2016 2008 CAPS NCS Geometry Gr 12 Memo 14 July

Western Cape Education Department Examination Preparation Learning Resource 2016 GEOMETRY MEMORANDUM MATHEMATICS Grade 12 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 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 2

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum QUESTION/VRAAG 9 A 1 2 3 B 1 F O 2 3 D 1 2 2 3 1 3 C G E 9.1 ABCD is a ||m [diags of quad bisect each other/ R hoekl v vh halveer mekaar] (1) 9.2 ED FE S  R = [Prop Th/Eweredigh st; DF | | BA] DB AF ED GE S  R = [Prop Th/Eweredigh st; DG | | BC] DB CG (4) 9.3 FE GE S = [proved/bewys] AF CG ∴AC | | FG [line divides two sides of ∆ in prop/  S R lyn verdeel 2 sye van ∆ eweredig] S Ĉ 2 = F̂2 [alt/verw ∠s/e; AC | | FG] S Â 1 = Ĉ 2 [alt/verw ∠s/e; AB | | CD] (5) ∴ Â 1 = F̂2 9.4 Â 1 = Â 2 [diags of rhombus/hoekl v ruit] S Â 2 = F̂2 [ Â 1 = F̂2 ] S ∴ ACGF = cyc quad/kdvh [∠s in the same seg =/ R ∠e in dies segm =] (3) OR/OF Ĉ 2 = Â 2 [∠s opp equal sides of rhombus/ S ∠e to gelyke sye v ruit] Â 2 = Ĝ 2 [alt/verw-∠s/e; AC | | FG] S ∴ Ĉ 2 = Ĝ 2 ∴ ACGF is a cyc quad/kdvh [∠s in the same seg =/ ∠e in dies segm =] R (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 3

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum QUESTION/VRAAG 10 A P 10.1 D E Q R B C 10.1.1 In ∆ADE and/en ∆PQR: AD = PQ [construction/konstr] all/al 3 S’s/e Â = P̂ [given/gegee] AE = PR [construction/konstr] ∴∆ADE ≡ ∆PQR [S∠S] reason/rede (2) 10.1.2 AD̂E = Q̂ [∆s ≡ ∴ corres/ooreenk ∠s/e =]  AD̂E = Q̂ But B̂ = Q̂ [given/gegee] ∴ AD̂E = B̂  AD̂E = B̂ ∴ DE | | BC [corres/ooreenk ∠s/e =] reason/rede (3) 10.1.3 AB AC = [Prop Th/Eweredigh st; DE | | BC] S/R AD AE But/Maar AD = PQ and/en AE = PR [construction/konstr] S AB AC ∴ PQ = PR (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 4

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum 10.2 R T O 2 2 1 1 2 P V 1 1 S 10.2.1 line from centre to midpt of chord/lyn van midpt na midpt van answ/antw koord (1) 10.2.2 OP | | VS [Midpt Theorem/Midpt-stelling] S  R In ∆ROP and/en ∆RVS: R̂ = R̂ [common/gemeen] S Ô 2 = V̂ [corresp/ooreenk ∠s/e; OP | | VS] S & ∠;∠;∠ OR/OF ∴∆ROP | | | ∆RVS [∠,∠,∠] 3 angles/hoeke (4) OR/OF In ∆ROP and/en ∆RVS: P̂2 = VŜR [corresponding ∠s/ ooreenkomstige ∠'e] S  R R̂ = R̂ [common/gemeen] S ∴∆ROP | | | ∆RVS [∠,∠,∠] S & ∠;∠;∠ OR/OF 3 angles/hoeke (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 5

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum 10.2.3 In ∆RVS and/en ∆RST: VŜR = ST̂R = 90° [∠ in semi-circle/∠ in halfsirkel] S  R R̂ is common/gemeen S & ∠;∠;∠ V̂ = TŜR OR/OF ∴∆RVS | | | ∆RST [∠,∠,∠] 3 angles/hoeke (3) 10.2.4 In ∆RTS and/en ∆STV: ∆RTS & ∆STV RT̂S = VT̂S = 90° [∠ s on straight line/∠e op rt lyn] S R̂ = 90° – TŜR S = TŜV S (with TŜR = V̂ justification/met ∴∆RTS | | | ∆STV [∠,∠,∠] motivering) RT TS ∴ = ∆RTS | | | ∆STV ST VT ∴ ST 2 = VT.TR ratio/verh (6) [21] Copyright reserved/Kopiereg voorbehou 6

Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum QUESTION/VRAAG 10 B 17 E 8 C F D A 10.1 BD̂C = 90° [∠ in semi circle/∠ in halfsirkel] S DC2 = 172 – 82 [Th of/stelling v Pythagoras]  using/gebruik = 225 Pyth korrek/ correctly ∴ DC = 15  answ/antw (3) 10.2.1 CF CE = [line | | one side of ∆/lyn | | een sy van ∆]  S/R CD CB CF 1 ∴ =  subst correctly/ 15 4 korrek ∴ CF = 3,75  answ/antw (3) 10.2.2 In ∆BAC and/en ∆ FEC: AB̂C = 90° [tan ⊥ diameter/raakl ⊥ middellyn] SR EF̂C = 90° [corresp ∠s/ooreenk ∠e; EF| |BD]  S/R Ĉ = Ĉ [common/gemeen] S ∴ ∆BAC | | | ∆ FEC [∠∠∠] R (5) OR/OF In ∆BAC and ∆ FEC: AB̂C = 90° [tan ⊥ diameter/raakl ⊥ middellyn] SR EF̂C = 90° [corresp ∠s/ooreenk ∠e; EF| |BD]  S/R Ĉ = Ĉ [common/gemeen] S BÂC = FÊC [∠ sum in ∆/∠ som van ∆] S ∴ ∆BAC | | | ∆ FEC (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 7

Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum ABEF = cyc quad/ kdvh [opp ∠s of quad supp/tos∠e v vh suppl)] ∴ FÊC = Â [ext ∠ of cyc quad/buite ∠ v kdvh] SR In ∆BAC and/en ∆ FEC:  S/R FÊC = Â [proven/bewys] Ĉ = Ĉ [common/gemeen] BÂC = FÊC [∠ sum in ∆/∠ som van ∆] S ∴ ∆BAC | | | ∆ FEC S (5) OR/OF ABEF = cyc quad/ kdvh [opp ∠s of quad supp/tos∠ SR e v vh suppl)]  S/R ∴ FÊC = Â [ext ∠ of cyc quad/buite ∠ v kdvh] In ∆BAC and ∆ FEC: FÊC = Â [proven/bewys] S Ĉ = Ĉ [common/gemeen] R ∴ ∆BAC | | | ∆ FEC [∠∠∠] (5) 10.2.3 1  length of/lengte v EC = × 17 = 4,25 4 EC AC BC = [||| ∆s/e] EC FC S AC 17 =  subst correctly/ 4,25 3,75 korrek 4 ∴AC = 19,27 or/of 19  answ/antw 15 (4) OR/OF CF BC cos Ĉ = =  correct ratios/ CE AC korrekte verh's 3,75 17 ∴ =  subst correctly/ 4,25 AC korrek 4  answ/antw ∴ AC = 19,27 or/of 19 15 (4) 10.2.4 AC is diameter of the circle [chord subtends 90° ]  S/R AC is middellyn van die sirkel [koord onderspan 90°] 1 19  answ/antw ∴radius = × 19,27 = 9,63 or/of 9 2 30 (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 8

Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum QUESTION/VRAAG 11 11.1 equiangular or similar/gelykhoekig of gelykvormig  answ/antw (1) P • 2 1,5 ♦ M ♠ 2,5 K ♦ 1,25 1 ♠ R • 0,75 N ♦ Q 11.2.1 KP 1,5 PM 2 KM 2,5 = =2 ; = =2 ; = =2 all 3 statements/ RN 0,75 NM 1 RM 1.25 al 3 bewerings KP PM KM ∴ = = RN NM RM ∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig]  R (3) OR/OF OR/OF RN 0,75 1 NM 1 RM 1,25 1 = = ; = ; = = all 3 statements/ KP 1,5 2 PM 2 KM 2,5 2 al 3 bewerings RN NM RM ∴ = = KP PM KM ∴ ∆KPM | | | ∆RNM [Sides of ∆ in prop/sye v ∆ eweredig]  R (3) 11.2.2 PK̂M = R̂ S ∴ P̂ is common/gemeen ∴ ∆RPQ | | | ∆KPM [∠∠∠]  ∆RPQ|||∆KPM RP RQ S = [||| ∆s] KP KM 3,25 RQ  subst correctly/ ∴ = 1,5 2,5 korrek 2,5 × 3,25 5 5 ∴ RQ = = 5,42 or 5  RQ = 5 1,5 12 12 2 ∴ NQ = 5,42 – 0,75 = 4,67 or 4  NQ = answ/antw 3 (6) [10] Copyright reserved/Kopiereg voorbehou 9

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 S die derde sy. (1) B F D C A 12 E 8 10.2.1 AE 12 3 S = = AC 20 5 AD 3 = AF 5 AE AD ∴ = S AC AF ∴ DE || FC (line divides two sides of ∆ in prop/ R lyn verdeel twee sye v ∆ in dieselfde verh) (3) 10.2.2 BF 8 = (prop theorem/eweredigh st; BC || FE)  S/R BA 20 8 ∴ BF = (14)  substitute 14/ 20 stel 14 in 28 3 ∴ BF = OR/OF FB = 5 OR/OF FB = 5,6  answer/antw 5 5 (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 10

Mathematics/P2/Wiskunde V2 DBE/2015 NSCNSS – Memorandum B D 1 2 1 C 2 1 2 E 1 2 A P 11.2.1 DĈA = 2 x (EC bisector) P̂ = x (∠ at centre = 2 ×∠ at circumference/ SR midpts∠ = 2 × omtreks∠) Â 1 = P̂ = x (tangent-chord theorem/rkl-kd st) SR In ∆BAD and ∆BCE: B̂ = B̂ (common/gemeen) S Â1 = Ĉ1 (proven above)  S(with justification) ∴∆BAD | | | ∆BCE (∠∠∠) R (7) OR/OF DĈA = 2 x (EC bisector) P̂ = x (∠ at centre = 2 ×∠ at circumference/ SR midpts∠ = 2 × omtreks∠) Â 1 = P̂ = x (tangent-chord theorem/rkl-kd st) SR In ∆BAD and ∆BCE: B̂ = B̂ (common/gemeen) S Â1 = Ĉ1 (proven above)  S(with justification) D̂1 = Ê 1 S ∴∆BAD | | | ∆BCE (7) Copyright reserved Please turn over/Blaai om asseblief 11

Mathematics/P2/Wiskunde V2 DBE/2015 NSCNSS – Memorandum 11.2.2(a) BÂC = 90° (tangent/raakl ⊥ radius) R ∴BC2 = 82+ 62 = 100 (Pythagoras theorem/stelling)  substitution into BC = 10 Pyth theorem AC = DC = 6 (radii)  BC = 10 ∴ BD = 10 – 6 = 4 units/eenhede  DC = 6  BD = 4 (5) 11.2.2(b) BA BD S = (∆BAD | | | ∆BCE) BC BE 8 4  substitution/ ∴ = 10 BE substitusie ∴ BE = 5 units/eenhede  BE = 5 (3) 11.2.2(c) AE = 3 In ∆ACE: 3  correct trig ratio/ tan x = korrekte trigvh 6  correct trig eq/ ∴x = 26,57° korrekte trigvgl  answer/antw OR/OF (3) 8  correct trig ratio/ sin 2x = korrekte trigvh 10  correct trig eq/ korrekte trigvgl ∴ 2x = 53,1301... (2x < 90°)  answer/antw ∴ x = 26,57° (3) [24] Copyright reserved 12

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum 9.3 P T 1 V 2 R 1 2 30° 3 4 50° 4 3 2 Q W 1 1 2 S Z 9.3.1 Equal chords subtend equal ∠s/Gelyke koorde onderspan gelyke ∠e R (1) 9.3.2 Ŵ4 = 30° (tan chord theorem/rkl-koordst)  answer/antw  reason/rede Ŵ1 = 30°  answer/antw (3) 9.3.3(a) R̂ 4 = Ŵ2 = 50° (tan chord theorem/rkl-koordst)  S R Ŝ 2 = R̂ 3 + Ŵ2 (ext∠ of ∆/buite ∠ v ∆) ∴ Ŝ 2 = 80° S (3) OR/OF R̂ 2 = R̂ 3 = 30° (= chords subtend =∠s /= kde onderspan=∠e)  S R R̂ 4 = Ŵ2 = 50° (tan chord theorem/rkl-koordst) ∴ Ŝ 2 = 80° S (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 13

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum 9.3.3(b) T̂2 = Ŝ 2 = 80° (ext ∠ of cyclic quad/buite∠ van koordevh)  SR V + Ŵ4 = T̂2 (ext∠ of ∆/buite∠ van ∆)  S ∴ V̂ = 50°  S (4) 9.3.4 In ∆RVW and/en ∆RWS:  using the correct ∆s/ gebruik korrekte ∆e R̂ 2 = R̂ 3 = 30° (proven/bewys in 9.3.1) S V̂ = Ŵ2 = 50° (proven/bewys in 9.3.3) S VŴR = Ŝ1 (3rd ∠ in ∆) R ∴∆RVW | | | ∆RWS (∠∠∠) (3rd ∠ in ∆) or (∠∠∠ ) WR RS ∴ = (∆RVW | | | ∆RWS) S RV WR ∴ WR 2 = RV . RS (5) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 14

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 10 T N 1 1 R P M 10.1.1 corresponding ∠s/ooreenkomstige∠e; PN | | RT  answer/antw (1) 10.1.2 ∠; ∠; ∠ OR/OF ∠; ∠  answer/antw (1) 10.2 PM PN = ( ΔPNM | | | ΔRTM ) RM RT S PN =  S 3PN 1 = 3 (2) 10.3 PM 1 RP 2 = ∴ =  Use of Pyth. for RM 3 RM 3 RN2 and PN2 RN² – PN² = (RM² + NM²) – (PM² + NM²) (Pyth) 3  RM = RP = RM² – PM² 2 3  1 2  2 1 =  RP  −  RP   PM = RP 2 2  2  9 1 9 1  RP 2 & RP 2 = RP 2 − RP 2 4 4 4 4 (4) = 2RP² OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 15

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum RN2 – PN2 = (RM2 + NM2) – (PM2 + NM2) (Pyth)  Use of Pyth. for = RM2 – PM2 RN2 and PN2 = (3PM)2 – PM2 RM = RP + PM = 8PM2  (3PM)2 – PM2 = 2(2PM)2  RP = 2PM = 2RP2 (4) OR/OF RN2 – PN2 = (RM2 + NM2) – (PM2 + NM2) (Pyth)  Use of Pyth. for = RM2 – PM2 RN2 and PN2 = (RP + PM)2 – PM2 RM = RP + PM = RP2 + 2RP.PM + PM2 – PM2  expansion/ 1 uitbreiding = RP2 + 2RP. RP 2 1 2  PM = RP = 2RP 2 (4) [8] Copyright reserved/Kopiereg voorbehou 16

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 9 9.1 A k h D 1 E B C 9.1.1 Same base (DE) and same height (between parallel lines)  same base/dies Dieselfde basis (DE) en dieselfde hoogte (tussen ewewydige lyne) basis between | | lines/ tussen | | lyne (1) 9.1.2 AD S DB 1 AE × k  S 2 1 EC × k 2 But/Maar area ∆DEB = area ∆DEC S (Same base and same height/dieselfde basis en dieselfde hoogte) R area ∆ADE area ∆ADE ∴ = area ∆DEB area ∆DEC  S AD AE ∴ = (5) DB EC Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 17

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum 9.2 A F D G E M B C 9.2.1 EM FD = (Line parallel one side of ∆ AM AD  S R OR prop th; EF | | BD) (Lyn ewewydig aan sy v ∆ EM 3 = OF eweredigst; EF ||BD)  answer/antw AM 7 (3) 9.2.2 CM = AM (diags of parm bisect/hoekl parm halv)  S R CM AM 7 = = (from 9.2.1/vanaf 9.2.1)  answer/antw ME ME 3 (3) 9.2.3 h of ∆FDC = h of ∆BDC (AD | | BC)  AD | | BC 1 FD.h area ∆FDC 2  subst into area = form/ subst in area ∆BDC 1 BC.h opp formule 2 FD = (opp sides of parm =) S AD (tos sye v parm =) 3  answer/antw = 7 (4) OR/OF area ∆FDC FD 3 = = (same heights)  S R area ∆ADC AD 7 (dieselfde hoogtes) But Area ∆ADC = Area ∆BDC (diags of parm bisect area) S (hoekl v parm halv opp) area ∆FDC 3  answer/antw = area ∆BDC 7 (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 18

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 10 P 1 2 S 1 2 1 2 T 3 4 X 1 W 2 1 x 2 Q y 3 4 3 2 R 1 Y 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 R ∠e onderspan deur dieselfde koord OF ∠e in dieselfde segment (1) 10.1.5 alternate ∠s/verwisselende ∠e ; WT | | SP R (1) 10.2 RW RT = (Line parallel one side of ∆ OR SR RS RP prop th; WT | | SP) ∴ RT = WR.RP (Lyn ewewydig aan sy v ∆ OF (2) RS eweredighst: WT | | SP) OR/OF ∆RTW | | | ∆RPS (∠; ∠; ∠) S RW RT ∴ = (∆RTW | | | ∆RPS) S RS RP RW.RP ∴ RT = (2) RS 10.3 y = T̂2 = R̂ 3 (tan chord theorem/Rkl-koordst) SR y = R̂ 3 = Q̂1 (∠s in same segment/∠e in dieselfde SR segment) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 19

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum P 1 2 x S 1 y 2 1 y 2 T 4 X 1 x3 W 2 y 1 x 2 Q y y 3 4 3 2 R 1 Y 10.4 Q̂ 3 = PŜR (ext ∠ of cyc quad/buite∠ v kdvh) S R PŜR = Ŵ2 (corresp∠s/ooreenk ∠e ; WT | | SP) S ∴ Q̂ 3 = Ŵ2 (3) OR/OF Q̂ 2 = x (∠s in same segment/∠e in dies segment) R Q̂ 3 = 180° − ( x + y ) (∠s on straight line/∠e op reguitlyn) S Ŵ2 = 180° − ( x + y ) (∠s of ∆WRT/∠e v ∆WRT ) S (3) ∴ Q̂ 3 = Ŵ2 10.5 In ∆RTS and ∆RQP: R̂ 3 = R̂ 2 = y (proven above/hierbo bewys) S Ŝ 2 = P̂2 (∠s in same segment/∠e in dies segment)  S/R RT̂S = RQ̂P (3 angle of ∆) rd  S OR/OF ∴∆RTS | | | ∆RQP (∠; ∠; ∠) (∠; ∠; ∠) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 20

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum 10.6 RT RS  S = (∆RTS | | | ∆RQP) RQ RP RS RS RS RT RS × on both × = × RP RP RP RQ RP sides 2  RS   RT  RS    =    RP   RP  RQ   RT  RS   RW  RS      =   (proven in 10.2/bewys in 10.2)  RP  RQ   RS  RQ  (3) RW = RQ OR/OF RT RS = (∆RTS | | | ∆RQP) S RQ RP WR.RP WR.RP But RT = (proven in 10.2/bewys in 10.2)  RT = RS RS RT WR.RP RS ∴ = = RQ RQ.RS RP multiplication/ WR.RP 2 = RQ.RS 2 vermenigvuldig WR RS 2 ∴ = RQ RP 2 (3) OR/OF RT RQ S = (∆RTS | | | ∆RQP) RS RP RT.RP RQ = RS RT.RS RT.RS  WR = and WR = (proven in 10.2/bewys in 10.2) RP RP RT.RS WR = RP RQ RT.RP RS  simplification/ RT.RS RS = × vereenvoudiging RP RT.RP 2 RS = RP 2 (3) [20] Copyright reserved/Kopiereg voorbehou 21

Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum VRAAG 9 9.1 D̂ = Â = x (rkl-koordstelling)  Â = x 4  rede Â = D̂ = x (∠e tos gelyke sye)  Â = D̂ = x 2 2 (S/R) (3) 9.2 M̂1 = 2x (buite ∠v∆) OF (∠ by midpt = 2∠ by omtr)  M̂ = 2x (S/R) 1 MD̂E = 90° (radius ⊥ rkl)  MD̂E = 90° M̂ 2 = 90° − 2 x (S/R) ∴ Ê = 180° − (90° + 90° − 2 x) (som v ∠e in ∆MDE)  Ê = 2 x = 2x ∴ CM is ‘n rkl (omgek rkl-koordst)  rede (4) 9.3 M̂ = 90° (EM ⊥ AC)  M̂ 3 = 90° 3 AD̂B = 90° (∠ in halfsirkel)  AD̂B = 90° (S/R) ∴ FMBD is koordevh (buite∠ v vh = tos binne ∠)  rede OF (3) EM̂C = 90° (EM ⊥ AC)  EM̂C = 90° AD̂B = 90° (∠ in halfsirkel)  AD̂B = 90° (S/R) ∴ FMBD is koordevh (tos ∠e v vh suppl)  rede (3) 9.4 DC2 = MC2 – MD2 (Pythagoras)  Pythagoras = (3BC)2 – (2BC)2 (MB = MD = radii)  substitusie = 9BC2 – 4BC2  9BC2 – 4BC2 = 5BC2 (3) 9.5 In ∆DBC en ∆DFM: D̂ 4 = D̂ 2 = x (bewys in 9.1)  D̂ 4 = D̂ 2 B̂1 = F̂2 (buite ∠ v koordevh)  B̂ = F̂ 1 2  rede Ĉ = M̂ 2 ∴∆DBC | | | ∆DFM (∠; ∠; ∠)  Ĉ = M̂ 2 of (∠; ∠; ∠) (4) 9.6 DM DC = (∆DBC | | | ∆DFM) S FM BC 5BC = BC  antwoord = 5 (2) [19] Kopiereg voorbehou Blaai om asseblief 22

Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum VRAAG 10 10.1 A h k D E B C Konstruksie: Verbind DC en BE en trek hoogtes k en h  konstruksie 1 opp ΔADE AD opp ΔADE 2 .AD.k AD  = = = (gelyke hoogtes) opp ΔDEB DB opp ΔDEB 1 .DB.k DB 2  rede 1 .AE.h opp ΔADE AE opp ΔADE 2 AE  = = = (gelyke hoogtes) opp ΔDEC EC opp ΔDEC 1 .EC.h EC 2 Maar Opp ∆DEB = Opp ∆DEC (dies basis, dies hoogte)  Area ∆DEB = Area ∆DEC (S/R) opp ΔADE opp ΔADE  ∴ = opp ΔDEB opp ΔDEC opp ΔADE opp ΔADE = AD AE opp ΔDEB opp ΔDEC ∴ = DB EC (6) Kopiereg voorbehou Blaai om asseblief 23

Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum 10.2.1 AB AC AB AC = (Ewered st; BC | | ED)  = BE CD BE CD 1 3 (S/R) =  substitusie 3 CD ∴ CD = 9 eenhede  antwoord (3) 10.2.2 DG FD DG FD = (Ewered st; FG | | EA)  = (S/R) GA FE GA FE 9− x 3 =  substitusie 3+ x 6 54 – 6x = 9 + 3x  vereenvoudig –9x = –45 x=5  antwoord (4) 10.2.3 In ∆ABC en ∆AED: Â is gemeen  Â is gemeen AB̂C = Ê (ooreenk ∠s; BC | | ED)  AB̂C = Ê (S/R) AĈB = D̂ (ooreenk ∠s; BC | | ED)  AĈB = D̂ (S/R) ∆ABC | | | ∆AED (∠, ∠, ∠) of (∠; ∠; ∠) BC AC BC AC ∴ =  = ED AD ED AD BC 3 = 9 12 1 BC = 2 eenhede  antwoord 4 (5) 10.2.4 1  gebruik v opp opp ΔABC 2 AC.BC.sin AĈB = reël opp ΔGFD 1 GD.FD.sin D̂  korrekte sye en 2 1 1 ∠e (3)(2 ) sin D̂  substitusie v = 2 4 (ooreenk ∠s; BC | | ED) waardes 1 (4)(3) sin D̂  sin AĈB = sin D̂ 2 9 (S/R) =  antwoord 16 (5) [23] Kopiereg voorbehou 24

Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 9 A 2x B x C F E D 9.1 AF : FE =2:1 (Prop Th; FB || EC)  answer (Eweredigheid St; FB || EC)  reason (2) 9.2 AF 2 = FE 1 AF 8 FE = = = 4 cm  FE = 4 cm 2 2  AE = 12 cm AE = 12 cm ED 1 ED 1 = (BE || DC; Prop Th) / (BE || DC; Eweredigheid St)  = AE 2 AE 2 ED 1 = 12 2  answer ED = 6 cm (4) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 25

Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 12 K 1 2 3 x x 1 E 2 2 1 O3 4 x T 2 3 x 1 1 x 2 Y W 12.1 K̂ 3 = x (tan ch th)  K̂ 3 = x (raaklyn koord)  tan ch th Ŷ2 = x (∠s in same seg)  Ŷ2 = x (∠e in selfde seg)  ∠s in same seg K̂ 2 = x (∠s opp = radii) / (= chs subt = ∠s)  K̂ 2 = x (∠e oork = radiusse) / (= koorde = ∠e)  reason Ŵ2 = x (∠s in same seg) / (= chs subt = ∠s)  Ŵ2 = x (∠e in selfde seg) / (= koorde = ∠e)  reason (8) 12.2 Ô1 + Ô 4 = 180° − 2 x (sum of int ∠’s of ∆) / (opp∠ cyclic quad)  Ô1 + Ô 4 = 180° − 2 x (som van binne∠e ∆) / (oorst ∠ koordevierhoek)  reason T̂ = 90° − x (∠ at circ cent = 2∠ at circumference)  reason (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief 26

Mathematics P3/Wiskunde V3 DBE/November 2013 NSC/NSS – Memorandum 12.3 Ê 2 = 90° (sum of int ∠’s of ∆) / (som van binne∠e ∆)  Ê 2 = 90° KE = ET (⊥ from centre to chord bisects chord)  sum of int ∠’s of ∆ (⊥ van middelpunt tot koord halveer koord)  ⊥ from centre to chord bisects chord OR (3) K̂1 + K̂ 2 + K̂ 3 = 90° − x (sum of int ∠’s of ∆)  K̂1 + K̂ 2 + K̂ 3 = 90° − x In ∆KWE and ∆TEW  sum of int ∠’s of ∆ 1. K̂1 + K̂ 2 + K̂ 3 = T̂ = 90° − x (proven above) 2. Ŵ1 = Ŵ2 = x (Proven in 12.1) 3. WE is common ∴ ∆KWE ≡ ∆TEW (∠∠S)  ∆KWE ≡ ∆TEW KE = ET (3) 12.4 In ∆KOE and ∆WTE  ∆KOE and ∆WTE i. K̂ 3 = Ŵ2 = x (proven)  K̂ 3 = Ŵ2 = x ii. Ê 2 = Ê 1 = 90° (∠s on str line / sum of int ∠’s of ∆)  Ê 2 = Ê 1 = 90° (∠e op reguit lyn / som van binne∠e ∆) iii. Ô 2 = T̂ = 90° − x (3rd ∠ of ∆)  Ô 2 = T̂ = 90° − x ∆KOE ||| ∆WTE (∠∠∠) KE OE KE OE = (||| ∆s)  = WE TE WE TE KE = TE (proven) KE.TE = OE.WE  KE = TE KE 2 = OE.WE (6) OR  ∆KOE and ∆WTE In ∆KOE and ∆KWE  K̂ 3 = Ŵ1 = x i. K̂ 3 = Ŵ1 = x (proven)  Ê1 is common ii. Ê1 is common iii. Ô 2 = K̂1 + K̂ 2 + K̂ 3 (3rd ∠ of ∆)  ∠∠∠ ∆KOE ||| ∆WKE (∠∠∠) KE OE KE OE = (||| ∆s)  = WE KE WE KE KE 2 = OE.WE (6) [20] Copyright reserved/Kopiereg voorbehou 27

Mathematics/P3 DBE/November 2012 NSC – Memorandum QUESTION 7 F L P Q M N G H 7.1 Draw a point P on FG such that FP = LM and a point Q on FH  construction such that FQ = LN. Note: No construction In ∆FPQ and ∆LMN constitutes a breakdown, hence no marks All three statements 1. F̂ = L̂ (given) 2. FP = LM (construction) must be given 3. FQ = LN (construction) ∴∆FPQ ≡∆LMN (SAS) ∆FPQ ≡∆LMN (SAS) FP̂Q = LM̂N (≡∆s)  FP̂Q = LM̂N But FĜH = LM̂N (given) FP̂Q = FĜH  FP̂Q = FĜH PQ || GH (corresponding angles =) FP FQ  PQ || GH = (PQ || GH ; Prop Th) FP FQ FG FH  = FG FH LM LN = (7) FG FH Copyright reserved Please turn over 28

Mathematics/P3 DBE/November 2012 NSC – Memorandum V 2x – 10 4 2x – 10 T P 66 99 9 K R 7.2 VP VT VP VT = (PT || RK;Prop Th)  = PR TK PR TK 2 x − 10 4  (PT || RK; Prop Th) =  substitution 9 6 2 x − 10 = 6 2 x = 16 x=8  answer (4) OR VP VT VP VT = (PT || RK; Prop Th)  = VR VK VR VK 2 x − 10 4  (PT || RK; Prop Th) = 2 x − 1 10  substitution 20 x − 100 = 8 x − 4 12 x = 96  answer x=8 (4) [11] Copyright reserved Please turn over 29

Mathematics/P3 DBE/November 2012 NSC – Memorandum QUESTION 9 C x B 1 4x T 2 3 360° – 2x x 1 O 180° – x2 2x K 1 2 3 A 9.1 AÔB = 2 x (∠circ centre = 2 ∠ circumference)  AÔB = 2 x T̂ = 180° − 2 x (opp∠ cyclic quad suppl) ∠circ centre = 2 ∠ circumference opp∠ cyclic quad suppl (3) 9.2 CÂT = x (∠ sum ∆)  CÂT = x K̂ 1 = x (ext∠ cyclic quad) ∠ sum ∆ CÂT = K̂ 1  K̂1 = x BK || AC (corresponding ∠s =) ext∠ cyclic quad  corresponding ∠s = OR (5) K̂1 = Ĉ = x (ext∠ cyclic quad)  K̂1 = Ĉ = x B̂ 4 = x (∠ sum ∆) ext∠ cyclic quad  B̂ 4 = x B̂ 4 = Ĉ = x ∠ sum ∆ BK || CA (corresponding ∠s =)  corresponding ∠s = (5) OR  CÂT = x CÂT = x (∠ sum ∆) ∠ sum ∆ BK̂A = 180° − x (opp∠ cyclic quad)  BK̂A = 180° − x CÂT + BK̂A = 180° opp∠ cyclic quad BK || AC (coint∠s supp) co-int∠s supp (5) Copyright reserved Please turn over 30

Mathematics/P3 DBE/November 2012 NSC – Memorandum 9.3 In ∆BKT and ∆CAT 1. CÂT = K̂ 1 (= x)  CÂT = K̂ 1 2. T̂ is common  T̂ is common 3. AĈT = B̂4 (∠ sum ∆) ∆BKT ||| ∆CAT (∠∠∠) ∠∠∠ (3) 9.4 AC AT AC AT = (||| ∆s)  = KB KT KB KT AC 7  ||| ∆s =  answer KB 2 (3) [14] Copyright reserved Please turn over 31

Mathematics/P3 DBE/November 2011 NSC – Memorandum QUESTION 9 9. Ĉ = 90° (∠s in semi circle) 9 Ĉ = 90° OÊA = 90° (corres ∠s; OD || BC) 9 OÊA = 90° AE = 8 cm (line from circ cent ⊥ ch bis ch) B 9 line from circ OE = 6 cm (Pythagoras) cent ⊥ ch bis ch ED = 10 – 6 O 9 OE = 6 cm = 4 cm 9 ED = 4 cm A OR Ĉ = 90° (∠s in semi circle) E C OÊA = 90° (corres ∠s; OD || BC) 9 Ĉ = 90° 9 OÊA = 90° D OE || BC (given) OA = OB (radii) 9 midpoint AE = EC = 8cm (midpoint theorem) theorem OE = 6 cm (Pythagoras) 9 OE = 6 cm ED = 10 – 6 9 ED = 4 cm = 4 cm OR Ĉ = 90° (∠s in semi circle) BC2 = (20)2 – (16)2 9 Ĉ = 90° BC2 = 144 9 BC = 12 BC = 12 9 reason 1 9 OE = 6 cm OE = BC (midpoint theorem) 9 ED = 4 cm 2 OE = 6 cm OD = 10cm [5] ED = 10 – 6 = 4 cm OR Ĉ = 90° (∠s in semi circle) 9 Ĉ = 90° BC = (20) – (16)2 2 2 9 BC = 12 BC2 = 144 9 reason BC = 12 1 OE = BC (midpoint theorem) 9 OE = 6 cm 2 9 ED = 4 cm OE = 6 cm ED = 4cm [5] Copyright reserved Please turn over 32

Mathematics/P3 DBE/November 2011 NSC – Memorandum QUESTION 10 A x F B 1 3 2 3 H 2 1 G 3 1 2 1 y x x2 E 3 2 x 4 1y C D 10.1 Â = D̂ 4 = x (tan ch th) 9 Â = x Ê 2 = x (tan ch th) OR (∠s in same seg) 9 tan ch th 9 Ê 2 = x D̂ 2 = Â = x (alt ∠s; CA || DF) 9 reason 9 D̂ 2 = x 9 alt ∠s; CA || DF (6) 10.2 In ΔBHD and Δ FED 1. B̂ 2 = F̂ (∠s in same seg) 9 B̂ 2 = F̂ 2. D̂ 3 = D̂1 (= chs subt = ∠s) 9 ∠s in same seg 9 D̂ 3 = D̂1 ΔBHD ||| Δ FED (∠∠∠) 9 = chs subt = ∠s 9 ∠∠∠ (5) 10.3 FE FD FE FD = (||| Δs) 9 = BH BD BH BD But FE = AB (given) 9 FE = AB AB FD (2) = BH BD AB.BD = FD.BH [13] Copyright reserved Please turn over 33

Mathematics/P3 DBE/November 2011 NSC – Memorandum QUESTION 11 B A F E C D P Q 11.1 AF = FC (diags of parallelogram bisect) 9 AF = FC FE || CD AE = ED (Prop Th; FE || CD) OR (Midpoint Theorem) 9 reason (2) 11.2 AC 1 = (given) CP 2 AD 1 = (given) DQ 2 AC AD 9 ratios equal = CP DQ CD || PQ (converse proportionality theorem) 9 CD || PQ CD || FE (given) 9 reason: converse ∴ PQ || FE prop th and conclusion (3) OR AC 1 = AP 3 AD 1 = AQ 3 AC AD 9 ratios equal = AP AQ CD || PQ (converse proportionality theorem) 9 CD || PQ CD || FE (given) 9 reason: converse ∴ PQ || FE prop th and conclusion (3) OR AF 1 AF 1 = 9 = AP 6 AP 6 AE 1 = AQ 6 AF AE AF AE 9 = = AP AQ AP AQ 9 conv prop theorem ∴ PQ || FE (converse proportionality theorem) Copyright reserved Please turn over 34

Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2 C 2 D 1 4 B 3 1 2 1 2 O 1 3 2 F x A E 8.2.1 B̂ 4 = x (tan chord theorem) 9 B̂ 4 = x Note: 9 tan chord theorem If start with Â = x and 9 Â = B̂4 = x with Â = B̂4 = x (corres ∠; BD || AO) do not use tan ch th: reason max 2 marks B̂ 2 = x (BO = EO = radii) 9 B̂ 2 = x (4) 8.2.2 DB̂E = 90° (∠ in semi-circle) 9 DB̂E = 90° CB̂E = 90° + x 9 ∠ in semi-circle 9 CB̂E = 90° + x (3) OR CB̂O = 90° (rad ⊥ tan) 9 CB̂O = 90° CB̂E = 90° + x 9 rad ⊥ tan 9 CB̂E = 90° + x OR (3) Ô1 = 2 x (∠ circ cent) B̂3 = D̂1 = 90° − x (radii) 9 Ô1 = 2 x 9 ∠ circ cent CB̂E = x + (90° − x) + x 9 CB̂E = 90° + x = 90° + x (3) 8.2.3 DB̂E = 90° (proved in 8.2.2) 9 DB̂E = 90° BF̂O = 90° (co-int angles supp; BD || AO) 9 BF̂O = 90° and BF = FE (line from circ cent ⊥ ch bisect ch) reason F is the midpoint of EB 9 BF = FE 9 line from circ cent ⊥ ch bisect ch) (4) 35

Mathematics/P3 DBE/November 2010 NSC - Memorandum OR OD = OE (radii) 9 OD = OE BF = FE (BD || AO) 9 radii F is the midpoint of EB 9 BF = FE 9 BD || AO (4) OR BF̂O = EF̂O = 90° (BD || AO) 9 BF̂O = EF̂O = 90° OF is common (BD || AO) BO = OE (radii) 9 BO = OE ΔBOF ≡ ΔEOF (90°HS) 9 ΔBOF ≡ ΔEOF BF = FE (≡ Δs) 9 BF = FE (4) OR B̂2 = Â = x (proven) Ô 2 is common 9 ΔAOB ||| ΔBOF ΔAOB ||| ΔBOF (AAA) AB̂O = BF̂O 9 AB̂O = BF̂O AB̂O = 90° (proven) AB̂O = BF̂O = 90° 9 BF = FE BF = FE (line from circ cent ⊥ ch bisects ch) 9 line from circ cent ⊥ ch bisects ch (4) OR DB̂E = 90° (∠ in semi-circle) 9 DB̂E = 90° B̂3 = 90° − x 9 F̂1 = 90° 9 BF = FE Ô 2 = 90° − x (alt ∠s; BD || FO) 9 line from circ cent F̂1 = 90° (∠ sum Δ) ⊥ ch bisects ch BF = FE (line from circ cent ⊥ ch bisects ch) (4) OR In ΔOBF and ΔOEF 9 OB = OE 1. OB = OE (radii) 9 BF̂O = EF̂O = 90° 2. BF̂O = EF̂O = 90° (BD || AO) (BD || AO) 3. B̂2 = Ê (radii) 9 ΔOBF ≡ ΔOEF 9 BF = FE ΔOBF ≡ ΔOEF (AAS) (4) BF = FE 8.2.4 In ΔCBD and ΔCEB 9 Ê = B̂ 4 = x 1. Ê = B̂ 4 = x (proven in 8.2.1) 9 Ĉ is common 2. Ĉ is common Or 3. D̂ 4 = CB̂E = 90° + x 9 ΔCBD ||| ΔCEB (AAA) D̂ 4 = CB̂E = 90° + x Any two of the above (2) 36

Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2.5 EB CE EB CE = (sim Δs ∴ sides in proportion) 9 = BD CB BD CB EB.CB = CE.BD 9 EB.CB = CE.BD but EB = 2EF (F is the midpoint of BE) 9 EB = 2EF 2EF.CB = CE.BD (3) [21] QUESTION 9 9. Â = D̂ (∠ in same seg) 9 Â = D̂ S/R B̂ = Ĉ (∠ in same seg) 9 B̂ = Ĉ S/R AÊB = DÊC (vert opp ∠s) 9 AÊB = DÊC ΔDEC ||| ΔAEB (∠∠∠) D 999 ΔDEC ||| DE EC DC ΔAEB = = (sides in prop) AE EB AB (∠∠∠) x 8 Let AC = 11a A 4a x 7a = E 7a 4a 8 C [6] 28a 2 x= y 8 8 x = 3,5a 2 y 8 = (sides in prop) B 8 7a 64 y= 7a 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. 37

Mathematics/P3 DBE/November 2010 NSC - Memorandum QUESTION 10 F E M A || || D B C 10.1 MÊC = 90° (tan ⊥ rad) 9 MÊC = 90° MD̂C = 90° (line from cent bisects ch) (tan ⊥ rad) MÊC + MD̂C = 180° 9 MD̂C = 90° ∴ MDCE a cyclic quad (opp ∠s of quad supplementary) 9 opp ∠s of quad NOTE: supplementary OR If the word (3) MÊC = 90° (tan ⊥ rad) cyclic is used in 9 MÊC = 90° MD̂A = 90° (line from cent bisects ch) the last reason: (tan ⊥ rad) MÊC = MD̂A max 2 / 3 marks 9 MD̂A = 90° ∴ MDCE a cyclic quad (ext ∠ quad = int opp) 9 ext ∠ quad = int opp (3) 10.2 MD2 = MB2 – DB2 (Pythagoras; Δ MBD) 9 MD = MB2 2 – DB2 MC2 = MD2 + DC2 (Pythagoras; Δ MDC) 9 Pythagoras = MB2 – DB2 + DC2 9 MC2 = MD2 + DC2 (3) 10.3 DB = 30 (given) MB = 40 (radii) 9 MB = ME MC = (40) + (50) − (30)2 2 2 2 9 DB = 30 = 3 200 9 MC2 = 3200 MC = 40 2 = 56,57 or MC = 40 2 MC2 = ME2 + CE2 (Pythagoras) or MC = 56,57 CE2 = 3 200 − 1 600 CE2 = 1 600 CE = 40 mm 9 answer (4) OR MC2 = CE2 + ME2 – 2CE.ME.cos MÊC 9 cosine rule 3200 = CE 2 + (40) 2 − 2CE.(40).cos 90° 9 ME = 40 = CE 2 + 1600 9 MC2 = 3200 CE 2 = 1600 9 answer CE = 40 (4) [10] 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. QUESTION 9 O B A M C 9.1 BĈA = 90° (∠’s in a semi-circle) 9 answer (1) 9.2.1 AC = 10 2 − 8 2 (Pythagoras) 9 diameter = 10 = 36 =6 9 AC AM = 3 (line from circle centre ⊥ chord bisects chord 9 AM (3) OR midpoint theorem) 9.2.2 OM = 5 2 − 32 (Pythagoras) 9 OM =4 (OR midpoint theorem) Area ΔAOM : Area ΔABC 9 substitution 1 1 = .4.3 : .8.6 2 2 = 6 : 24 9 answer =1:4 (3) [7] OR Area ΔAOM : Area ΔABC 1 1 = .AM.OM.sin OÂM : .AC.AB.sin BÂC 2 2 1 1 = .4.3 : .8.6 2 2 = 6 : 24 Copyright reserved Please turn over 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. QUESTION 10 A 2 2 4 G H 1 1 B F 2 E C 1,5 D 10.1.1 AH 2 9 statement = (GHB || FEC) 9 reason HE 1 AH = 2y HE = y AE 2 = (BE || CD) 9 ED = 1,5y ED 1 ED = 1,5 y AH 2 = If learner stops at 2 : 1,5 : ED 1,5 no penalty AH 4 9 answer = (4) ED 3 10.1.2 BE 4 9 answer = (ΔAEB ||| ΔADC) 9reason (2) CD 6 2 = 3 10.2 HE = 2 cm (given) AH = 4 cm 9 AH and ED ED = 3 cm AD.HE = (AH + HE + ED).HE 9AD = AH + = (4 + 2 + 3).(2) HE + ED = 18 (2) [8] Copyright reserved Please turn over 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. QUESTION 11 1 A E 2 2 1 3 4 2 3 1F 4 O• B 2 3 1 C 2 1 D 11.1 D̂1 = Â 4 (tan-chord theorem) 9 Statement 9 Reason = Ĉ 2 (alt ∠’s , BA || CE) 9 S/R (3) OR Ĉ 2 = D̂ 2 (∠'s in same seg) 9 Statement = Â1 (tan-chord theorem) 9 Reason 9 S/R = Ê 2 (alt ∠’s , BA || CE) = D̂1 (∠'s in same seg) (3) OR Â 3 + Â 4 = 90° (tan ⊥ rad) F̂1 = 90° (AB || EC; coint ∠s) In ΔAFC: Ĉ 2 = 90° − Â 3 (∠ sum Δ) Ĉ1 + Ĉ 2 = 90° (∠s in semi circle) In ΔADC: D̂1 = 90° − Â 3 (∠ sum Δ) D̂1 = Ĉ 2 Copyright reserved Please turn over 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. 11.2 In ΔACF and ΔADC 1. Â 3 is common 9 Â 3 is 2. Ĉ 2 = D̂1 (proved) common ΔACF ||| ΔADC (∠∠∠) 9 Ĉ 2 = D̂1 9 Reason OR (3) In ΔACF and ΔADC 1. Â 3 is common 2. Cˆ = Dˆ 2 1 (proved) 3. F̂1 = AĈD (remaining ∠s in triangles) ΔACF ||| ΔADC 11.3 AF AC = (sim Δ’s ∴ sides in proportion) 9 statement AC AD AC.AC AF = AD 1 9 Statement AC = AO = AD (2radius = diameter) 2 1 AD. 2 AD 1 9 Simplification AF = 2 AD AD 9 Substitution AF = 4 4AF = AD (4) OR ΔAOC is equilateral 9 S/R ∴ AÔC = Â 3 = 60° AF 1 9 Statement cos 60° = = AC 2 9 Simplification 1 1 AF = AC = AO 2 2 9 Substitution 1 1 AF = ( AD) (2radius = diameter) 2 2 1 AF = AD 4 (4) AD = 4AF [10] Copyright reserved Please turn over 42

Mathematics/P3 DoE/November 2008 NSC – Memorandum QUESTION 9 P Q 1 2 2 x 1 W 1 1 2 2 1 T 3 2 R angle in a S ∧ semi-circle (1) 9.1 R 1 = 90° …( angle in a semi-circle) ∧ ∧ P 2 = 90° − x 9.2 P 2 = 90° − x …( angle between radius and tangent) ∧ ∧ ∧ ∧ S = 90° − P2 S = 90° − P2 …( ext. angle of Triangle)(sum of angles of triangle) 90° -( 90° - x ) = x = 90° – ( 90° – x ) = x (3) ∧ ∧ ∴ P1 = S = x ∧ ∧ ∧ ∧ 9.3 W 2 = P1 = x …( angles in the same segment) Q W R = P1 = x ∧ ∧ ∧ Also S = x …( proved 9.2) QWR = S ∧ ∧ reason (3) W2 = S ∴ SRWT is a cyclic quad…(ext angle = int. opposite angle) ∧ ∧ 9.4 In ∆ QWR ; ∆ QST Q W R = QST ∧ ∧ ∧ W 2 = S ….( proved 9.3) R Q W is common ∧ Q1 is common ∧ ∧ AAA or ∠∠∠ or W R Q = T 2 ….(remaining angles) equiangular or 3rd angle ∆ QWR ||| ∆ QST (AAA) or (∠∠∠) or equiangular equal (3) Copyright reserved Please turn over 43

Mathematics/P3 DoE/November 2008 NSC – Memorandum TS QT TS QT 9.5.1 = ….. ∆ QWR ||| ∆ QST = RW QR RW QR TS 8 TS 8 ∴ = = 2 4 2 4 4TS = 16 TS = 4 cm (3) ∴ TS = 4 cm 9.5.2 SQ TS = SQ TS WQ RW = WQ RW 4×5 SQ = = 10 cm 2 10 cm ∴ SR = SQ − RQ = 6 cm 6 cm (3) [16] Copyright reserved Please turn over 44

Mathematics/P3 DoE/November 2008 NSC – Memorandum QUESTION 10 A T F C E D B 10.1 CE CT 1 answer (1) = = ED TA 2 CE 1 10.2 From 10.1 = use of ratio ED 2 But DC = 9 cm ∴ DE = 6 cm DE = 6 cm (2) = BD. ∴D is the midpoint of BE. 10.3 FD BD proportion = TE BE 2 6 = answer (2) TE 12 6 × TE = 24 TE = 4 cm ALTERNATIVE proportion D is the midpoint of BE. (from 10.2) Then F is the midpoint of BT. … (sides in proportion) answer (2) ∴ TE = 2FD (midpoint theorem) = 4 cm Copyright reserved 45

2016 2008 CAPS NCS Geometry Gr 12 Memo 14 July

Study this subject

Recommended for this subject

Mathematics Functions and Finance

Mathematics Grade 12 P1 Eastern Cape Mock 2025 Addendum (Afr)

Mathematics Grade 12 Paper 2 Eastern Cape Mock 2025 Memo

Mathematics Grade 12 Paper 2 Gauteng Mock 2025 Question Paper

Mathematics Grade 12 P2 Limpopo Mock 2025 Question Paper (Afr)

Mathematics Grade 12 P2 Northern Cape Mock 2025 Memo (English)

Mathematics Grade 12 Paper 1 Eastern Cape Mock 2016 Memo

Mathematics Grade 12 Paper 1 EC June 2016 Question Paper

Mathematics Grade 12 Paper 1 Gauteng Mock 2016 Question Paper

Mathematics Grade 12 P2 Gauteng Mock 2016 Question Paper (Afr)

Mathematics Grade 12 Paper 1 Mpumalanga November 2016 Memo

Mathematics Grade 12 Paper 1 NSC June 2016 Question Paper

More from Grade 12 Mathematics

Mathematics Grade 12 P2 Northern Cape Mock 2025 QP (Eng)

Mathematics Grade 12 P2 Eastern Cape June 2024 Addendum (English)

Mathematics Grade 12 P2 Eastern Cape June 2024 Memo (English)

Mathematics Grade 12 P2 Eastern Cape June 2024 QP (Eng)

Mathematics Grade 12 P1 Eastern Cape Prelim 2024 Memo (Afrikaans)

Mathematics Grade 12 P2 Eastern Cape Prelim 2024 QP (Afr)