# Mathematics Grade 12 Paper 2 NSC June 2013 Memo

> Grade 12 Mathematics Paper 2 memorandum from the NSC June 2013, 31 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/mathematics-p2-hg-2013-memo/4449737b-fb27-4b41-a928-84fe8424d68a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2013
- Exam period: NSC June
- Paper: 2
- Curriculum: CAPS
- Pages: 31

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## About this document

Mathematics · Grade 12 · NSC June · 2013. Memorandum, 31 pages. Read online or download the PDF.

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## Document text

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### Page 1

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basic education

Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

SENIOR CERTIFICATE EXAMINATION

MATHEMATICS P? |
HIGHER GRADE | ™:
2013

POSSIBLE ANSWERS _

MARKS: 200

This memorandum consists of 31 pages.

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Mathematics/HG/P2 2 DBE/2013
SCE-Final Memorandum 14 May 2013

QUESTION 1 [26]

AC1;4)t

B(p;-2)
C5 ; 4)
v

1.1 | tan 135° =-1 “A 1A tan 135° =-

vM VA

4-€2) 1) vM
-l-p

l+p=6 VCA
p=5
OR
tan 135° = -1%A
YM v
y-4 = -1@-C1) VA
a.
Sub: forB: y=-2
-2=-x+3yce4
fs Jee! b)

OR

1M gradient formula

1 correct subs

1M equating the
gradient

simplification

tan 135° =—

value of m

equation of line
substitution into the
line

simplification

hh

l4-05-1Y

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Mathematics/HG/P2 3

SCE-Final Memorandum 14 May 2013

DBE/2013

11 Assume B is (5 ; — 2) Assumption
Cont. vA
—~ 4-¢2)_ 6 _ Gradient of line
mB" 1-5 6 “1 WM VA correct gradient
YM vA Og
inclination of AB is: tan(— 1) = 135° calculating inclination
conclusion
fd So)
G)
12 |AC = BC YM the distance
AC2 = BC2 vA LHS sub
C5-CbDP + @- 4 = 5-5 +@-C2DY VA | RHS sub
16+ @—-8q+16 =100+q’+4q+4vS Multiplication
—8q-4q =104-32v§ Simplification
—12q = 72
q=-6YCA (6) | value of g
13 : 7 ett at
(xy sv u) = ( 2° 2
oe ae 4+(2) vM correct substitution
id 2 2 into mid-point formula
M(2;1) “CA VCA correct x and y values
vw)
_ 6-1 © he YCA correct subst into
=e) gradient formula
_ gradient CM
map=—lva gradient AB
=-1 YCA
mem * map =—I product of gradients
CM Ll AB
OR
os 1+5 ~4+¢2) vM correct substitution
te! 2 7 2 into mid-point formula
M(2;1) YCA VYCA correct x and y values
AC = (-5-(-1)? + (6 - 47
= 116VA Correct length
CM = (5-27 + - 17
=98 VCA Correct length
AM’ = (-1-27 + (4-17
= 18 vCA Correct length
TER 2 ¥CA
—L EDUCATION SIC ] Pythagoras
i CML AB
on0i3 -06- 24

bene  ETORIA 8004
PUBLIOE

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i XAMINATIONS

PRIVATE BAG Kk 40

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Llir L

14-05-13

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Mathematics/HG/P2 4 DBE/2013
SCE-Final Memorandum 14 May 2013
1.3
Cont. —14+5 4+(-2) vM correct substitution
= 3 a into mid-point formula
M(2;1) VCA VCA correct x and y values
BC?=116 VA Correct length
CM? =98 VCA Correct length
2
BM? = 18VCA Correct length
BC? = CM? +BM? VCA Pytha
goras
“CM 1 AB
7)
1.4 | Ax=xa-xc = xp—xB
-1+5=xp-5 YM change in x
9 =xp YCA x value of D
Ay=ya-Ye = W-ya v :
446 = yt? M change in y
8 =y VCA D(9;8) y value of D
OR
Ax=xp—Xc =Xp—XA vM OR
Oa oe Gy) change in x
: =9 YCA gt
Sw x value of D
Ay=ya—Yo=Yo-Ya
ie ~ a OAR (4) “Mi change in y
“Yo y value of D
OR
M is the point of intersection of the vertices of the parm.
Let D be (xp; yp)
= *%c+%p
xy 2
—~54+x, Sub into midpoint
2= ee vM formula
*> = 9 VCA x value of D
— Yct+¥d
ym 2
i= 29+2> ve Sub into midpoint
2 formula
yo = 8 VCA y value of D
YVA
Answer bint DQ;8) “%A
; EDUCATION (4)
2613 -06- 24

—PUGIC ExaMinarions polo \t3

1

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Mathematics/HG/P2 5 DBE/2013
SCE-Final Memorandum 14 May 2013
1.5 -6-4 5 VCA gradient of AC
Mc = TG
-5-(-l) 2
5 mM
y- 62) = > &-5) va .
2 sub into line formula
y= 5 x- 25 _ 2 correct gradient used
2 2
5 29 y
any oe simplification
OR
AC //BD {ACBD is a parm)
-8-©2) _5 vea gradient of BD
1D
9-5 2
v)
| bY sub into line formula
y-(2) = 3 @-5) vA correct gradient used
5 25
=2y- 2-2
4 2 . 2
=35 oS 29 V¥CA simplification
2 2
(4)

pore

EDUCATION

BASIC

2013 -05- 24

PRIVATE BAG x 179
PRETORIA o004

—LUBLIC EXAMINATIONS]

ran Satnemencens

ow

np”

Hort

14-05-13

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Mathematics/HG/P2 6 DBE/2013
SCE-Final Memorandum 14 May 2013
QUESTION 2.- [25]
2.1.1
v4
A (6; 3)
B(C-8;1)
| \ .
Cc
v
2.1.1 | Let centre of circle be M
M= (* + xy data)
2 2 ere
Using midpoint
formula
- (2 #1) = (-1;2)VA VM Coordinates of M.
2 2
vM
R= (6—CDY + (3-2 OR RB’ = C8-CY + (1-2) | Use of formula to
= 50VCA calc radius
Equation of circle:
Correct substitution
(x41P + (y-2P = 50 YCA into equation of circle
6) formula

LEPARTMENT OF BASIC |
EDUCATION i

2013 -06- 2

PUMLIC EXAMINATIONS

WA 3

14-05 -(B

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Mathematics/HG/P2 Z
SCE-Final Memorandum 14 May 2013

DBE/2013

2.1.2

For y-intercepts,x=0 “A
(OF1P +(y-2P =50 “A
(y-2" = 49 YEA
y-2=7 0 ory = 7 YCA
y=9 or y =-5
C= (0;-5) CA
OR
YA
Let C be (0 ; &)
mac X mpc =~]
k-3 _k-1 YA
0-6 0-(-8)
R-4k+3-48=0VvCA
 -'4k — 45 = 0
(k-9)(k+5) = 0 YCA
k=-5 or9

C= (0;-5) YEA

OR
“A
Let C be (0 ; &)
AC? + BC? = AB?
(6-0)? + 3-H? + (8-0) + (1-wY = 200 “A
36+9-6k+h +64+1-2k+K -200=0 VYCA
2h —8k -90=0
BP - 4k -45=0
(k-9) (+5) = 0
=-5 or 9 YCA

C= (0;-5) YCA

6)

Concept
substitution
simplification

calculation

coordinates of C

Concept

substitution

simplification

calculation

coordinates of C

Concept

substitution
simplification

calculation

coordinates of C

__PUELIG E PRAM NAVIONS

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Mathematics/HG/P2 8 DBE/2013
SCE-Final Memorandum 14 May 2013
2.1.3 Mean = YorYm _ ~5- 2 -7M YCA Correct snbst into
Xo —x, O+1 gradient form.
i ‘ M value of gradient
7 vCA Gradient of tang.
y=oxtec VCA OR (v-y) = m(@&-x) Correct subst into line
formula
1 1
y= zs YCA Y-5))= 5(4-9 Equation
1
= >x-5
25
OR
Equation of circle: x? + 2x + —-4y-45=0 VCA aii equation of
circle.
Point of contact of tangent C (0 ; -— 5)
Equation of tangent:
YM YCA Correct subst into form.
x0) + WS) + 1@+0)-2(y-5)-45=0 Correct values used
-Sy+x-2y+10-45=0 VCA oe :
-ly+x-35=0 simplification
y= axes YCA Equation
(5)

14-05-13

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Mathematics/HG/P2 9 DBE/2013
SCE-Final Memorandum 14 May 2013
2.2.1 Mgs = Mts
k-2_ _ k-(-4) MIA concept of collinear
1-(3)—s1-5
K-2 _kt4 vA Correct substitution
4 -4
vw
k-2 = -k-4 — simplification
2k = 2
k =-1VCA value of k
OR VMIA
Mar = Mrs concept of collinear
4-2 _ k-C) vA Correct substitution
5-(-3) 1-5
id ues
8 4
8k + 32=24 VYCA simplification
8 = -8
k= -1 CA value of k
OR
Mpr=mgs YM concept of collinear
2-(-4) _ 2-k
Si Gn A are
(-3)-5--3-1 Correct substitution
6 _2-k
3 4
—16+8k =—24 “CA Simplification
8 = -8
k =-1V%CA value of k
OR
2-(4)_ 6 3
Mer = ae tgs gM gradient of RT
eq of line:
Yeas ~4e +3) Correct substitution
3 1 into equation of line
ys=-2x-+ VCA
4 4 Simplification
subst (1;):
3 1 v
Later Vere a value of k
j PERARTMENT OF Basie (4)
= ERUCATION i
i 2013 -06- 21
: PRIVATE GAG x 149 b \ v3 \y
oo ETORIA 6004 4-05-13
eS a Rnetaee ntact rec acca 1?) 4-0
‘__EMBLIC EXAMINATIONS

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Mathematics/HG/P2 10

SCE-Final Memorandum 14 May 2013

DBE/2013

2.2.2

PO;
RP = 2PT “A RE3;2) ad

TS 3-4)

. YM
@-C3Y +Q-2)7 =4e-sP+O-Cyyy VA
P+OX+9 + —4y+4 =4(" — 10x +25 + + By + 16)

YCA
XP +Oxty—4yt 13 =42 —40x4+ 16444 +32y YCA

3x? 46x +37 +36 +151=0 YS
(6)

correct ratio

distance formula

correct substitution
LHS multiplication
RHS multiplication

Simplification

14-05-13

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Mathematics/HG/P2 W DBE/2013
SCE-Final Memorandum 14 May 2013
QUESTION 3 [22]
3.1.1 | 2tana +k =0
-tana=—* vA value of tana
Pythagoras 27+ (-k? =" VM subst into pyth.

wre fP+4 VCA
evs YCA

calculation of r

value of seca

seca =
OR
k vA
(oe. value of tana,
‘sec? a=1+ ten’ a’¥M identity
. KY vca
zu (-£ substitution
2 2
sca =.jtt4 = K+4 von simplification
4 2
(4)
3.2
sin (900°—x) . sec(180°+ x) . sin 58°
tan(180°—x) . sin 61°. sin 29°. cos420°
v AY A numerator
= sin x .(-sec x) . sin 58° VAYAYA
(-tan x) . cos29°. sin 29°.cos 60° denominator
PER
sinx 1. sinsse CA identity
x
SS CA identity
Si 69829°, sin29°4 ;
cos x 2 YCA cos60° = 3
_ 4sin29°.cos29 YCA identity

cos 29°. sin29°
=4

YCA Simplification

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Mathematics/HG/P2 : 2 DBE/2013
‘ SCE-Final Memorandum 14 May 2013
3.3.1
LHS: 08 *_ + tanx

sinx +1

= _cosx sinx ya identity
sinx+1  cosx
cos x(cosx) + sinx(sinx +1) YCA numerator

cos x(sinx +1) YCA denominator
YA

cos?x +sin?x + sinx ed

| =F multiplication

cos x(sinx + 1)
ee identity
cos x(sin x +1)

1 A

al = secx = RHS (6) | identity

cos x
3.3.2 | undefined when tan x is undefined or cos x = 0 or sinx =—1
vA VA
x = 90° OR 270° two angles
Q)

[4-05-13

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Mathematics/HG/P2 13 DBE/2013
SCE-Final Memorandum 14 May 2013
QUESTION 4 [24]
4.1 | sin2x =cos(x—-30°) [ ;
sin 2x = sin (90°-(x—30°) “4 YA co-function S
2x = 120°-x+h.360%A or 2x =180°—(120°—x) 1A solution 1" quad
3x = 120°+ &.360° 2x=60° +x +k360° 1 A solution 2™ quad.
x = 40° +4.120° ;keZ x =60°+k.360°
wv wv vi
x= BO" 40s 160° —- x =60° YCA 4A — correct angles
OR
sin 2x = cos (x—30°)
vA co-function

cos (90° — 2x) = cos (x — 30°)

° re? fe)
90° —2x =x-—30°+k.3: or 90°—2x =—(x-—30°)+ k.360° 1A solution 1° quad
— 3x =-120°+ £.360° —x=— 60° + 360° | 1 A solution 4” quad.

x = 40° +4120° keZ x =60° + k.360°
YCA v v
x = —80°; 40° Heo" a x = 60° YCA | 44 —comect angles
@
42 Ty
id ||
it |]
iy j
hs F " ~
ie a oN de ge ix
gq 8 o7 | 457 |o
is 3 5
=
os
+

f g All points to be visually

amplitudev’A Turning point (30°; 1)“A read.

shapevA shapevA Shape includes end-points

y-interceptVA y-interceptVA and turning points visually

x-intercepts“A x-interceptsY”A read.

(8)
each interval VVWCA

4.3.1 | x@(-80°; 40°) or (60°; 160°)

-80° < x < 40° or 60° < x < 160° @) | “A notation

4.3.2 | xe [-90° ; 30°) interval YCA
-90° < x < 30° (2)_| “A notation
each intervalYVVCA

4.3.3 | xe (—90°;-60°) or (0°; 90°) or (120°; 180°)

90° < x < -60° or 0° < x < 90° or 120° < x < 180° “A notation

(4)

ML

[4-05-13

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Mathematics/HG/P2 14 DBE/2013
SCE-Final Memorandum 14 May 2013
QUESTION 5 [14]
5.1
2 cos?x-3cosxt+1=0 “A quadratic equation
factorisation

(2 cosx-1)(cosx—1)=0 YCA

=tan 15° VA
= tan (45°— 30°) “CA

__tan45° ~ tan30°
1+ tan 45°,tan30°

YCA

YCA 3
ee YCA

v3

1+1.—
3

OR

3-3 . 3-V3
343 3-3

y¥M

= 9-6¥3+3 vou
6

=2-3

OR

1 “CA YCA 2 values for cos x
cosx = 3 or cosx = 1
vA YCA :
x =60° + &360° or 240° + £.360° OR x = 360°; ke Z| solutions
Penalty — 1 omitted on
oR oA VA full marks
x =60° + 360° OR x= 0°+k360°; ke Z
OR 300° + &.360°
(6)
5.2.1 _ tanA-—tanB
EG) Tea Rey vA recog of iden
)
5.2.2 | cot 255° =cot 75°
co-function

double angle

expansion

sub tan 45°; tan 30°

Rationalising denom.

Simplification

DEPARTMENT OF BASIC
EBUCATION

2613 -05- 21

PRIVATE BAG X 119
~ PRE! ORIA 8001
PUBLIC EXAMINATIO s\ob \

(G-0s~ 13

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Mathematics/HG/P2 IS DBE/2013
SCE-Final Memorandum 14 May 2013
5.2.2 | cot 255° =cot 75°
Cont.
=tan 15° VA co-function
= tan (60° - 45°) YCA double angle
— tan60°-tan45° vo,
Te tan? tana expansion
3-1
a 1+v3 4°" sub tan 60°; tan 45°
= 3 -1 7 V3 -lvM Rationalising denom.
V3 +1 v3 -1
YCA

-

OR

cot 255° = cot 75°

vA

=cot(45°+30°) = 1 YCA

tan(45°+30°)

_1—tan45°.tan30°
tan 45°+ tan 30°
YCA

(3)

¥3 -1)
4341

YCA

YCA

43-1 /M
3-1

YCA

- 3-2V3+1

”

Simplification

Use of inverse
double angle

expansion

sub tan 30°; tan 45°

Rationalising denom

Simplification

203 -05- 21
PRIVATE BAG X 140

nn PRETORIA DCO DS
PUBLIC EXAMINATIONS

Atul

(4-0S- [2

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Mathematics/HG/P2 16 DBE/2013
SCE-Final Memorandum 14 May 2013
[ QUESTION 6 {22}
6.1.1 1
D
D B .
Construction: h=AD.L BCD vc Construction

peu = sin(180°—B) = sinB ep = -cosB vA
e e

“AD = c. sinB BD = —c.cosB

By Pythagoras

b? =CD? + AD?

= (a + BD) + 2—BD?, YM VA
= @ 4+2.aBD+ BD?+ ¢?-BD? “A
=@ +? +2aBD VA

= a'+c’—2.ac.cosB

v
OR Construction: h=AD1 BCD .
BD
—=-cosB
c
BD = —c.cosB <
By Pythagoras

b? =CD?+ ap? “M

= (a+ BD)’+(ADY VA
@ +2aBD+ (BD?+AD?) YA
@+c? -2.accosB VA

CEPARTHENT OF

EDUCATION

BASIC

2013 -05- 21

H PRANYATE ¢
i PRET BAG X 119

Correct ratios

Use of Pythagoras
Substitution
Expansion
Identity

Construction

Correct ratios

Use of Pythagoras
Substitution
Expansion
Identity

Mu

[4-0y~~ f3

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DBE/2013

Mathematics/HG/P2 _
SCE-Final Memorandum 14 May 2013
yt
; If no axes drawn in
A(c.cosB;c. binBYM max 5/6
b
c
B x C(a;0)
vA
By distance formula .
“M Use of formula
b? = (cos B—a)* + (c.sin B—0)* vA Substitution
= c’cos’B — 2.ac.cosB + a” + c’sin’B VA Expansion
= @& + A(cos’B +sin’B)— 2.a.c.cosB YA _| Identity
= &+¢’-2.a.c.cosB
2
6) a ly
jag (S|
Ta o fs
le Zz 4 =siB!
6.1.2 GG al <sis
he Fe .| o2;3
iba @ Ss/8
Cc I ‘ el ms |
PB peg BE Sy Fae
A E k & ised
8 ; “* zt
B
@ | BCE =90°+9 “A Value of BCE
BE’ =BC’+ CE’ -2.BC.CEcos BCE VA Peas
= p+ p— 2p) p) 205(90° + 4) ee Us
= 2p? + 2p’sin 0 Si a ti
= js YCA VYCA implification
See ak oN, Common factor
BE = p,j2(i+sin@)
©)
6.1.2] (3p= pJj2i+sind) VM p#0 Equating
©) | 3 = 2+ 25indvA Squaring
: 1
sing 7)
0 =30°VCA (3) | Value of 8
Ne) MUS
nd\or\ Y (4-00-13

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Mathematics/HG/P2 18 DBE/2013
SCE-Final Memorandum 14 May 2013
6.2 A
h
¥y
B Cc
D
6.2.1 A =siny VA Correct ratio
Cc
Correct subject
- vea ’
sin y
OR
xe =cosecy VA Correct ratio
AC = hcosecy VCA (2) | Correct subject
6.2.2 a
DAC = 180°-2x “A angle DAC
In A ADC:
AC _ cD M/A subst into sine rule
sin x sin (180°-2x)
AC rsinx = YCA simplification
sin 2x
= rsinx 7 r
2 sinx.cos x, CA 2 cos x double angle formula
h r Aeron
—_— = simplification
sin y 2eosx “CA E
, = 7siny
2cosx
OR
- 1 Es TOF Ease
i EOUCATION
i 208 -05- |
; :
j PRIVATE BAG XY ‘a
f PRETORIA o00, ’ tb \ St bin 2

A ES3 ¢ ag ~ pe
LUPUSLIC EXAM!

### Page 19

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Mathematics/HG/P2 19 DBE/2013
SCE-Final Memorandum 14 May 2013
622 | DAC = 180°-2e v
Cont | in & ADC: “ angle DAC
DC? = AC? + AD? — 2,AC.AD cos (180° — 2x)
a r 2h?
= (-cos2x) “M/A subst into cos rule

? sin’? y =2h( 1 +cos 2x) YCA

~ Ginyy? * Giny?  Giny)

r’ sin? y

simplification

w= 2(1+2cos* x - 1) CA double angle formula
r’ sin? y cap :
w= SD YVCA simplification
4cos” x
h rsiny
2cosx
(5)
6.2.3 — 500. sin27,4°vM Substitution
2cos 65,3° _
= 275326 VA Simplification
=275m YCA (3)_| Rounding for paper

ware BAG y
an DAG X14,
TORIA Dany :

AA MINATIONS

[4-05-12

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Mathematics/HG/P2 20 DBE/2013

SCE-Final Memorandum 14 May 2013

[QUESTION7 fig) __ |

ys

D

TAA! a vs YR
M; = 90° (Line from centre to midpoint of chord is L)

N, = 90° (given)

M, + N, = 180° vg

YR
ANOM is acyclic quad (opp Z’s supplementary)
(4)
TA2 Me ite

““) Oy = 180° — A (opp Z’s of cyclic quad suppl)

C = 180° A (opp 2’s of cyclic quad suppl) YSIR

O=C

@)

7.1.3 | In A ABM and A DBM
AM = MD (given) vs
BM = BM (common) vs
M, = M,_ (Line from midpt of chord is 1)
A ABM = ADBM (S,A,S) VYS/R
B, = B, — (corresp Z’s of = A’s) YS/R

(4)

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2)
SCE-Final Memorandum 14 May 2013

DBE/2013

72

BCD = 180° - A = 180° — x - svR
0, =2y VS (Z at centre = 22 at circum)

6; = BCD (opp 2’s of rhombus =) WS/R

y= 180° -y YS
3y = 180°
y = 60° vS

OR

C = 180°—y (opp Z’s of cyclic quad) YSR
Bi = 90° -y (sumZ’sofA) YS

Bi = Bz = 90° — y (proved)

Bs = Ba (rhombus)

€ + OBC = 180° (co-int Z’s)_ vg
180°— y +180° —2y= 180° VS
y= 60° VS 6)

OR
B, = 90°-y (sum 2Z’s A)
BCD = 180° -y

YS

vw
diagonals bisect angles and opposite angles are equal:

: YSIR
4(90° — y) + 2(180° —y) = 360° (sum 2’s of rhombus)
6y = 360° VS
y=60° vg

(5)

ES eee
| DEBARTHENT OF BASIE 7
EDUCATION

PRIVATE BAG X 119
PRETORIA GUOT

PUBLIC EXARI NATIONS

|
i
2013 -05- 21 |
j

ns sy)

(4-05-13

### Page 22

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Mathematics/HG/P2 22 DBE/2013
SCE-Final Memorandum 14 May 2013
[QUESTION 8 21) |
8.1
M
2
1
9)
R
2
1
A P
Construction: Draw POMandQM “YC
Proof: P;+ P,=90° “S tant radius YR
A A
P,=90°-P,
A a vs A YR
Qi t+ Q, =90° angle in the semi circle
M =90°-P, sum Z’s A
M = R vs ’s in same segment yp
APQ= PRQ
OR
: DERE DRERF Se
j IMENT OF BASIC 7
: ERUCATION |
3
i
2613 -08- 21
j PRIVATE BAG X 110
I ermine PRETORIA 0001
Lm MGIC EXAMINATIONS I
Vole 14. -05-[3

Jeo

### Page 23

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DBE/2013

Mathematics/HG/P2 23
SCE-Final Memorandum 14 May 2013
8.1
Cont. Be
O 1
2
1
A P
: xC
Construction: Draw POM and MR
YR

_
Proof: P,;+ P,=90° YS (tan 1 radius)

A a
P,= 90°-P

; ws vR
= 90°-Ri (2’s in same segment)

. A ; YR
R,+ R,=90° YS (angle in the semi circle)

R, = 90°-Ri

APQ= PRQ

(¢-0s- 13

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DBE/2013

Mathematics/HG/P2 24
SCE-Final Memorandum 14 May 2013
8.1cont
oO
1 R
2
1\\3

A P
Construction : Draw radii QO, RO and PO vc

A A vS
Proof: Pi+ P,=90° tan 1 radius VR

P, =90°-P,
Boek Ae .
Q,+Q,+Ri+R2+P3+P2 =180° Zsof AYSR
AQOP and AQOR and APOR are isosceles equal radii VS
BoB :
2R,+2R2+2P2 =180° “8
2R, +2R2 = 180° - 2P2
Doe le
R,+R2=90°-P2 VS
Boa : 7
R=Pi=APQ -. APQ= PRQ
OR

1

A P
Constr: Join OQ and OPY'C

2P,+ QOP=180° (sum angles in triangle) VS/R

P,+ R=90°,

But P,+ P2=90° S (tangent L radius) “R

a

Pi=R-. APQ= PRO

R = 596? YS (ang at centre = 2x ang at circum) VR
_
Q,=P2 (equal radii) YS/R

1G)

if ple?

pL
(y-0S- /3

### Page 25

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Mathematics/HG/P2 25 DBE/2013
SCE-Final Memorandum 14 May 2013
8.2
8.2.1 A i.) v
C, = x ™ (tan-chord theorem) YR
§= <x YS/Rtan-chord theorem)
* v)
T= xV¥S (2’s in same segment/Sin-chord)
(5)
822) 7 = 6, “8 (both = x)
@) VR
BC is a tangent to circle ACT (Converse tan-chord
theorem) 2
Q)
a DEPARTMENT OF BASIC
: EDUCATION
2013 -06- 21
i PHVAT
i PRETO!
| PUBLIC EXAMINATIONS
ig-os- 1%

ot \v \7

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DBE/2013

Mathematics/HG/P2 26

SCE-Final Memorandum 14 May 2013
8.2.2 B, = A; +A, o (tan-chord theorem) S/R
(b)

SRA = A, + A, (tan-chord theorem) “S/R
- * YR
B, = D,vYS_ (ext Z cyclic quad= inter opp 2)

SRA = BD, = CDT.

OR

§ =¢, =x (proved) “S
CB //RS_ (corresp Z’s=) VS

SRA =B, (correspond. 2’s;//lines) “S/R

YS
B, = Dd, (ext Z cyclic quad = inter opp 2) yR
SRA = D,

OR

A, = ART (tan-chord theorem) YSIR

A, = 6 (tan-chord theorem) “S/R

A, = SRT (2’s in same segment) YSIR

2
D, = €, + SRT (@xtZ ofA) VSR

SRA = ART + SRT vs

i

SRT

(5)

TRENT OF BASIC
EDUCATION

y-os- 1%

### Page 27

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Mathematics/HG/P2 27
SCE-Final Memorandum 14 May 2013

DBE/2013

sy

[ QUESTION 9

9.1

B ar"

R Q

Construction : Draw BQ and CR. Draw heights k and h YC

1
—PBA YS
Proof : Areaof APBC 2

Areaof BRC I pp,

2

| a

BR

eck
Areaof APBC_ 2. ~

Area of ABCQ — leas
762

= PC yg
CQ
but PB _ PC
BR CQ

AreaofAPBC _ Area of APBC
Areaof ABCR Area of ABCQ
Area of ABCR= Area of ABCQ , they have same

base and must have same height.

They lie between parallel lines vg

BC/RQ

ieee BUCATION

vs

OR

### Page 28

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DBE/2013

Mathematics/HG/P2 28

SCE-Final Memorandum 14 May 2013
9.1
cont

vC
Construction : On PQ mark off point D such that BD // RQ

Proof: = . to (given) VS/R
vs
PB = PD (BD //RQ)YR
BR CQ
=PC vg
CQ

vS
But this is impossible unless D and C are the same point.

But BD/RQVYC
BC/RQ

)

ri

i GEPARTHENT OF BASIC 71

i .

j 2013 ~08- 21

i PRI AG X10

i: 4 ORIA 9004

| PUsLIC EXAMINATIONS

### Page 29

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Mathematics/HG/P2 29 DBE/2013
SCE-Final Memorandum 14 May 2013
92
R
s
12
r r ar
M re) N o Pp
ee oe ws (proportionality theoremy R
SR _ jr YS
12 2r
SR = 4units “S (4)
9.2.21 SN MN smn ///ARMP) YSIR
RP MP
SNL
1 fr YS
SN = ox 12 = 9units vs (3)
9.2.3 A
InAMSN, MSN = 90° (Z’sinasemicircle) “S/R
SM = 12; SN = 9
MN? = MS? +SN? (pythagoras) VS/R
= 12? + 9 = 225
MN = 15 =2rvs
= 7,5 units | PAPRRTRENT OF Bisre—
EDucation |= }
OR - |
2 =
MR? + RP? = MP? ; ve i 2013 -86- 24 |
16? +12? = (22r)? (pythagoras) i PRivere i
al
= (22r YS } PRETORIA G
400 = (22r) Ppp een
20 = 22r Amend XAMINATIONS
15 =7 “8S

line

(4-0S-1>

### Page 30

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Mathematics/HG/P2 30 DBE/2013

SCE-Final Memorandum 14 May 2013

QUESTION 10 {12]
10 “
T
mm N
oO
R
} i
B ! Pp } Cc
10.1
RC = NR (Ris midpt of NC) VS

BE = PC (Pismidpt of BC) VS

v
-. BT || PR (converse proportionality theorem)

-. PR {| ON
OR
RC=NR=x (given)
AN =NC=2x (medians ofa A are concurrent) YS
AO _ 2 (median AABC VSR
OP 1
AN _2
NR 1

- AO SR

oP

PR || ON (sides in proportion) “S/R (4)

Pome SBUSATIO rm

sflolvr>

Pesce

i

### Page 31

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Mathematics/HG/P2

3f DBE/2013

SCE-Final Memorandum 14 May 2013

10.2 vs

AN = NC (medians ofa A are concurrent)

ON =NT (given) “S/R
AOCT is a parm (diagonals bisect each other) “S/R

YR

4

10.3 | Area of A AON

4x AO x AN x sinPAC YS

Area of A APC

OR “8

4

4

1

AAON = r AAPR

(2 arc) vs
4

| AreaAAON 3A4PC

“"AreahAPC  AAPC 3

4x APx ACx sin PAC YS

AO | AN

—— x ——

AP AC
2p 1

fx

3p 2
1

3

vs

vs (median concurrency point)

vs

YS
@

TOTAL:

200

2613 ~85- 21

Fea

. OEPARTHENT OF BASIC 7
i
PRETORIA

EDUCATION }
y |
u\v |
j 3X 140
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