Mathematics Grade 12 Paper 2 NSC November 2021 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2021 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 24 pages. Hierdie nasienriglyne bestaan uit 24 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE:  If a candidate answers a question TWICE, only mark the FIRST attempt.  If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version.  Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.  Assuming answers/values in order to solve a problem is NOT acceptable. NOTA:  As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.  As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na.  Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout.  Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE. GEOMETRY  MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 10 11 13 14 14 15 16 18 18 19 19 20 21 35 35 37 40 41 1.1.1 396  396 x Answer only: Full marks 18 Slegs antw: Volpunte x  22  answer (2) 1.1.2  = 10,1707  10,17  answer (1) 1.1.3 x    32,17  32,17  5 days  5 (2) 1.2 22 18  396 ordered/bestel 20 18  360 sold/verkoop  18x1 and 18x 2 Total not sold/Totaal nie verkoop nie: 36  answer (2) OR/OF 22–20 = 2  x1  x2 2  18  36  answer (2) 1.3.1 Option B/Opsie B  B Any one of the following reasons/Enige een van die vlg redes:  Median/Mediaan = 18,5  reason  Q1  14  IQR = 21 (2)  Mean > Median, therefore the data is skewed to the right 1.3.2 Data is positively skewed/skewed to the right  answer Data is positief skeef/skeef na regs (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 Price of milk in rands per 5-litre container (x) Prys van melk in rand, 26 32 36 28 40 33 29 34 27 30 per 5 liter-houer (x) Number of 5-litre containers of milk sold (y) 48 30 26 44 23 32 39 29 42 33 Aantal 5 liter-houers melk verkoop (y) 2.1 SCATTER PLOT 50 45 1 mark: 3 to 5 points plotted 40 correctly Units sold 35 2 marks: 6 to 9 points plotted 30 correctly 25 3 marks: all points plotted 20 correctly 20 25 30 35 40 Price per 5 litre (3) 2.2 a  90,478...  90,48 a b  1,773...  1,77 Answer only: Full marks b yˆ  90,48  1,77 x Slegs antw: Volpunte  equation (3) 2.3 y = 23,069…  23,07 units/eenhede (calculator/sakrekenaar)  answer (2) OR/OF y  90,48  1,77(38)  substitution y = 23,22 units/eenhede  answer (2) 2.4 r = – 0,94  value of r OR/OF The value of r indicates a strong relationship between the cost per 5 litre strong relationship/ and the number of units sold  there is a good chance of the prediction sterk verwantskap being accurate./ Die waarde van r dui 'n sterk vewantskap tussen die koste per 5 liter en  accurate/akkuraat die aantal eenhede verkoop aan  daar is 'n goeie kans dat die (2) voorspelling akkuraat is [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y A(–2; 10) 81,87° H F G E(12; 0) x C(4; –2) B(k; k) 3.1.1 0  (2) 20  substitution C & E mBE  mCE  OR/OF mBE  mCE  12  4 4  12 1 1    answer 4 4 (2) 3.1.2 mAB  tan 81,87 Answer only: Full marks  substitution mAB  7 Slegs antw: Volpunte  answer (2) 3.2 y  mx  c y  y1  m( x  x1 ) 1 1 0  (12)  c or y  0  ( x  12)  substitution of E 4 4 1 c  3 y  x3 4 1  answer y x3 4 (2) OR/OF y  mx  c y  y1  m( x  x1 )  substitution of C 1 1  2  (4)  c or y  (2)  ( x  4) 4 4 1 c  3 y  x3  answer 4 (2) 1 y  x3 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3.1 1 y x3 4 1  substitution k  k 3 4 3 k  3 4 k  4  answer B(–4; –4) (2) OR/OF 1 1 mBE  OR/OF mBE  4 4  substitution 0k 1 k 1   12  k 4 k  12 4  4k  12  k 4k  k  12 k  4 k  4  answer B(–4; –4) (2) OR/OF mAB  tan 81,87 mAB  7  substitution 10  k mAB  2k 7(2  k )  10  k –14 – 7k = 10 – k – 6k = 24 k =–4  answer B(–4; –4) (2) OR/OF EB: y  14 x  3 and AB: y  7 x  24  equating EB & AB 1 x  3  7 x  24 4 4 x  27 27  answer x  k  4 (2)  B(4; 4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3.2 In AFG: 10  (2) mAC   2 24  mAC  2 tan   mAC  2  tan   2   180  63,43...   116,57    116,57  Â  116,57  81,87 [ext  of  ]  Â  34,70  answer (4) OR/OF In ABC: a  BC  2 17; b  AC  6 5; c  AB  10 2  all 3 lengths a  b  c  2bc.cos A 2 2 2  2 17    6 5   10 2   2(6 5)(10 2).cos A 2 2 2  substitution into the correct cosine rule  6 5   10 2    2 17  2 2 2 cos A=  cos A subject 2  6 5 10 2   0,822...  answer  A  34, 7 (4) 3.3.3  12   2 10  (0)  M ;   2 2  Diagonals intersect at the point (5 ; 5)  x-value  y-value (2) 3.4.1 BE = ET 4 17  (12  p) 2  (0  p) 2  substitution of E & T  equating 4 17    (12  p)  (0  p)  2 2 2 2 272  144  24 p  p 2  p 2 p 2  12 p  64  0  standard form ( p  16)( p  4)  0  factors p = 16 or p = – 4 (n.a.)  p = 16 T(16; 16) (5)    272 2 3.4.2a ( x  12)2  y 2  4 17  LHS  RHS (2) 3.4.2b 1 m radius  4 mtangent   4  mtangent y  4 x  c OR/OF  4  4(4)  c  substitution of B c = – 20 y = – 4x – 20  equation (3) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y C 6 D N(–1 ; 3) B(–4 ; 2) E O x A(–1 ; –1) 4.1 Radius = 4 units/eenhede  answer (1) 4.2.1 CD  CN  C(  1 ; 7) x value  y value (2) 4.2.2 CD = 6 units  D(5 ; 7) x value  y value (2) 4.2.3  h  5 units   h  5 units DC = 6 units 1  substitution into Area ΔBCD  (6)(5) Area formula 2  15 units 2  answer (3) OR/OF  h  5 units   h  5 units DC = 6 units 1  substitution into Area ΔBCD  [Area of || m ] 2 Area formula  (5)(6) 1 2  15 units 2  answer (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF Let angle of inclination of BC =  5 tan   3   59,036... BĈD 180   BĈD  180  59,036...  BĈD  120,96 BĈD  120,96  substitution into Area ΔBCD  1 2   34 (6) sin 120,96 Area rule  answer = 15 units2 (3) 4.3.1 M(3 ; –1) [reflection of N(–1 ; 3) about the line y = x]  coordinates of M (A)  MN  (3  (1)) 2  (1  3) 2  substitution of M&N MN  32  4 2  5,66 units  answer (3) 4.3.2 M(3 ; –1) 3  (1) mMN   1 1 3 MN:  1  (3)  c or c2  y  x  2  equation of MN x  x  2  equating AF & MN 2x  2 x 1  y 1 x value  y value midpoint (1 ; 1) (4) OR/OF N(–1; 3) F(3; 3) N(1 ; 3) yF  y N  3 Reflected about y = x A(–1; –1)  F(3 ; 3)  coordinates of F  1 3 1 3 midpoint  ;   (1 ; 1) x value  y value  2 2  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF N(–1; 3) F(3; 3) A(–1; –1) M(3;–1) NAMF is a square (NA=NF=AM=MF and NA  AM)  NAMF = square Midpoint NM = (1 ; 1) x  y of midpt NM  Midpoint of AF  midpt AF (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 sin 140. sin(360  x) cos 50. tan( x) sin 40( sin x)  sin40°   sin x  sin 40 tan x   co-ratio   tan x  sin x  sin x sin x   tan x  cos x cos x  cos x  answer (6) 5.2  2 sin x  cos x  1 2 LHS  RHS  2 cos x  1 1  cos(540  x) 2(1  cos 2 x)  cos x  1  identity i. t. o. cos x LHS  1  ( cos x)  cos(540°  x)  cos x  2  2 cos 2 x  cos x  1 LHS  1  cos x 2 cos x  cos x  1 2 LHS   standard form 1  cos x (2 cos x  1)(cos x  1)  factors LHS  1  cos x LHS  2 cos x  1 LHS = RHS (4) 5.3.1 sin 36  1  p 2 1 p2 1  method tan 36  p 36° p  value of p OR/OF  answer (3) cos 2 36  1  sin 2 36  method cos 36  1  (1  p ) 2  cos 36° = p p sin 36 tan 36  cos 36 1 p2   answer p (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3.2 cos 108   cos 72  reduction = –cos (2×36°)  double angle  (2 cos 2 36  1)  expansion  2 p 2  1  answer i. t. o. p (4) OR/OF cos 108   cos 72  reduction = –cos (2×36°)  double angle  (1  2 sin 2 36)  expansion   1  2 1  p 2  2  answer i. t. o. p  1  2(1  p ) 2 (4)  2 p 2  1 OR/OF cos 108   cos 72  reduction = –cos (2×36°)  double angle  (cos 2 36  sin 2 36)  expansion     p 2  1  p 2     2  answer i. t. o. p    p  (1  p ) 2 2 (4)  2 p 2  1 OR/OF cos108  cos(2  54)  double angle  2cos 54  1 2  expansion  2(1  p 2 )  1  1  2 p2  answer i. t. o. p OR/OF (4) cos108  cos(72  36)  cos 72 cos 36  sin 72 sin 36  expansion  (2 cos 2 36  1) cos 36  (2sin 36 cos 36) sin 36  2 cos3 36  cos 36  2 cos 36 sin 2 36  both double angle identities  1 p  2  2 p3  p  2 p 2  2 p3  p  2 p  2 p3  value of sin 36°  answer i. t. o. p  4 p3  3 p (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1.1 cos(   )  cos(  ( ))  cos(  ( ))  cos  cos( )  sin  sin( )  expansion  cos cos   sin  ( sin  )  reduction  cos  cos   sin  sin  (3) 6.1.2 2 cos 6 x cos 4 x  cos 10 x  2 sin 2 x  cos10x = cos(6x + 4x)  2 cos 6 x cos 4 x  cos(6 x  4 x)  2 sin 2 x  2 cos 6 x cos 4 x  (cos 6 x cos 4 x  sin 6 x sin 4 x)  2 sin 2 x  expansion of cos(6x + 4x)  cos 6 x cos 4 x  sin 6 x sin 4 x  2 sin 2 x  cos 2 x  2 sin 2 x  cos 2 x  1  2 sin 2 x  2 sin 2 x  1  2 sin 2 x =1  answer (5) 6.2 tan x  2 sin 2 x sin x  2(2sin x cos x)  quotient identity cos x sin x  4 sin x cos 2 x  double angle identity 4 sin x cos 2 x  sin x  0 sin x(4 cos 2 x  1)  0  factors 1 sin x  0 or cos 2 x  4  both equations 1 cos x   2 x = 180° + k.360°; k  Z or x = 120° + k.360°; k  Z  x = 180° x = 240° + k.360°; k  Z  x = 120°& 240° OR/OF x = ±120° OR/OF  k.360°; k  Z tan x  2sin 2 x (7) sin x  4sin x cos x  quotient identity cos x sin x  4sin x cos 2 x 4sin x cos 2 x  sin x  0  identity 4sin x(1  sin 2 x)  sin x  0 3sin x  4sin 3 x  0 sin x(3  4sin 2 x)  0  factors 3 sin x  0 or sin 2 x  4  both equations 3 3 sin x  or sin x    x = 180° 2 2  x = 120°& 240° OR/OF x  180  k .360 , k  Z or x  120  k .360 , k  Z x = ±120° or x  240  k .360 , k  Z  k.360°; k  Z (7) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 y 2 1 g x  both –180° –90° 0 90° 180° turning points  both –1 f x intercepts (–30° & 150°) –2  shape (3) 7.2 Period = 120°  answer (2) 7.3 x   30  answer (1) 7.4 Range of/waardeversameling van g: y [1 ; 1] 1  1 1 Range of/Waardeversameling van g: y   ;  2  2 2 1 1 3  critical values Range of/Waardeversameling van g + 1: y   ;  2 2 2  correct notation OR/OF (2) 1 1 3 Range of/Waardeversameling van g + 1:  y   critical values 2 2 2  correct notation (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 T S x 90° + x x Q 5 R 8.1 In ∆SQR: QS QR  sinx sin 90  x   correct use of sine rule QS 5  sinx cosx  sin90  x   cos x 5sin x 5sin x QS   QS  cos x cos x QS  5 tan x (3) 8.2 QT TS  correct use of sine rule  sin 180  2 x  sin x QT 5tanx   TS = QS = 5tanx sin 2 x sin x 5tan x sin 2 x QT   QT  5tan x sin 2 x sin x sin x  sin x  5  (2 sin x cos x) QT   cos x  sin x  tan x  sin x cos x 5sin x (2sin x)  sin 2 x  2sin x cos x QT  sin x QT 10 sin x (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF QT 2  QS2  TS2  2QS.TScosQŜT  correct use of cos rule QT  (5 tan x)  (5 tan x)  2(5 tan x).(5 tan x)cos(180 - 2 x)  TS = QS = 5tanx 2 2 2 QT 2  50 tan 2 x  50 tan 2 x(cos2 x) QT 2  50 tan 2 x(1  cos2 x)  cos 2 x  2 cos 2 x  1 & QT 2  50 tan 2 x(1  2 cos 2 x  1) reduction QT 2  50 tan 2 x(2 cos 2 x) sin 2 x sin x QT  100 2 (cos 2 x)  tan x  2 cos x cos x QT 2 = 100 sin 2 x  QT = 100 sin 2 x 2 QT = 10 sin x (5) OR/OF TS2 =QS2 +TQ 2  2QS.TQ.cos x  correct use of cos rule  TS = QS = 5tanx (5 tan x) 2  (5 tan x) 2  TQ 2  2(5 tan x).TQ.cos x  quadratic equation ito 0  TQ 2  2(5 tan x).TQ.cos x TQ 0  TQ  TQ  10 tan x.cos x  TQ  10 tan x.cos x (TQ  0)  TQ = 10tan x . cos x sin x sin x  tan x   10 .cos x cos x cos x  10sin x (5) 8.3 1 Area of Δ TQR  .TQ.QR sin TQ̂R 2  10 sin 255( sin70) 1  correct substitution into 2 the area rule  9,93 unit 2  answer (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 Q 93° P 1 R 2 1 2 S 3 78° T W 9.1 tangents from same(common) point/raaklyne vanaf dieselfde punt R (1) 9.2.1 Ŝ 2  SR̂T [s opp equal sides/e teenoor gelyke sye] R  Ŝ2  51 [sum of s in ∆/som van e in ] S (2) 9.2.2 Ŝ2  Ŝ3  93 [ext  of cyclic quad/buite van koordevh] R Ŝ3  42  answer (2) OR/OF Ŝ1  87 [opp s of cyclic quad/teenoorst e v kdvh] R Ŝ3  180  (87  51) Ŝ3  42 [s on a str line/e op reguitlyn]  answer (2) [5] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 C E x F 1 1 2 M 2 3 A 3 B D 10.1 line from centre  to chord/lyn vanaf middelpunt  op koord R (1) 10.2  Â1  90  x [sum of s in ∆/som van e in ] S  M̂1  180  2 x [at centre=2× at circumf/midpts= 2×omtreks] SR (3) 10.3 CÂD  90 [ in semi circle/ in halfsirkel ] SR Â 2  90  (90  x) Â 2  x S  Â 2  Ĉ  x  AD is a tangent [converse tan-chord theorem/omgek rkl-kd st. ] R (4) OR/OF ˆ  2x [adj suppl s/aanligg suppl e] EMD S Aˆ x [at centre= 2× at circumf/midpts= 2 ×omtreks] 2 SR  ˆ A C x 2  AD is a tangent [converse tan-chord theorem/omgek rkl-kd st. ] R OR/OF (4) M̂3  180  2 x [vert. opp/ regoorstaande e]  Â3  90  x [at centre= 2× at circumf/midpts= 2 S ×omtreks] R ˆ  90 BAE [ in semi-circle / in halfsirkel] S  ˆA C x 2 R  AD is a tangent [converse tan-chord theorem/omgek rkl-kd st. ] (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne CD || AB [midpt. Thm/ middelpuntst.] ˆ  90 BAE [ in semi-circle / in halfsirkel] ˆ D ˆ  90  x S A 3 [alt.s; CD||AB/verwiss e] R  Â2  x  C  AD is a tangent [converse tan-chord theorem/omgek rkl-kd st. ] S R OR/OF (4) ˆ  90 CAD [ in semi circle/ in halfsirkel ] AC = diameter [converse  in semi circle/omgek  in halfsirkel] SR  AD is a tangent [converse radius  tangent/omgek radiusrkl] S R (4) 10.4 AF = FE and BM = ME [given & radii] 1  FM = AB = 12 units [Midpt Theorem/middelptstelling]  FM = 12  R 2 EM = MB = CM = 18 units [radii]  EB = 36 units [diameter = 2 radius]  EB = 36  AE2 = (36)2 – (24)2 [Pythagoras]  using Pyth correctly AE = 12 5 or 26,83 units  answer (5) OR/OF AF = FE and BM = ME [given & radii] 1  FM = 12  R  FM = AB = 12 units [Midpt Theorem/middelpuntstelling] 2 EM = MB = CM = 18 units [radii]  EM = 18  FE2 = (18)2 – (12)2 [Pythagoras]  using Pyth correctly FE = 6 5  answer AE = 12 5 or 26,83 units (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 11.1 B E D O C F K Construction: Draw diameter BF and draw BE Constr Konstruksie: Trek middellyn BF en verbind BE ˆ BFK  90 or DF̂K  90  BF̂D [radius  tangent/raaklyn] SR BÊF  90 [ in semi-circle/semi-sirkel] S ˆ = 90  BED  DEF ˆ = 90° – BF̂D [s same segment/e dieselfde segment]  S/R ˆ  DEF  DFK ˆ (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 21 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne E D O K F C Construction: Draw radii DO and OF  construction Konstruksie: Trek radii DO en OF ˆ  90 or DFK OFK ˆ  90  OFD ˆ radius tangent/raaklyn]  S R ˆ  OFD ODF ˆ [s opp = sides/e teenoor = sye]  S ˆ  180  2OFD  DOF ˆ [s of /e van ] ˆ  90  OFD DEF ˆ [ at centre = 2× circumf/  S/R midpts = 2× omtreks] ˆ  DEF  DFK ˆ (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 22 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne B E D O K F C Construction: Draw diameter BF and join BD.  construction Konstruksie: Trek middellyn BF en verbind BD. ˆ  90 or DFK BFK ˆ  90  BFD ˆ [radius  S /R tangent/raaklyn] ˆ S FDB  90 [ in half circle/semi-sirkel] ˆ  90  BFD B ˆ ˆ B  DFK ˆ but Bˆ  Eˆ [s same segment/e dieselfde segment]  S/R  DF̂K  Ê (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 23 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne 11.2 L K 4 1 3 2 1 2 3 T N 2 1 1 1 2 2 S 3 1 M P 11.2.1(a) K̂ 4  Ŝ1 [tan chord theorem/raaklynkoordstelling] S R S R M̂ 2  M̂ 3  Ŝ1 [corresp s; / ooreenk s; MN || KS]  K̂ 4  M̂ 2  M̂ 3  NM̂L (4) 11.2.1(b)  K̂  M̂  M̂  NM̂L 4 2 3  KLMNis a cyclic quad [ext  of quad = opp int  /  R buite  van vh = teenoorst binne ] (1) OR/OF N1  K̂ 1  K̂ 2  NK̂S [corresp s; / ooreenk s; MN || KS] NK̂S  KL̂S [tan chord theorem / raaklynkoordstelling] N̂1  KL̂S  KLMNis a cyclic quad [ext  of quad = opp int  /  R buite  van vh = teenoorst binne ] (1) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 24 DBE/November 2021 NSC/NSS – Marking Guidelines/Nasienriglyne NKL  180  K 4 [adj. suppl.]  NKL  180  NML [proved]  KLMN is a cyclic quad [opp.s supplementary]  R (1) 11.2.2 In LKN ||| ΔKSM : N̂ 3  M̂ 3 [s in the same seg / e in dieselfde sirkel segm] SR L̂1  M̂ 2 [s in the same seg / e in dieselfde sirkel segm] S  K̂ 2 [alt s; / verwe; MN||KS]  S/R ˆ ˆ  MSK NKL [s of  / e van ] S LKN ||| ΔKSM (5) OR/OF In LKN ||| ΔKSM : N̂ 3  M̂ 3 [s in the same seg / e in dieselfde sirkel segm] SR NKL ˆ =ˆ M [ext  of cyclic quad/buite van koordevh]  S/R 1 [corresp s/ooreenk e; KS || NM]  Ŝ2 S LKN ||| ΔKSM [,,]  R (5) OR/OF In LKN ||| ΔKSM : N̂ 3  M̂ 3 [s in the same seg / e in dieselfde sirkel segm] SR Kˆ + ˆ 4 NKL=Sˆ +Sˆ [s on straight line/e op reguitlyn] 1 2 ˆ ˆ  S/R  NKL=S2 [ K̂ 4 =Sˆ 1 ] S LKN ||| ΔKSM [,,]  R (5) 11.2.3 LK KN  [ LKN ||| ΔKSM ] KS SM SR 12 4   KS 3  substitution KS  9 units  answer (4) 11.2.4 4SM = 3KN 3(8) SM  4 SM = 6  SM = 6 LT LS  [line || one side of  / lyn || een sy v ∆] NL M L SR LT 13  16 19 208  answer LT   10,95 19 (4) [23] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou