MATHEMATICS Grade 12

Western Cape Education Department

Examination Preparation Learning Resource 2016

GEOMETRY MEMORANDUM

Razzia Ebrahim

Senior Curriculum Planner for Mathematics

E-mail: Razzia.Ebrahim@wced.info

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

6. 2014 Exemplar Paper 2 22 – 24

10. 2010 November Paper 3 35 – 38

11. 2009 November Paper 3 39 – 42

12. 2008 November Paper 3 43 – 45

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

QUESTION/VRAAG 9

9.1 ABCD is a ||m [diags of quad bisect each other/

hoekl v vh halveer mekaar] R

9.2 AFFE

= [Prop Th/Eweredigh st; DF | | BA]

= [Prop Th/Eweredigh st; DG | | BC]

S R S R

(4) 9.3

= [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 =

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 =]

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 =]

(3) [13]

2 2 3 3

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

= [Prop Th/Eweredigh st; DE | | BC] But/Maar AD = PQ and/en AE = PR [construction/konstr]

∴ PRAC

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

Copyright reserved/Kopiereg voorbehou

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

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

∴ VT.TRST 2 =

∆RTS & ∆STV S S S (with justification/met motivering) ∆RTS | | | ∆STV ratio/verh

(6) [21]

Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum

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

= [line | | one side of ∆/lyn | | een sy van ∆]

∴ 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

S R S/R S S

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

S R S/R S R

10.2.3 EC = 25,41741

= [||| ∆s/e]

3,7517

4,25AC

∴AC = 19,27 or/of 19154

CECFCcos ==

∴ AC17

25,475,3

∴ AC = 19,27 or/of 19154

length of/lengte v EC S subst correctly/ korrek answ/antw

correct ratios/ korrekte verh's subst correctly/ korrek answ/antw

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]

QUESTION/VRAAG 11 11.1 equiangular or similar/gelykhoekig of gelykvormig

answ/antw

11.2.1

275,05,1

== ; 212

== ; 225.15,2

∴ RMKM

∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig] OR/OF

5,175,0

== ; 21

= ; 21

5,225,1

∴ KMRM

∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig]

all 3 statements/ al 3 bewerings

OR/OF all 3 statements/ al 3 bewerings

11.2.2 RMKP = ∴ P is common/gemeen ∴ ∆RPQ | | | ∆KPM [∠∠∠]

= [||| ∆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

∆RPQ|||∆KPM S subst correctly/ korrek

RQ = 5125

NQ = answ/antw

(6) [10]

1 1,25

♦ ♦

Mathematics/P2/Wiskunde V2 DBE/2015 NSCNSS – Memorandum

QUESTION/VRAAG 10 10.1 then the line is parallel to the third side/is die lyn ewewydig aan

die derde sy. S

10.2.1

=∴ DE∴ || FC (line divides two sides of ∆ in prop/

lyn verdeel twee sye v ∆ in dieselfde verh)

(3) 10.2.2

(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]

E 12 8

Copyright reserved Please turn over/Blaai om asseblief

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

S S(with justification) S

Copyright reserved

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)

= (∆BAD | | | ∆BCE)

∴BE4

∴ 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) [24]

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum

9.3.1 Equal chords subtend equal ∠s/Gelyke koorde onderspan gelyke ∠e

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 =

0°3RR 32 == (= chords subtend =∠s /= kde onderspan=∠e)

0°5WR 24 == (tan chord theorem/rkl-koordst) ∴ 0°8S2 =

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

RWS) | | |RVW( ∆∆

∴ RS.RVWR 2 =

using the correct ∆s/ gebruik korrekte ∆e S S R

)in 3rd( ∆∠ or ) (∠∠∠ S

(5) [22]

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

= ( ΔRTM|||ΔPNM )

313PNPN

(2) 10.3

RN² – PN² = (RM² + NM²) – (PM² + NM²) (Pyth) = RM² – PM²

RP21RP

= 22 RP41RP

= 2RP²

Use of Pyth. for RN2 and PN2

RM = RP23

RP21PM =

2RP49 & 2RP

R P M 1

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.

= 2RP2

Use of Pyth. for RN2 and PN2 RM = RP + PM (3PM)2 – PM2 RP = 2PM

Use of Pyth. for RN2 and PN2 RM = RP + PM expansion/ uitbreiding

RP21PM =

(4) [8]

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum

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

But/Maar area ∆DEB = area ∆DEC (Same base and same height/dieselfde basis en dieselfde hoogte)

DECareaADEarea

DEBareaADEarea

∆∆

=∆∆

S S S R S

= (Line parallel one side of ∆

OR prop th; EF | | BD) (Lyn ewewydig aan sy v ∆

= OF eweredigst; EF ||BD)

S R answer/antw

9.2.2 CM = AM (diags of parm bisect/hoekl parm halv)

== (from 9.2.1/vanaf 9.2.1)

S R answer/antw

(3) 9.2.3 h of ∆FDC = h of ∆BDC (AD | | BC)

)parmofsides(oppADFD

BDCareaFDCarea

=∆∆

heights) (same 73

ADCareaFDCarea

==∆∆

ADCArea ∆ = BDCArea ∆ (diags of parm bisect area)

BDCareaFDCarea

=∆∆

AD | | BC subst into area form/ subst in opp formule S answer/antw

S R S answer/antw

(4) [16]

(tos sye v parm =)

(dieselfde hoogtes)

(hoekl v parm halv opp)

QUESTION/VRAAG 10

10.1.1 Tangent chord theorem/Raaklyn-koordstelling

10.1.2 Tangent chord theorem/Raaklyn-koordstelling

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

10.1.5 alternate ∠s/verwisselende ∠e ; WT | | SP R (1)

10.2 RPRT

= (Line parallel one side of ∆ OR

prop th; WT | | SP)

WR.RPRT =

OR/OF ∆RTW | | | ∆RPS (∠; ∠; ∠)

=∴ (∆RTW | | | ∆RPS)

RW.RPRT =

10.3 32 RTy == (tan chord theorem/Rkl-koordst)

13 QR ==y (∠s in same segment/∠e in dieselfde segment)

S R (4)

1 2 3 4

(Lyn ewewydig aan sy v ∆ OF eweredighst: WT | | SP)

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 =

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 OR/OF (∠; ∠; ∠)

1 2 3 4

= (∆RTS | | | ∆RQP)

But RS

WR.RPRT = (proven in 10.2/bewys in 10.2)

RQ.RSWR.RP

= (∆RTS | | | ∆RQP)

RPRT.RS WRand

RSRT.RPRQ

(proven in 10.2/bewys in 10.2)

RT.RPRS

RPRT.RS

RSRT.RP

RPRT.RS

RPRS =

× on both

WR.RPRT =

multiplication/ vermenigvuldig

RT.RS WR =

simplification/ vereenvoudiging

(3) [20]

(proven in 10.2/bewys in 10.2)

Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum

Kopiereg voorbehou Blaai om asseblief

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 −°=

)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

= (∆DBC | | | ∆DFM)

S antwoord

(2) [19]

Kopiereg voorbehou Blaai om asseblief

VRAAG 10 10.1

Konstruksie: Verbind DC en BE en trek hoogtes k en h

ΔDEBoppΔADEopp

k (gelyke hoogtes)

ΔDECoppΔADEopp

h (gelyke hoogtes)

Maar Opp ∆DEB = Opp ∆DEC (dies basis, dies hoogte)

∴ ΔDECoppΔADEopp

ΔDEBoppΔADEopp

∴ ECAE

konstruksie

ΔDEBoppΔADEopp

ΔDECoppΔADEopp

Area ∆DEB = Area

∆DEC (S/R)

ΔDECoppΔADEopp

ΔDEBoppΔADEopp

D E k h

Kopiereg voorbehou

10.2.1

= (Ewered st; BC | | ED)

= ∴ CD = 9 eenhede

(S/R) substitusie antwoord

(3) 10.2.2

= (Ewered st; FG | | EA)

54 – 6x = 9 + 3x –9x = –45 x = 5

= (S/R)

substitusie vereenvoudig antwoord

(4) 10.2.3 In ∆ABC en ∆AED:

∴ADAC

BC = 241 eenhede

A is gemeen ECBA = (S/R) DBCA = (S/R) of (∠; ∠; ∠)

antwoord

(5) 10.2.4

DGD.FD.sin

BCAAC.BC.sin

ΔGFDoppΔABCopp

Dsin)3)(4(21

Dsin)412)(3(

gebruik v opp reël korrekte sye en ∠e substitusie v waardes DsinBCAsin =

(S/R) antwoord

(5) [23]

(ooreenk ∠s; BC | | ED)

QUESTION/VRAAG 9 9.1 AF : FE

= 2 : 1 (Prop Th; FB || EC) (Eweredigheid St; FB || EC)

answer reason

(2) 9.2

2AFFE === cm

AE = 12 cm

= (BE || DC; Prop Th) / (BE || DC; Eweredigheid St)

ED = 6 cm

FE = 4 cm AE = 12 cm

answer

(4) [6]

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

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

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 (∠∠∠)

= (||| ∆s)

KE = TE (proven)

OE.WEKEOE.WEKE.TE

OR In ∆KOE and ∆KWE i. x== 13 WK (proven)

ii. 1E is common iii. =2O 321 KKK ++ (3rd ∠ of ∆) ∆KOE ||| ∆WKE (∠∠∠)

= (||| ∆s)

OE.WEKE2 =

∆KOE and ∆WTE x== 23 WK

°== 90EE 12 x−°== 90TO2

KE = TE

∆KOE and ∆WTE x== 13 WK

1E is common ∠∠∠

(6) [20]

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 =)

= (PQ || GH ; Prop Th)

construction All three statements must be given ∆FPQ ≡∆LMN (SAS) PQ || GH

NMLQPF =

HGFQPF =

Note: No construction constitutes a breakdown, hence no marks

= (PT || RK;Prop Th)

8162610264

===−

= (PT || RK; Prop Th)

4810020104

−=−

(PT || RK; Prop Th) substitution answer

(4) [11]

2x – 10

4 2x – 10

T 1 2 3

360° – 2x

180° – 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

9.3 In ∆BKT and ∆CAT

1. 1KTAC = (= x) 2. T is common 3. 4BTCA = (∠ sum ∆)

∆BKT ||| ∆CAT (∠∠∠)

1KTAC = T is common ∠∠∠

9.4 KTAT

= (||| ∆s)

||| ∆s answer

(3) [14]

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

°= 90C BC = 12 reason

OE = 6 cm ED = 4 cm

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

= (||| Δs)

But FE = AB (given)

AB.BD = FD.BH

FE = AB (2)

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

= (given)

CD || PQ (converse proportionality theorem) CD || FE (given) ∴ PQ || FE OR

∴ 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)

conv prop theorem

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

x2O1 = (∠ circ cent) x−°== 90DB 13 (radii)

+°=+−°+=

90)90(EBC

°= 90EBD ∠ in semi-circle x+°= 90EBC

°= 90OBC rad ⊥ tan x+°= 90EBC

x2O1 = ∠ circ cent x+°= 90EBC

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)

Note: If start with x=A and do not use tan ch th: max 2 marks

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

°== 90OFEOFB (BD || AO)

BO = OE ΔBOF ≡ ΔEOF BF = FE

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

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

x+°== 90EBCD4 Any two of the above

Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2.5

= (sim Δs ∴ sides in proportion)

EB.CB = CE.BD but EB = 2EF (F is the midpoint of BE) 2EF.CB = CE.BD

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 (∠∠∠)

== (sides in prop)

Let AC = 11a

8= (sides in prop)

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 (∠∠∠)

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

°−+= ..

MB = ME DB = 30 MC2 = 3200

or MC = 240 or MC = 56,57

answer (4)

cosine rule ME = 40 MC2 = 3200

answer

(4)[10]

NOTE: If the word cyclic is used in the last reason: max 2 / 3 marks

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.

QUESTION 9

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

AM (3)

9.2.2 22 35OM −= (Pythagoras) = 4 (OR midpoint theorem) Area ΔAOM : Area ΔABC

= 3.4.21 : 6.8.

= 6 : 24 = 1 : 4 OR Area ΔAOM : Area ΔABC

= MAOsin.OM.AM.21 : CABsin.AC.AB.

= 3.4.21 : 6.8.

= 6 : 24

substitution

answer (3)

QUESTION 10 10.1.1

= (GHB || FEC)

AH = 2y

HE = y

= (BE || CD)

ED = 1,5 y

statement reason

ED = 1,5y

answer (4)

10.1.2 64

=CDBE (ΔAEB ||| ΔADC)

If learner stops at 2 : 1,5 : no penalty

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)

QUESTION 11

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

11.2 In ΔACF and ΔADC

1. is common 3A is common

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

= (sim Δ’s ∴ sides in proportion) statement

ADAF44

ADADAD.

AD21AOAC

ADAC.ACAF

Statement

Simplification

Substitution

(2radius = diameter)

OR ΔAOC is equilateral S/R

°==∴ 60 A COA 3 Statement

AD41AF

)AD21(

AO21AC

ACAF60cos

==° Simplification

Substitution

(4) [10]

(2radius = diameter)

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==∧∧

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)

is common

2TQRW∧∧

= ….(remaining angles) ∆ QWR ||| ∆ QST (AAA) or (∠∠∠) or equiangular

angle in a semi-circle (1)

x−°=∧

∧∧

−°= 2P90S 90° -( 90° - x ) = x

x==∧∧

∧∧

= SRWQ reason (3)

TSQRWQ∧∧

WQR∧

is common

AAA or ∠∠∠ or equiangular or 3rd angle equal

9.5.1 QRQT

= ….. ∆ QWR ||| ∆ QST

cm 4TS

164TS48

6RQSQSR

=−=∴

cm 4TS = (3)

cm6 (3)

Copyright reserved

QUESTION 10 10.1

10.2 From 10.1 21

But DC = 9 cm ∴ DE = 6 cm = BD. ∴D is the midpoint of BE.

cm 4 TE24TE6126

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)

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