 23 a)
re
oe
Oo x
5.1.1 x+y=r? [Pythagoras]
(x)? +(15)? $17? v subst in pyth/ verv
x =64 in pyth
x=-8  (P isin quadrant 2/ is in kwadrant 2) Y answer! antwoord
(2)
5.1.2 15
tand-~— ¥ answer! antwoord
QQ)
5.1.3 cos(180°-8)
=—cos@ ¥Y -cos@
-_ (#
17
8
= ¥ answer/ antwoord
(2)
5.1.4 sin °0
= (S) ¥ substitution/
17 vervanging
= 72) v / antwoord
289 answer/ antwoor

(2)

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Mathematics/P2/Wiskunde V2 12 DBE/November 2018
CAPS/KABV — Grade/Graad 11 —- Memorandum
5.1.5 |
tan@= = 15 Y tanga
—2 -8 -2 |
a 15 : |
TT 18 Y equating/ stel gelyk
a5 ¥ answer/ antwoord
4 (3)
OR/OF
_15
8 15
15 v yar
y=—x _
-8
15 (2) Y substitution of
~_gt P(-2;a)/
15 vervanging van
a P(-2;a)
¥ answer! antwoord
@)
5.2 LH — 82 !20°.cos 210° tan 315°.cos27°
~ cos 540°.sin 63° Y sin 60°/ cos 30°
vo Oo
sin60°,(—cos30°)(-tan45)°.sin63° ¥ oe
= ~tan
cos180°.sin 63° .
¥ sin 63°/cos27°
3-3.) Y cos180°
= 22 Y special angle
3 tatios/ spesiale hoeke
= 4 verhoudings
¥ answer/ antwoord
(7)

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Mathematics/P2/Wiskunde V2 13 DBE/November 2018
CAPS/KABV ~ Grade/Graad 11 — Memorandum
53 LHS = 1 cosd
cos@ 1+sin0 Y common
: 2
_ 1+sin @-cos* 6 denominator/ gemene
cos al +sin 6) noemer
_1+sin@-(1-sin? 9) -
cos O(1+sin9) v1-sin’@
_sin@(1+sin 6) ¥ simplification/
cos al +sin 6) vereeny.
sin@
= cos | ¥ factors/ faktore
=tan@ | :
“Rus y Sin Gi
~ cos
OR/OF ()
LHS= 1 _ cos6
cos@ 1+sind y
common
: 2
_ l+sin@—cos" @ denominator/ gemene
cosO(1+sin 0) noemer
_ (1-cos” 9 +sind ;
cosO(1+sin8) v1-cos' 0
__ sin? O+sind ¥ simplification/
cos@(1+sin 6) vereeny.
_ sin@(1+ sind) v
cosd(14sind) @(t-+sind) factors/ faktore
_ sind v sind
cos@ cosé
=tand
=RHS ©)
OR/OF

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Mathematics/P2/Wiskunde V2 14 DBE/November 2018
CAPS/KABV — Grade/Graad 11 - Memorandum
LUs= 1 __ cos
cos@ 1+sin0
_ l+sin@-cos’ 8
cos@(1+sin@)
_ sin? 0+cos’ 0 +sin@-cos’ 0
cos al +sin 0)
_ sin’ O+sin0
cos@(1+sin A)
sin O(1+ sin)
———Ss ¥ common
cos@(1+sin 8) :
. denominator/ gemene
sind
= noemer
cos@
=tan@ ¥ sin? @+cos’ 0
=RHS
¥ simplification/
vereeny.
¥ factors/ faktore
v sind
cos?
(5)
54 3 sinx= aun
3sinx=2 x 2% sinx
cos x v
3sin xcosx=2sin x cos
3sin xcosx—2sinx=0 Y factors/ faktore
| .
sin x(3cosx—2)=0
sinx=0 :

3 Y both equations/
x= 360°K, kez beide vergelykings
or
x =180°+360°k, kEZ

¥ both general
or solutions/ beide
| algemene oplossings
|

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Mathematics/P2/Wiskunde V2 15 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum
cosx ==

3
x =48,19° + 360°.K, KEZ Y both general
or solutions/ beide

_ ° ° algemene oplossings
x =311,81°+360°A, kKEZ VkeZ
(6)
OR/ OF
3sinx= 2tanx.
sin x
3sin x=2 .
sin x=2 x / sinx
3sin xcosx=2sin x cosx
3sin xcosx—2sin x=0
sin x(3cosx—2)=0 Y¥ factors/ faktore
sinx=0 | .
x=180°k, ke Z | 7 both equations/

4 beide vergelykings
cosx = 3 Y general solution/
x=+48,19°+360°K, ke Z algemene oplossing

Y¥ both general
solutions/ beide
algemene oplossings
VkeEZ
(6)
[28]

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Mathematics/P2/Wiskunde V2 16 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

QUESTION/VRAAG 6

a
|
tal
N
99

Q
=F .
|
Y end points/
eindpunte
Y notation/ notasie
(2)
6.2 | c=1 Y answer/ antwoord
(1)
63 | a=3, b=1 Yan3
Vb=l1
(2)
6.4 | Q(122°;-1,6) Y x- value/ waarde
¥ y- value/ waarde
(2)
6.5 K(-45°: 3V2 Y coordinates of/
c >? koGrdinate van K.
M(—45°:-1)
KM 02 +1
3 2 +2 - tenets off lengte
2
=312 2)
6.6 | f(x)=3cos(@-45°) 3
v —45°
(2)
{11]

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Mathematics/P2/Wiskunde V2

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Memorandum

QUESTION/VRAAG 7

3cm,

67°

92cm

R

Q

7A

QR? =PR? + PQ? —2PR.PQcosP
QR? =(3) +(9,2) ~2(3X9,2)cos67°

QR = (3) + (9,27 -26)9,2)c0s67°
QR =8,49em

¥ using cos rule/
gebruik cos reél

¥ substitution/
vervanging

¥ answer/ antwoord

GB)

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Mathematics/P2/Wiskunde V2 18 DBE/November 2018
CAPS/KABV — Grade/Graad 11 - Memorandum

A
Up
2
()
a
Cc B
72.1) ACD=0-a ¥ answer/antw. (1)
7.2.2 D. =90° A
D, =90° +@ YD, =90' +a
sin(90°+a@) _ sin (0 - a)
h AD v sin (90° +a)
cosa _ sin(9-«) h
h AD in (0 -«)
. sin (9 -a@
AD= hsin(6-a) v AD
cosa
¥ sin (90° + a@)=cosa
(4)
oF Oo _ ° vo
723 | yy — 17sin (58° —23°) subst/ very.
cos 23° v J
AD = 10,59 units answer/antw.
(2)
ts Areaof AADC = 4x ADx ACxsinA Y correct area rule/
; 2 korrekte area reél
-—_— 1 °
= 5 %10,59x17 xsin32 v302
= it?
= 47,70 unit Y answer/antw. (3)
OR/ OF

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Mathematics/P2/Wiskunde V2 19 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

sin58° = AB
17

AB =17sin58°
=14,41682....
BD =14,41682....-10,59289... = 3,82393...

_ 3,82393...
CD
3,82393... Vlength of BD /
De sin 23° lengte van BD
= 9,78660...

sin 23°

1 . “length of CD/
Areaof AADC = 3% CD x AC x sin35°

1 lengte van CD
= 5 x 9,78660...x 17 x sin35°

= 47,71 unit?

Y answer/antw. (3)

[28]

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Mathematics/P2/Wiskunde V2 20

CAPS/KABV — Grade/Graad 11 — Memorandum

QUESTION/VRAAG 8

DBE/November 2018

8.1

r? =13?—h? (Pythagoras)

Y using theorem of

r =169—h? pythagoras/ gebruik
stelling van
pythagoras

v= xan

i Yr? =169-h°
=-arr’h
3 Y¥ substitution/
=30(169—h? Yh vervanging
169h—ah> Y simplification/
= 3° ~~” vereenvoudig
(4)
8.2 r= 13-12? (Pythagoras) | v value off waarde
=5 van r
Total surface area/ buite oppervlakte = nr? +21 s
= (5°) +2(5)(13)
=90r ¥ subst. / verv.
= 282,74 cm? Y answer/ antwoord

(3)
[7]

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Mathematics/P2/Wiskunde V2 21 DBE/November 2018
CAPS/KABV — Grade/Graad 11 —Memorandum

QUESTION/VRAAG 9

91.1 6, = 78° [angle at centre = 2x Z at circumference ] YSVYR
[middelpuntshoek = 2x omtrekshoek | (2)

9.1.2 0, =78° [equalchords; equal 2*/ gelyke koorde; gelyke hoeke] YSYR
(2)

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Mathematics/P2/Wiskunde V2 22

CAPS/KABV — Grade/Graad 11 —- Memorandum

DBE/November 2018

R

9.2

angle at centre = 2x Z at circumference/
x —8°+180°= 2(2x-40°)
middelpuntshoek = 2x omtrekshoek
4x—80° =172°+x
3x = 252°

x = 84°
OR/OF

JoinT and R/ verbind T enR

- opp Z's of cyclic quad/
T=180° —(2x-40°)
teenoorst.Z° van koordevierhoek

R=T=220°-2x la opp.=sides /Z* teeoor gelyke sye ]

sumof int Z° of A
x —8°+220°—2x +220°-2x =180°

som binne Z° van A
—3x=- 252°
x=84°

YSVYR

v simplification/
vereenvoudiging

¥ answer/
antwoord

(4)
Y¥SVYR

vs

Y answer/
antwoord

(4)

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Mathematics/P2/Wiskunde V2 23 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

9.3
AB=BC=4x line fromcentre | to chord/ YSYR
lynvanmiddelpunt . aankoord
OA? = (4x)? +(2x)? [Pythagoras] ¥ Substitut
_ ; 5 ubstitution/
OA=Vi6x? +4x vervanging
= 20x?
=2V5x
v
OD=O0A=2V5x _ [radii) length of OA /
lente van OA
BD=2¥5 x -2x
v
=2x(/5-1
*WS5-1) BD=2V5 x -2x
(5)
(13]

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Mathematics/P2/Wiskunde V2 24 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

QUESTION/VRAAG 10

10.1 Construction: Draw radii OR and OT Y construction/
Konstruksie: teken raduise OR en OT konstruksie
Let/ Jaat: =x and/en B = y
A A it =2ti i
6, =28 angle at centre = 2 times angle at circumference/ YS VR
middelpuntshoek = 2 keer omtrekshoek
6, =2x
Similarly/ in die selfde manier: 0, =2y vs
2x+2y = 360° [anlgesarounda pt/hoeke om'n punt] v SIR
x+y =180°
-.8+B=180°
(3)
L_ _

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Mathematics/P2/Wiskunde V2 25
CAPS/KABV — Grade/Graad 11 - Memorandum

DBE/November 2018

10.2.1(a) 6=8, =48° ext Z of cyclic quad/ VS YR
at buite Z van 'n koodervierhoek
Q, =Q, =24° [QS bisects/ halveer PQN] VS
(3)
10.2.1(6) | R= Q, = 24° [za in the same segment/ in dieselfde segment | YS YR
(2)
10.2.1(c) | Mf, = Q = 48° [corresp/ ooreenkomst Z*, PQ|| SR] YSYR
OR/OF (2)
8, = Q, = 24° [alt Z°/ ooreenkomst Z°, PQ|| SR]
¥S/R
M, = 48° [ext Zof A/buite Zvan A] Y answer/ antwoord
(2)
10.2.2 M, =§, =4g° vs
..STisa tangentto circle MNS.{converse tan—chordtheorem]| yp
«ST is 'n raaklyn aan MNS lomgekrd raaklyn-koord st]
(2)
[14]

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Mathematics/P2/Wiskunde V2 26 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

QUESTION/VRAAG 11

1d fi =x [tan — chord theorem/ raaklyn-koord st| v SIR
4 Ox Z atcentre= 2x Z at circumference/ vS
' middelpuntshoek = 2 keer omtrekshoek VS
P, =90°—x [tan L diameter/raaklyn 1 middellyn]
R, =90°- x|Z opp.=sides/ Z° teeoor gelyke sye| VR, =90°—x
Rak, Us

PR bisects /halveer ORN

6)

OR/ OF

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Mathematics/P2/Wiskunde V2 27 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum
RPT =90° [in the semi circle/ in dieselfide segment]
P= x =x [tan—chord theorem/ raaklyn-koord st|
R= =90' 0° ~ x[sum int Z* of A/ som binne Z° van A]
R, =90' 0° —x [sum int Z° of A/ som binne Z! van A] vs
wR =R, Y SR
PR bisects/ halveer ORN vs
vs
OR/ OF
r, =x {tan — chord theorem/ raaklyn-koord st] vS
Pp, = x[Z opp.=sides/ Z* teeoor gelyke sye |
6, =2x [ext Z of A/buiteZ van A] (5)
s 5  180°- . v SIR
R, =P, = ——— 2k Ls opp.=sides/ Z* teeoor gelyke sye ]
2 v
A Ss
R, =90°-x
F) ways . s vs
R, =90°-x [sum int Z* of A/ som binne Z* van Al
wR = R, “Ss
PR bisects/ halveer ORN
vs
G)
11.2
PAT =90° + x | OPP Z's of cyclic quad/ VS VR
teenoorst. hoeke van koordevierhoek
A or: vS YR
S, =90° [Linefrom centre to chord / dyn van mdpt | aanko
ROS=90° +x [ext Zof A/buite ZvanA] VSIR
-, ROS =PAT
OR/ OF
(5)

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Mathematics/P2/Wiskunde V2 28 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Memorandum

RPT =90° [Zin the semi circle/ in dieselfde segment]

NPT = 90° + x

PAT =NPT = 90° + x [tan — chord theorem / raaklyn-koord s
B, =x|2* opp.=sides / 2° teeoor gelyke sye]

s, =90°[Line from centre L to chord / lyn van mdpt |.aankoord

0, =90°—x|sum of int 2° of A]

ROS= 6, +0,

=2x+90°-x

=90°+ x v8

“. ROS =PAT
VS VR
YSIR
vs

6)

{10}

TOTAL/TOTAAL: 150

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