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Western Cape Education Department
Examination Preparation Learning Resource 2016
GEOMETRY MEMORANDUM
MATHEMATICS Grade 12
Razzia Ebrahim
Senior Curriculum Planner for Mathematics
E-mail: [email protected]
Website: http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835-
grade-12?Itemid=-1
Website: http://wcedeportal.co.za
Tel: 021 467 2617
Cell: 083 708 0448
Index Page
1. 2016 Feb-March Paper 2 3 – 6
2. 2015 November Paper 2 7 – 9
3. 2015 June Paper 2 10 – 12
4. 2015 Feb-March Paper 2 13 – 16
5. 2014 November Paper 2 17 – 21
6. 2014 Exemplar Paper 2 22 – 24
7. 2013 November Paper 3 25 – 27
8. 2012 November Paper 3 28 – 31
9. 2011 November Paper 3 32 – 34
10. 2010 November Paper 3 35 – 38
11. 2009 November Paper 3 39 – 42
12. 2008 November Paper 3 43 – 45
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Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum
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QUESTION/VRAAG 9
9.1 ABCD is a ||m [diags of quad bisect each other/
hoekl v vh halveer mekaar] R
(1)
9.2 AFFE
DBED
= [Prop Th/Eweredigh st; DF | | BA]
CGGE
DBED
= [Prop Th/Eweredigh st; DG | | BC]
S R S R
(4) 9.3
CGGE
AFFE
= [proved/bewys]
∴AC | | FG [line divides two sides of ∆ in prop/ lyn verdeel 2 sye van ∆ eweredig]
22 FC = [alt/verw ∠s/e; AC | | FG]
21 CA = [alt/verw ∠s/e; AB | | CD] ∴ 21 FA =
S S R
S
S (5)
9.4 21 AA = [diags of rhombus/hoekl v ruit]
22 FA = [ 21 FA = ] ∴ ACGF = cyc quad/kdvh [∠s in the same seg =/ ∠e in dies segm =]
OR/OF
22 AC = [∠s opp equal sides of rhombus/ ∠e to gelyke sye v ruit]
22 GA = [alt/verw-∠s/e; AC | | FG] ∴ 22 GC = ∴ ACGF is a cyc quad/kdvh [∠s in the same seg =/ ∠e in dies segm =]
S
S R
(3)
S
S R
(3) [13]
B
C G E
F
1
1 1
1
2 3
3 2
2 2 3 3
O
D
A
3
Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum
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QUESTION/VRAAG 10 10.1 10.1.1 In ∆ADE and/en ∆PQR:
AD = PQ [construction/konstr] PA = [given/gegee]
AE = PR [construction/konstr] ∴∆ADE ≡ ∆PQR [S∠S]
all/al 3 S’s/e reason/rede
(2) 10.1.2
QEDA = [∆s ≡ ∴ corres/ooreenk ∠s/e =] But QB = [given/gegee] ∴ BEDA =
∴ DE | | BC [corres/ooreenk ∠s/e =]
QEDA =
BEDA = reason/rede
(3) 10.1.3
AEAC
ADAB
= [Prop Th/Eweredigh st; DE | | BC] But/Maar AD = PQ and/en AE = PR [construction/konstr]
∴ PRAC
PQAB
=
S/R
S
(2)
E
A
C B
D R Q
P
4
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10.2
10.2.1 line from centre to midpt of chord/lyn van midpt na midpt van
koord answ/antw
(1) 10.2.2 OP | | VS [Midpt Theorem/Midpt-stelling]
In ∆ROP and/en ∆RVS: RR = [common/gemeen] VO2 = [corresp/ooreenk ∠s/e; OP | | VS]
∴∆ROP | | | ∆RVS [∠,∠,∠]
OR/OF In ∆ROP and/en ∆RVS:
RSVP2 = [corresponding ∠s/ ooreenkomstige ∠'e] RR = [common/gemeen]
∴∆ROP | | | ∆RVS [∠,∠,∠]
S R S S & ∠;∠;∠ OR/OF 3 angles/hoeke
(4)
S R S S & ∠;∠;∠ OR/OF 3 angles/hoeke
(4)
S
V
O T
R
P
1
1
1
1 2
2 2
5
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10.2.3 In ∆RVS and/en ∆RST:
°== 90RTSRSV [∠ in semi-circle/∠ in halfsirkel] R is common/gemeen
RSTV = ∴∆RVS | | | ∆RST [∠,∠,∠]
S R S & ∠;∠;∠ OR/OF 3 angles/hoeke
(3)
10.2.4 In ∆RTS and/en ∆STV: °== 90STVSTR [∠ s on straight line/∠e op rt lyn]
R = 90° – RST = VST
VRST = ∴∆RTS | | | ∆STV [∠,∠,∠]
∴ VTTS
STRT
=
∴ VT.TRST 2 =
∆RTS & ∆STV S S S (with justification/met motivering) ∆RTS | | | ∆STV ratio/verh
(6) [21]
6
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QUESTION/VRAAG 10
10.1 °= 90CDB [∠ in semi circle/∠ in halfsirkel]
DC2 = 172 – 82 [Th of/stelling v Pythagoras] = 225 ∴ DC = 15
S using/gebruik Pyth korrek/ correctly answ/antw
(3) 10.2.1
CBCE
CDCF
= [line | | one side of ∆/lyn | | een sy van ∆]
∴41
15CF
=
∴ CF = 3,75
S/R subst correctly/ korrek answ/antw
(3) 10.2.2 In ∆BAC and/en ∆ FEC:
°= 90CBA [tan ⊥ diameter/raakl ⊥ middellyn] CFE = 90° [corresp ∠s/ooreenk ∠e; EF| |BD]
CC = [common/gemeen] ∴ ∆BAC | | | ∆ FEC [∠∠∠] OR/OF
In ∆BAC and ∆ FEC: °= 90CBA [tan ⊥ diameter/raakl ⊥ middellyn]
CFE = 90° [corresp ∠s/ooreenk ∠e; EF| |BD] CC = [common/gemeen]
CEFCAB = [∠ sum in ∆/∠ som van ∆] ∴ ∆BAC | | | ∆ FEC OR/OF
S R S/R S R
(5)
S R S/R S S
(5)
B
C
E
A
D
F
17
8
7
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ABEF = cyc quad/ kdvh [opp ∠s of quad supp/tos∠e v vh suppl)] ∴ ACEF = [ext ∠ of cyc quad/buite ∠ v kdvh] In ∆BAC and/en ∆ FEC:
ACEF = [proven/bewys] CC = [common/gemeen]
CEFCAB = [∠ sum in ∆/∠ som van ∆] ∴ ∆BAC | | | ∆ FEC OR/OF ABEF = cyc quad/ kdvh [opp ∠s of quad supp/tos∠ e v vh suppl)] ∴ ACEF = [ext ∠ of cyc quad/buite ∠ v kdvh] In ∆BAC and ∆ FEC:
ACEF = [proven/bewys] CC = [common/gemeen] ∴ ∆BAC | | | ∆ FEC [∠∠∠]
S R S/R S S
(5)
S R S/R S R
(5)
10.2.3 EC = 25,41741
=×
FCBC
ECAC
= [||| ∆s/e]
3,7517
4,25AC
=
∴AC = 19,27 or/of 19154
OR/OF
ACBC
CECFCcos ==
∴ AC17
25,475,3
=
∴ AC = 19,27 or/of 19154
length of/lengte v EC S subst correctly/ korrek answ/antw
(4)
correct ratios/ korrekte verh's subst correctly/ korrek answ/antw
(4)
10.2.4 AC is diameter of the circle [chord subtends 90° ] AC is middellyn van die sirkel [koord onderspan 90°]
∴radius = 27,1921
× = 9,63 or/of 93019
S/R answ/antw
(2) [17]
8
Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum
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QUESTION/VRAAG 11 11.1 equiangular or similar/gelykhoekig of gelykvormig
answ/antw
(1)
11.2.1
275,05,1
RNKP
== ; 212
NMPM
== ; 225.15,2
RMKM
==
∴ RMKM
NMPM
RNKP
==
∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig] OR/OF
21
5,175,0
KPRN
== ; 21
PMNM
= ; 21
5,225,1
KMRM
==
∴ KMRM
PMNM
KPRN
==
∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig]
all 3 statements/ al 3 bewerings
R (3)
OR/OF all 3 statements/ al 3 bewerings
R (3)
11.2.2 RMKP = ∴ P is common/gemeen ∴ ∆RPQ | | | ∆KPM [∠∠∠]
KMRQ
KPRP
= [||| ∆s]
∴ 2,5RQ
1,53,25
=
∴ RQ = 5,1
25,35,2 × = 5,42 or 5125
∴ NQ = 5,42 – 0,75 = 4,67 or 432
S
∆RPQ|||∆KPM S subst correctly/ korrek
RQ = 5125
NQ = answ/antw
(6) [10]
P
Q
N
R
M
K
1,5 2
2,5
1 1,25
0,75
•
•
♦ ♦
♦
♠
♠
9
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QUESTION/VRAAG 10 10.1 then the line is parallel to the third side/is die lyn ewewydig aan
die derde sy. S
(1)
10.2.1
53
2012
ACAE
==
53
AFAD
=
AFAD
ACAE
=∴ DE∴ || FC (line divides two sides of ∆ in prop/
lyn verdeel twee sye v ∆ in dieselfde verh)
S S R
(3) 10.2.2
208
BABF
=
(prop theorem/eweredigh st; BC || FE)
)14(208BF =∴
528BF =∴ OR/OF
535FB = OR/OF 6,5FB =
S/R substitute 14/ stel 14 in
answer/antw (3) [7]
A
B
C
D
F
E 12 8
10
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11.2.1 x2ACD = (EC bisector)
x=P (∠ at centre = 2 ×∠ at circumference/ midpts∠ = 2 × omtreks∠)
x== PA1 (tangent-chord theorem/rkl-kd st) In ∆BAD and ∆BCE:
BB = (common/gemeen)
1C1A = (proven above) ∴∆BAD | | | ∆BCE (∠∠∠)
OR/OF x2ACD = (EC bisector)
x=P (∠ at centre = 2 ×∠ at circumference/ midpts∠ = 2 × omtreks∠)
x== PA1 (tangent-chord theorem/rkl-kd st) In ∆BAD and ∆BCE:
BB = (common/gemeen)
1C1A = (proven above)
11 ED = ∴∆BAD | | | ∆BCE
S R S R
S S(with justification) R
(7)
S R
S R
S S(with justification) S
(7)
1
A
B
C
P
D
E
1
1
1
2
2
2
2
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11.2.2(a) °= 90CAB (tangent/raakl ⊥ radius)
∴BC2 = 82+ 62 = 100 (Pythagoras theorem/stelling) BC = 10 AC = DC = 6 (radii) ∴ BD = 10 – 6 = 4 units/eenhede
R substitution into Pyth theorem BC = 10 DC = 6 BD = 4
(5) 11.2.2(b)
BEBD
BCBA
= (∆BAD | | | ∆BCE)
∴BE4
108
=
∴ BE = 5 units/eenhede
S substitution/ substitusie BE = 5
(3) 11.2.2(c) AE = 3
In ∆ACE:
tan x = 63
∴x = 26,57° OR/OF
sin 2x = 108
∴ 2x = 53,1301... (2x < 90°) ∴ x = 26,57°
correct trig ratio/ korrekte trigvh correct trig eq/ korrekte trigvgl answer/antw
(3)
correct trig ratio/ korrekte trigvh correct trig eq/ korrekte trigvgl answer/antw
(3) [24]
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9.3
9.3.1 Equal chords subtend equal ∠s/Gelyke koorde onderspan gelyke ∠e
R
(1)
9.3.2 30°W4 = (tan chord theorem/rkl-koordst) 30°W1 =
answer/antw reason/rede
answer/antw (3)
9.3.3(a) 0°5WR 24 == (tan chord theorem/rkl-koordst)
232 WRS += (ext∠ of ∆/buite ∠ v ∆) ∴ 0°8S2 =
OR/OF
0°3RR 32 == (= chords subtend =∠s /= kde onderspan=∠e)
0°5WR 24 == (tan chord theorem/rkl-koordst) ∴ 0°8S2 =
S R S
(3)
S R S
(3)
2 1
P
R
Q
V T
W S
Z
1 2
1
1
2 3 4
2 3 4
30°
50°
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9.3.3(b) 0°8ST 22 == (ext ∠ of cyclic quad/buite∠ van koordevh)
24 TWV =+ (ext∠ of ∆/buite∠ van ∆) ∴ 0°5V =
S R S S
(4) 9.3.4 In ∆RVW and/en ∆RWS:
0°3RR 32 == (proven/bewys in 9.3.1)
0°5WV 2 == (proven/bewys in 9.3.3)
1SRWV = )in 3rd( ∆∠ ∴∆RVW | | | ∆RWS (∠∠∠)
∴WRRS
RVWR
=
RWS) | | |RVW( ∆∆
∴ RS.RVWR 2 =
using the correct ∆s/ gebruik korrekte ∆e S S R
)in 3rd( ∆∠ or ) (∠∠∠ S
(5) [22]
14
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QUESTION/VRAAG 10
10.1.1 corresponding ∠s/ooreenkomstige∠e; PN | | RT
answer/antw
(1) 10.1.2 ∠; ∠; ∠ OR/OF ∠; ∠ answer/antw
(1) 10.2
RTPN
RMPM
= ( ΔRTM|||ΔPNM )
313PNPN
=
=
S S
(2) 10.3
31
RMPM
= 32
RMRP
=∴
RN² – PN² = (RM² + NM²) – (PM² + NM²) (Pyth) = RM² – PM²
= 22
RP21RP
23
−
= 22 RP41RP
49
−
= 2RP²
OR/OF
Use of Pyth. for RN2 and PN2
RM = RP23
RP21PM =
2RP49 & 2RP
41
(4)
T
N
R P M 1
1
15
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RN2 – PN2 = (RM2 + NM2) – (PM2 + NM2) (Pyth)
= RM2 – PM2 = (3PM)2 – PM2 = 8PM2 = 2(2PM)2 = 2RP2
OR/OF RN2 – PN2 = (RM2 + NM2) – (PM2 + NM2) (Pyth) = RM2 – PM2 = (RP + PM)2 – PM2 = RP2 + 2RP.PM + PM2 – PM2
= RP2 + 2RP.
RP21
= 2RP2
Use of Pyth. for RN2 and PN2 RM = RP + PM (3PM)2 – PM2 RP = 2PM
(4)
Use of Pyth. for RN2 and PN2 RM = RP + PM expansion/ uitbreiding
RP21PM =
(4) [8]
16
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QUESTION/VRAAG 9 9.1
9.1.1 Same base (DE) and same height (between parallel lines)
Dieselfde basis (DE) en dieselfde hoogte (tussen ewewydige lyne) same base/dies basis between | | lines/ tussen | | lyne
(1) 9.1.2
DBAD
k
k
×
×
EC21
AE21
But/Maar area ∆DEB = area ∆DEC (Same base and same height/dieselfde basis en dieselfde hoogte)
DECareaADEarea
DEBareaADEarea
∆∆
=∆∆
∴
ECAE
DBAD
=∴
S S S R S
(5)
A
B
C
D E
h1
k
17
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9.2
9.2.1
ADFD
AMEM
= (Line parallel one side of ∆
OR prop th; EF | | BD) (Lyn ewewydig aan sy v ∆
73
AMEM
= OF eweredigst; EF ||BD)
S R answer/antw
(3)
9.2.2 CM = AM (diags of parm bisect/hoekl parm halv)
37
MEAM
MECM
== (from 9.2.1/vanaf 9.2.1)
S R answer/antw
(3) 9.2.3 h of ∆FDC = h of ∆BDC (AD | | BC)
73
)parmofsides(oppADFD
.B21
.FD21
BDCareaFDCarea
=
==
=∆∆
hC
h
OR/OF
heights) (same 73
ADFD
ADCareaFDCarea
==∆∆
But
ADCArea ∆ = BDCArea ∆ (diags of parm bisect area)
73
BDCareaFDCarea
=∆∆
AD | | BC subst into area form/ subst in opp formule S answer/antw
(4)
S R S answer/antw
(4) [16]
A
B C
D
M
E
F
G
(tos sye v parm =)
(dieselfde hoogtes)
(hoekl v parm halv opp)
18
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QUESTION/VRAAG 10
10.1.1 Tangent chord theorem/Raaklyn-koordstelling
R (1)
10.1.2 Tangent chord theorem/Raaklyn-koordstelling
R (1)
10.1.3 Corresponding angles equal/Ooreenkomstige ∠e gelyk R (1)
10.1.4 ∠s subtended by chord PQ OR ∠s in same segment ∠e onderspan deur dieselfde koord OF ∠e in dieselfde segment
R (1)
10.1.5 alternate ∠s/verwisselende ∠e ; WT | | SP R (1)
10.2 RPRT
RSRW
= (Line parallel one side of ∆ OR
prop th; WT | | SP)
∴RS
WR.RPRT =
OR/OF ∆RTW | | | ∆RPS (∠; ∠; ∠)
RPRT
RSRW
=∴ (∆RTW | | | ∆RPS)
∴RS
RW.RPRT =
S R
(2)
S
S
(2)
10.3 32 RTy == (tan chord theorem/Rkl-koordst)
13 QR ==y (∠s in same segment/∠e in dieselfde segment)
S R
S R (4)
Y
X
R
W
S
P
Q
T
1 2 3
4
1 2
1 2
3
1 2
1 2
1 2 3 4
X
R
W
S
P
T
y
1 2 3
4
1 2
1 2
3
1 2
1 2
1 2 3 4
x
(Lyn ewewydig aan sy v ∆ OF eweredighst: WT | | SP)
19
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10.4 RSPQ3 = (ext ∠ of cyc quad/buite∠ v kdvh)
2WRSP = (corresp∠s/ooreenk ∠e ; WT | | SP) ∴ 23 WQ = OR/OF
x=2Q (∠s in same segment/∠e in dies segment) )(180Q3 yx +−°= (∠s on straight line/∠e op reguitlyn)
)(180W2 yx +−°= (∠s of ∆WRT/∠e v ∆WRT ) ∴ 23 WQ =
S R S
(3)
R S S
(3)
10.5 In ∆RTS and ∆RQP: y== 23 RR (proven above/hierbo bewys)
22 PS = (∠s in same segment/∠e in dies segment) PQRSTR = (3rd angle of ∆)
∴∆RTS | | | ∆RQP (∠; ∠; ∠)
S S/R
S OR/OF (∠; ∠; ∠)
(3)
Y
X
R
W
S
P
Q
T
1 2 3
4
1 2
1 2
3
1 2
1 2
1 2 3 4
X
R
W
S
P
T
y
1 2 3
4
1 2
1 2
3
1 2
1 2
1 2 3 4
x
x
x
y
y
y
y
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10.6
RPRS
RQRT
= (∆RTS | | | ∆RQP)
RQRW
RQRS
RSRW
RQRS
RPRT
RPRS
RPRS
RQRT
RPRS
RPRS
2
=
=
=
×=×
OR/OF
RPRS
RQRT
= (∆RTS | | | ∆RQP)
But RS
WR.RPRT = (proven in 10.2/bewys in 10.2)
2
2
22
RPRS
RQWR
RQ.RSWR.RP
RPRS
RQ.RSWR.RP
RQRT
=∴
=
==∴
OR/OF
RPRQ
RSRT
= (∆RTS | | | ∆RQP)
RPRT.RS WRand
RSRT.RPRQ
=
=
(proven in 10.2/bewys in 10.2)
RT.RPRS
RPRT.RS
RSRT.RP
RPRT.RS
RQWR
×=
=
2
2
RPRS =
S
RPRS
× on both
sides
RQRS
RPRT
(3) S
RS
WR.RPRT =
multiplication/ vermenigvuldig
(3)
S
RP
RT.RS WR =
simplification/ vereenvoudiging
(3) [20]
(proven in 10.2/bewys in 10.2)
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VRAAG 9 9.1 x== AD4 (rkl-koordstelling)
x== 2DA (∠e tos gelyke sye)
x=A rede x== 2DA
(S/R) (3)
9.2 1M = 2x (buite ∠v∆) OF (∠ by midpt = 2∠ by omtr)
EDM = 90° (radius ⊥ rkl) x290M 2 −°=
∴x
x2
)290(90 180E=
−°+°−°= (som v ∠e in ∆MDE)
∴ CM is ‘n rkl (omgek rkl-koordst)
1M = 2x (S/R)
EDM = 90° (S/R)
x2E = rede
(4) 9.3 °= 90M3 (EM ⊥ AC)
BDA = 90° (∠ in halfsirkel) ∴ FMBD is koordevh (buite∠ v vh = tos binne ∠)
OF °= 90CME (EM ⊥ AC)
BDA = 90° (∠ in halfsirkel) ∴ FMBD is koordevh (tos ∠e v vh suppl)
°= 90M3
BDA = 90° (S/R) rede
(3) °= 90CME BDA = 90° (S/R) rede
(3) 9.4 DC2 = MC2 – MD2 (Pythagoras)
= (3BC)2 – (2BC)2 (MB = MD = radii) = 9BC2 – 4BC2 = 5BC2
Pythagoras substitusie 9BC2 – 4BC2
(3) 9.5 In ∆DBC en ∆DFM:
x== 24 DD (bewys in 9.1)
21 FB = (buite ∠ v koordevh)
2MC = ∴∆DBC | | | ∆DFM (∠; ∠; ∠)
24 DD = 21 FB = rede 2MC = of (∠; ∠; ∠)
(4) 9.6
BCDC
FMDM
= (∆DBC | | | ∆DFM)
BCBC5
=
5=
S antwoord
(2) [19]
22
Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum
Kopiereg voorbehou Blaai om asseblief
VRAAG 10 10.1
Konstruksie: Verbind DC en BE en trek hoogtes k en h
DBAD
.DB.
.AD.
ΔDEBoppΔADEopp
2121
==k
k (gelyke hoogtes)
ECAE
.E.
.AE.
ΔDECoppΔADEopp
2121
==hC
h (gelyke hoogtes)
Maar Opp ∆DEB = Opp ∆DEC (dies basis, dies hoogte)
∴ ΔDECoppΔADEopp
ΔDEBoppΔADEopp
=
∴ ECAE
DBAD
=
konstruksie
DBAD
ΔDEBoppΔADEopp
=
rede
ECAE
ΔDECoppΔADEopp
=
Area ∆DEB = Area
∆DEC (S/R)
ΔDECoppΔADEopp
ΔDEBoppΔADEopp
=
(6)
A
B C
D E k h
23
Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum
Kopiereg voorbehou
10.2.1
CDAC
BEAB
= (Ewered st; BC | | ED)
CD3
31
= ∴ CD = 9 eenhede
CDAC
BEAB
=
(S/R) substitusie antwoord
(3) 10.2.2
FEFD
GADG
= (Ewered st; FG | | EA)
63
39
=+−
xx
54 – 6x = 9 + 3x –9x = –45 x = 5
FEFD
GADG
= (S/R)
substitusie vereenvoudig antwoord
(4) 10.2.3 In ∆ABC en ∆AED:
A is gemeen ECBA = (ooreenk ∠s; BC | | ED) DBCA = (ooreenk ∠s; BC | | ED) ∆ABC | | | ∆AED (∠, ∠, ∠)
∴ADAC
EDBC
=
123
9BC
=
BC = 241 eenhede
A is gemeen ECBA = (S/R) DBCA = (S/R) of (∠; ∠; ∠)
ADAC
EDBC
=
antwoord
(5) 10.2.4
DGD.FD.sin
BCAAC.BC.sin
ΔGFDoppΔABCopp
21
21
=
169
Dsin)3)(4(21
Dsin)412)(3(
21
=
=
gebruik v opp reël korrekte sye en ∠e substitusie v waardes DsinBCAsin =
(S/R) antwoord
(5) [23]
(ooreenk ∠s; BC | | ED)
24
Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 9 9.1 AF : FE
= 2 : 1 (Prop Th; FB || EC) (Eweredigheid St; FB || EC)
answer reason
(2) 9.2
12
FEAF
=
428
2AFFE === cm
AE = 12 cm
21
AEED
= (BE || DC; Prop Th) / (BE || DC; Eweredigheid St)
21
12ED
=
ED = 6 cm
FE = 4 cm AE = 12 cm
21
AEED
=
answer
(4) [6]
A B C
D
E
F
2x x
25
Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 12
12.1 x=3K (tan ch th)
(raaklyn koord) x=2Y (∠s in same seg)
(∠e in selfde seg) x=2K (∠s opp = radii) / (= chs subt = ∠s)
(∠e oork = radiusse) / (= koorde = ∠e) x=2W (∠s in same seg) / (= chs subt = ∠s)
(∠e in selfde seg) / (= koorde = ∠e)
x=3K tan ch th x=2Y ∠s in same seg x=2K reason x=2W reason
(8) 12.2 x2180OO 41 −°=+ (sum of int ∠’s of ∆) / (opp∠ cyclic quad)
(som van binne∠e ∆) / (oorst ∠ koordevierhoek) x−°= 90T (∠ at circ cent = 2∠ at circumference)
x2180OO 41 −°=+ reason reason
(3)
K
E
T
Y
W
O
1 2 3
1 2
4 3
1 2
1 2
1
2 3 x
x
x
x
x
26
Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum
Copyright reserved/Kopiereg voorbehou
12.3 °= 90E2 (sum of int ∠’s of ∆) / (som van binne∠e ∆)
KE = ET (⊥ from centre to chord bisects chord) (⊥ van middelpunt tot koord halveer koord) OR
x−°=++ 90KKK 321 (sum of int ∠’s of ∆) In ∆KWE and ∆TEW 1. x−°==++ 90TKKK 321 (proven above)
2. x== 21 WW (Proven in 12.1) 3. WE is common ∴∆KWE ≡ ∆TEW (∠∠S)
KE = ET
°= 90E2 sum of int ∠’s of ∆ ⊥ from centre to chord
bisects chord (3)
x−°=++ 90KKK 321 sum of int ∠’s of ∆ ∆KWE ≡ ∆TEW
(3)
12.4 In ∆KOE and ∆WTE i. x== 23 WK (proven)
ii. °== 90EE 12 (∠s on str line / sum of int ∠’s of ∆) (∠e op reguit lyn / som van binne∠e ∆)
iii. x−°== 90TO2 (3rd ∠ of ∆) ∆KOE ||| ∆WTE (∠∠∠)
TEOE
WEKE
= (||| ∆s)
KE = TE (proven)
OE.WEKEOE.WEKE.TE
2 =
=
OR In ∆KOE and ∆KWE i. x== 13 WK (proven)
ii. 1E is common iii. =2O 321 KKK ++ (3rd ∠ of ∆) ∆KOE ||| ∆WKE (∠∠∠)
KEOE
WEKE
= (||| ∆s)
OE.WEKE2 =
∆KOE and ∆WTE x== 23 WK
°== 90EE 12 x−°== 90TO2
TEOE
WEKE
=
KE = TE
(6)
∆KOE and ∆WTE x== 13 WK
1E is common ∠∠∠
KEOE
WEKE
=
(6) [20]
27
Mathematics/P3 DBE/November 2012 NSC – Memorandum
Copyright reserved Please turn over
QUESTION 7
7.1 Draw a point P on FG such that FP = LM and a point Q on FH
such that FQ = LN. In ∆FPQ and ∆LMN
1. LF = (given) 2. FP = LM (construction) 3. FQ = LN (construction)
∴∆FPQ ≡∆LMN (SAS)
NMLQPF = (≡∆s) But NMLHGF = (given)
HGFQPF = PQ || GH (corresponding angles =)
FHFQ
FGFP
= (PQ || GH ; Prop Th)
FHLN
FGLM
=
construction All three statements must be given ∆FPQ ≡∆LMN (SAS) PQ || GH
(7)
NMLQPF =
HGFQPF =
FHFQ
FGFP
=
G H
F
P Q
L
M N
Note: No construction constitutes a breakdown, hence no marks
28
Mathematics/P3 DBE/November 2012 NSC – Memorandum
Copyright reserved Please turn over
7.2
TKVT
PRVP
= (PT || RK;Prop Th)
8162610264
9102
===−
=−
xx
x
x
OR
VKVT
VRVP
= (PT || RK; Prop Th)
89612
4810020104
12102
==
−=−
=−
−
xx
xxx
x
TKVT
PRVP
=
(PT || RK; Prop Th) substitution answer
(4)
VKVT
VRVP
=
(PT || RK; Prop Th) substitution answer
(4) [11]
9
2x – 10
6
9
K
T P
4 2x – 10
6
9
V
R
29
Mathematics/P3 DBE/November 2012 NSC – Memorandum
Copyright reserved Please turn over
C x
A
O
B
K
T 1 2 3
4
1 2
1 2 3
2x
360° – 2x
180° – x
x x
QUESTION 9
9.1 x2BOA = (∠circ centre = 2 ∠ circumference) x2180T −°= (opp∠ cyclic quad suppl)
x2BOA = ∠circ centre = 2 ∠ circumference opp∠ cyclic quad suppl
(3) 9.2 x=TAC (∠ sum ∆)
x=1K (ext∠ cyclic quad)
1KTAC = BK || AC (corresponding ∠s =) OR
x== CK1 (ext∠ cyclic quad) x=4B (∠ sum ∆)
x== CB4 BK || CA (corresponding ∠s =) OR
x=TAC (∠ sum ∆) x−°= 180AKB (opp∠ cyclic quad)
°=+ 180AKBTAC BK || AC (coint∠s supp)
x=TAC ∠ sum ∆ x=1K ext∠ cyclic quad corresponding ∠s =
(5) x== CK1 ext∠ cyclic quad x=4B ∠ sum ∆ corresponding ∠s =
(5) x=TAC ∠ sum ∆ x−°= 180AKB opp∠ cyclic quad co-int∠s supp
(5)
30
Mathematics/P3 DBE/November 2012 NSC – Memorandum
Copyright reserved Please turn over
9.3 In ∆BKT and ∆CAT
1. 1KTAC = (= x) 2. T is common 3. 4BTCA = (∠ sum ∆)
∆BKT ||| ∆CAT (∠∠∠)
1KTAC = T is common ∠∠∠
(3)
9.4 KTAT
KBAC
= (||| ∆s)
27
KBAC
=
KTAT
KBAC
=
||| ∆s answer
(3) [14]
31
Mathematics/P3 DBE/November 2011 NSC – Memorandum
Copyright reserved Please turn over
QUESTION 9
9. °= 90C (∠s in semi circle) °= 90AEO (corres ∠s; OD || BC)
AE = 8 cm (line from circ cent ⊥ ch bis ch) OE = 6 cm (Pythagoras) ED = 10 – 6 = 4 cm OR
°= 90C (∠s in semi circle) °= 90AEO (corres ∠s; OD || BC)
OE || BC (given) OA = OB (radii) AE = EC = 8cm (midpoint theorem) OE = 6 cm (Pythagoras) ED = 10 – 6 = 4 cm OR
°= 90C (∠s in semi circle) BC2 = (20)2 – (16)2 BC2 = 144 BC = 12
OE = 21 BC (midpoint theorem)
OE = 6 cm OD = 10cm ED = 10 – 6 = 4 cm OR
°= 90C (∠s in semi circle) BC2 = (20)2 – (16)2 BC2 = 144 BC = 12
BC21OE = (midpoint theorem)
OE = 6 cm ED = 4cm
°= 90C °= 90AEO line from circ
cent ⊥ ch bis ch OE = 6 cm ED = 4 cm
°= 90C °= 90AEO midpoint
theorem OE = 6 cm ED = 4 cm
°= 90C BC = 12 reason OE = 6 cm ED = 4 cm
[5]
°= 90C BC = 12 reason
OE = 6 cm ED = 4 cm
[5]
A
C
B
D
E
O
32
Mathematics/P3 DBE/November 2011 NSC – Memorandum
Copyright reserved Please turn over
QUESTION 10
10.1 x== 4DA (tan ch th) x=2E (tan ch th) OR (∠s in same seg)
x== AD 2 (alt ∠s; CA || DF)
x=A tan ch th
x=2E reason
x=2D alt ∠s; CA || DF
(6)10.2 In ΔBHD and Δ FED
1. FB2 = (∠s in same seg) 2. 13 DD = (= chs subt = ∠s) ΔBHD ||| Δ FED (∠∠∠)
FB2 = ∠s in same seg 13 DD = = chs subt = ∠s ∠∠∠
(5)10.3
BDFD
BHFE
= (||| Δs)
But FE = AB (given)
BDFD
BHAB
=
AB.BD = FD.BH
BDFD
BHFE
=
FE = AB (2)
[13]
D
A
B
C
E
F
4
1
1
1 1G
H 2
2 2
2
3
3
31 2
3
y
x
xxx
y
33
Mathematics/P3 DBE/November 2011 NSC – Memorandum
Copyright reserved Please turn over
QUESTION 11
11.1 AF = FC (diags of parallelogram bisect) FE || CD AE = ED (Prop Th; FE || CD) OR (Midpoint Theorem)
AF = FC
reason (2)
11.2 21
CPAC
= (given)
21
DQAD
= (given)
DQAD
CPAC
=
CD || PQ (converse proportionality theorem) CD || FE (given) ∴ PQ || FE OR
31
APAC
=
31
AQAD
=
AQAD
APAC
=
CD || PQ (converse proportionality theorem) CD || FE (given) ∴ PQ || FE OR
61
AQAE
61
APAF
=
=
AQAE
APAF
=
∴ PQ || FE (converse proportionality theorem)
ratios equal
CD || PQ reason: converse
prop th and conclusion (3)
ratios equal
CD || PQ reason: converse
prop th and conclusion (3)
61
APAF
=
AQAE
APAF
=
conv prop theorem
P Q
C D
B A
F E
34
Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2
8.2.1 x=4B (tan chord theorem)
x== 4BA (corres ∠; BD || AO) x=2B (BO = EO = radii)
x=4B tan chord theorem x== 4BA with
reason x=2B
(4)8.2.2 °= 90EBD (∠ in semi-circle)
x+°= 90EBC OR
°= 90OBC (rad ⊥ tan) x+°= 90EBC
OR
x2O1 = (∠ circ cent) x−°== 90DB 13 (radii)
xxxx
+°=+−°+=
90)90(EBC
°= 90EBD ∠ in semi-circle x+°= 90EBC
(3)
°= 90OBC rad ⊥ tan x+°= 90EBC
(3)
x2O1 = ∠ circ cent x+°= 90EBC
(3)
8.2.3 °= 90EBD (proved in 8.2.2) °= 90OFB (co-int angles supp; BD || AO)
BF = FE (line from circ cent ⊥ ch bisect ch) F is the midpoint of EB
°= 90EBD °= 90OFB and
reason BF = FE line from circ cent
⊥ ch bisect ch) (4)
A
O
E
F
B
D
C
x
1 234
1
2
1 2
32 1
Note: If start with x=A and do not use tan ch th: max 2 marks
35
Mathematics/P3 DBE/November 2010 NSC - Memorandum
OR OD = OE (radii) BF = FE (BD || AO) F is the midpoint of EB OR
°== 90OFEOFB (BD || AO) OF is common BO = OE (radii) ΔBOF ≡ ΔEOF (90°HS) BF = FE (≡ Δs) OR
x== AB2 (proven)
2O is common ΔAOB ||| ΔBOF (AAA)
OFBOBA = °= 90OBA (proven)
°== 90OFBOBA BF = FE (line from circ cent ⊥ ch bisects ch) OR
°= 90EBD (∠ in semi-circle) x−°= 90B3
x−°= 90O2 (alt ∠s; BD || FO) °= 90F1 (∠ sum Δ)
BF = FE (line from circ cent ⊥ ch bisects ch) OR In ΔOBF and ΔOEF
1. OB = OE (radii) 2. °== 90OFEOFB (BD || AO) 3. EB2 = (radii)
ΔOBF ≡ ΔOEF (AAS) BF = FE
OD = OE radii BF = FE BD || AO
(4)
°== 90OFEOFB (BD || AO)
BO = OE ΔBOF ≡ ΔEOF BF = FE
(4)
ΔAOB ||| ΔBOF
OFBOBA =
BF = FE line from circ cent
⊥ ch bisects ch (4)
°= 90EBD °= 90F1 BF = FE line from circ cent
⊥ ch bisects ch (4)
OB = OE °== 90OFEOFB
(BD || AO) ΔOBF ≡ ΔOEF BF = FE
(4)
8.2.4 In ΔCBD and ΔCEB 1. 4BE = x= (proven in 8.2.1) 2. C is common 3. x+°== 90EBCD4 ΔCBD ||| ΔCEB (AAA)
4BE = x= C is common
Or
x+°== 90EBCD4 Any two of the above
(2)
36
Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2.5
CBCE
BDEB
= (sim Δs ∴ sides in proportion)
EB.CB = CE.BD but EB = 2EF (F is the midpoint of BE) 2EF.CB = CE.BD
CBCE
BDEB
=
EB.CB = CE.BD EB = 2EF
(3)[21]
QUESTION 9 9. DA = (∠ in same seg)
CB = (∠ in same seg) CEDBEA = (vert opp ∠s)
ΔDEC ||| ΔAEB (∠∠∠)
ABDC
EBEC
AEDE
== (sides in prop)
Let AC = 11a
87
4a
ax=
2
2
5,38
28
ax
ax
=
=
ay
78
8= (sides in prop)
ay
764
=
If candidate proves similarity of two triangles: full marks. If candidate does not prove similarity max 3 marks. The triangles have to be in the correct order in order to be given 3 marks.
DA = S/R CB = S/R CEDBEA =
ΔDEC ||| ΔAEB (∠∠∠)
[6]
A
B
C
D
E
4a 7a
x 8
8y
37
Mathematics/P3 DBE/November 2010 NSC - Memorandum QUESTION 10
10.1 °= 90CEM (tan ⊥ rad) °= 90CDM (line from cent bisects ch)
°=+ 180CDMCEM ∴MDCE a cyclic quad (opp ∠s of quad supplementary) OR
°= 90CEM (tan ⊥ rad) °= 90ADM (line from cent bisects ch) ADMCEM =
∴MDCE a cyclic quad (ext ∠ quad = int opp)
°= 90CEM (tan ⊥ rad)
°= 90CDM opp ∠s of
quad supplementary
(3) °= 90CEM
(tan ⊥ rad) °= 90ADM ext ∠ quad =
int opp (3)
10.2 MD2 = MB2 – DB2 (Pythagoras; ΔMBD)
MC2 = MD2 + DC2 (Pythagoras; ΔMDC) = MB2 – DB2 + DC2
MD2 = MB2 – DB2
Pythagoras MC2 = MD2
+ DC2 (3)
10.3 DB = 30 (given) MB = 40 (radii) MC2 = (40)2 + (50)2 − (30)2 = 3 200 MC = 240 = 56,57 MC2 = ME2 + CE2 (Pythagoras) CE2 = 3 200 − 1 600 CE2 = 1 600 CE = 40 mm OR MC2 = CE2 + ME2 – 2CE.ME.cos CEM
40CE1600CE
1600CE90cos)40(CE2)40(CE3200
2
2
22
==
+=
°−+= ..
MB = ME DB = 30 MC2 = 3200
or MC = 240 or MC = 56,57
answer (4)
cosine rule ME = 40 MC2 = 3200
answer
(4)[10]
A
M
D B
C
E
F
|| ||
NOTE: If the word cyclic is used in the last reason: max 2 / 3 marks
38
Mathematics/P3 DoE/November 2009 NSC – Memorandum
• Consistent Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt.
Copyright reserved Please turn over
QUESTION 9
C
O B
M
A
9.1 °= 90ACB (∠’s in a semi-circle) answer (1) 9.2.1 22 810AC −= (Pythagoras)
36= = 6 AM = 3 (line from circle centre ⊥ chord bisects chord OR midpoint theorem)
diameter = 10
AC
AM (3)
9.2.2 22 35OM −= (Pythagoras) = 4 (OR midpoint theorem) Area ΔAOM : Area ΔABC
= 3.4.21 : 6.8.
21
= 6 : 24 = 1 : 4 OR Area ΔAOM : Area ΔABC
= MAOsin.OM.AM.21 : CABsin.AC.AB.
21
= 3.4.21 : 6.8.
21
= 6 : 24
OM
substitution
answer (3)
[7]
39
Mathematics/P3 DoE/November 2009 NSC – Memorandum
• Consistent Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt.
Copyright reserved Please turn over
QUESTION 10 10.1.1
12
HEAH
= (GHB || FEC)
AH = 2y
D
A
C
B
E
G
F
H
4
2
1
2
1
2
1,5
HE = y
12
EDAE
= (BE || CD)
ED = 1,5 y
34
EDAH
5,12
EDAH
=
=
statement reason
ED = 1,5y
answer (4)
10.1.2 64
=CDBE (ΔAEB ||| ΔADC)
If learner stops at 2 : 1,5 : no penalty
32
=
answer reason (2)
10.2 HE = 2 cm (given) AH = 4 cm ED = 3 cm AD.HE = (AH + HE + ED).HE = (4 + 2 + 3).(2) = 18
AH and ED
AD = AH +
HE + ED (2)
[8]
40
Mathematics/P3 DoE/November 2009 NSC – Memorandum
• Consistent Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt.
Copyright reserved Please turn over
QUESTION 11
B
C
D
O
11.1 41 AD = (tan-chord theorem)
(alt ∠’s , BA || CE) 2C= OR
22 DC = (∠'s in same seg) (tan-chord theorem) 1A=
(alt ∠’s , BA || CE) 2E=
(∠'s in same seg) 1D= OR
°=+ 90AA 43 (tan ⊥ rad)
°= 90F1 (AB || EC; coint ∠s) In ΔAFC: (∠ sum Δ) 32 A90C −°=
°=+ 90CC 21 (∠s in semi circle) In ΔADC: (∠ sum Δ) 31 A90D −°=
21 CD =
Statement Reason S/R (3)
Statement Reason S/R
(3)
•
E A
F
1 2
1 2
1
2 3
1
23 4
12
34
41
Mathematics/P3 DoE/November 2009 NSC – Memorandum
• Consistent Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt.
Copyright reserved Please turn over
11.2 In ΔACF and ΔADC
1. is common 3A is common
3A
2. (proved) 12 DC = 12 DC =ΔACF ||| ΔADC (∠∠∠) Reason
(3)OR In ΔACF and ΔADC 1. is common 3A
2. (proved) 12ˆˆ DC =
3. (remaining ∠s in triangles) DCAF1 =ΔACF ||| ΔADC
11.3 ADAC
ACAF
= (sim Δ’s ∴ sides in proportion) statement
ADAF44
ADAF
ADADAD.
AF
AD21AOAC
ADAC.ACAF
21
21
=
=
=
==
=
Statement
Simplification
Substitution
(4)
(2radius = diameter)
OR ΔAOC is equilateral S/R
°==∴ 60 A COA 3 Statement
AF4AD
AD41AF
)AD21(
21AF
AO21AC
21AF
21
ACAF60cos
=
=
=
==
==° Simplification
Substitution
(4) [10]
(2radius = diameter)
42
Mathematics/P3 DoE/November 2008 NSC – Memorandum
Copyright reserved Please turn over
QUESTION 9
9.1 °=∧
90R1 …( angle in a semi-circle)
9.2 x−°=∧
90P2 …( angle between radius and tangent)
∧∧
−°= 2P90S …( ext. angle of Triangle)(sum of angles of triangle) = 90° – ( 90° – x ) = x
∴ x==∧∧
SP1
9.3 x==∧∧
12 PW …( angles in the same segment)
Also x=∧
S …( proved 9.2) ∧∧
= SW2 ∴SRWT is a cyclic quad…(ext angle = int. opposite angle) 9.4 In ∆ QWR ; ∆ QST
∧∧
= SW2 ….( proved 9.3)
1Q∧
is common
2TQRW∧∧
= ….(remaining angles) ∆ QWR ||| ∆ QST (AAA) or (∠∠∠) or equiangular
angle in a semi-circle (1)
x−°=∧
90P2
∧∧
−°= 2P90S 90° -( 90° - x ) = x
(3)
x==∧∧
1PRWQ
∧∧
= SRWQ reason (3)
TSQRWQ∧∧
=
WQR∧
is common
AAA or ∠∠∠ or equiangular or 3rd angle equal
(3)
P Q
R
W T
S
x 1
1
2
2 12
31 2
12
43
Mathematics/P3 DoE/November 2008 NSC – Memorandum
Copyright reserved Please turn over
9.5.1 QRQT
RWTS
= ….. ∆ QWR ||| ∆ QST
cm 4TS
164TS48
2TS
=∴=
=∴
9.5.2
cm
cm
6RQSQSR
102
54SQ
RWTS
WQSQ
=−=∴
=×
=
=
QRQT
RWTS
=
48
2TS
=
cm 4TS = (3)
RWTS
WQSQ
=
cm10
cm6 (3)
[16]
44
Mathematics/P3 DoE/November 2008 NSC – Memorandum
Copyright reserved
QUESTION 10 10.1
21
TACT
EDCE
==
10.2 From 10.1 21
EDCE
=
But DC = 9 cm ∴ DE = 6 cm = BD. ∴D is the midpoint of BE.
10.3
cm 4 TE24TE6126
TE2
BEBD
TEFD
==×
=
=
ALTERNATIVE D is the midpoint of BE. (from 10.2)
Then F is the midpoint of BT. … (sides in proportion) ∴ TE = 2FD (midpoint theorem) = 4 cm
answer (1)
use of ratio
DE = 6 cm (2)
proportion
answer (2)
proportion
answer (2)
A
BD
EC
T
F
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