# Mathematics Grade 12 P2 Gauteng Mock 2022 Memo (Afrikaans) > Grade 12 Mathematics Paper 2 memorandum from the Gauteng Mock Exam 2022 (Afrikaans), 26 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr12-mathematics-p2-english-and-afrikaans-sep-2022-possible-answers/9c2c4ea7-3e18-475d-a041-f5b938ecc7d6 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2022 - Exam period: Gauteng Mock Exam - Paper: 2 - Curriculum: CAPS - Pages: 26 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 P2 Gauteng Mock 2022 Memo (Afrikaans)](https://www.hlayiso.com/library/gr12-mathematics-p2-english-and-afrikaans-sep-2022-possible-answers/9c2c4ea7-3e18-475d-a041-f5b938ecc7d6) ## About this document Mathematics · Grade 12 · Gauteng Mock Exam · 2022 · Afrikaans. Memorandum, 26 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN 2022 MARKING GUIDELINES/ NASIENRIGLYNE (10612) MATHEMATICS (PAPER 2)/WISKUNDE (VRAESTEL 2) 26 pages/bladsye ### Page 2 Downloaded from hlayiso.com 10612/22 NOTE: • If a candidate answers a question TWICE, mark only the FIRST attempt. • If a candidate has crossed-out an attempt of a question and has not redone the question, mark the crossed-out version. • Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. • Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: • As ʼn kandidaat ʼn vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As ʼn kandidaat ʼn antwoord van ʼn 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. • Aannames van antwoorde/waardes om ʼn 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 ʼn Punt vir ʼn korrekte bewering (ʼn Punt vir ʼn bewering is onafhanklik van die rede.) A mark for a correct reason (A reason mark may only be awarded if the statement is correct.) R ʼn Punt vir ʼn korrekte rede (ʼn Punt word slegs vir die rede toegeken as die bewering korrek is.) Award a mark if the statement AND the reason are both correct. S/R (Ken ʼn punt toe as beide die bewering EN die rede korrek is.) 2 ### Page 3 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 1 1.1 a = 38,26  a = 38,26 b = 2,46  b = 2,46 yˆ = 38,26 + 2,46 x  yˆ = 38,26 + 2,46 x answer only full NOT allowed to round to Units, but may round off to 3 decimal marks/ or 1 decimal place antwoord alleenlik Mag NIE afrond tot Ene nie,maar mag afrond tot 3 desimale of volpunte 1 desimale plek (3) 1.2 r = 0,97  answer/antwoord (1) 1.3 Very strong positive correlation/  srong positive/ Baie sterk positiewe korrelasie sterk positief (1) 1.4 x = (2 + 8 + 4 + 6 +12 +10 +11)  7 = 7,57  x = 7,57 y = 38,26 + 2,46(7,57 ) − 15,74 + 4,54 (7,57 ) substitute/vervang y = 6, 77 7,57  answer/antwoord The difference is/Die verskil is 6,77% (% not necessary to indicate/ % nie nodig om aan te dui) (3) 1.5 (2 ; 9)  answer/antwoord OR/OF x = 2; y = 9  answer/antwoord (1) 1.6 Lack of attendance by online learner/Gebrek aan bywoning van any valid aanlyn leerling answer/enige geldige OR/OF antwoord Section taught may have been difficult to grasp online/Afdeling wat onderrig was, was moeilik om te verstaan met aanlynklas OR/OF Poor connectivity/Slegte verbinding OR/OF Lack of resources/Gebrek aan hulpbronne OR/OF Insufficient ICT skills/Onvoldoende Rekenaarvaardighede (1) [10] 3 ### Page 4 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 2 2.1.1 2 500  w  3 000  answer/antwoord (1) 2.1.2 250 (1) + 750 (2) + 1 250 (3) +1 750 (8) + 2 250 (6) + 2 750 (15) + 3 250 (5) 40 90500 estimated/geskatte x =  90 500 40 40 = 2 262,5grams/gram  answer/antwoord If numerator is correct and denominator is incorrect 1/3 If numerator is incorrect and denominator is correct 2/3 (CA) Indien teller korrek is en noemer is verkeerd 1/3 Indien teller verkeerd is en noemer is korrek 2/3 (CA) (3) 2.2.1 Cumulative Frequency Curve (Ogive) Kumulatiewe Frekwensiekurwe (Ogief) 45  grounding point/ Cumulative frequency/Kumulatiewe frekwensie gegronde punt (4 ; 0) 40 and/en (3 504 ; 40) 35 shape/vorm (if ruler is used to connect 30 points, NO mark for 25 shape/indien liniaal gebruik word om 20 punte te verbind 15 GEEN punt vir vorm 10  (1 mistake - 1 5 mark; 2 mistakes - 0 no mark/1 fout – 1 4 504 1004 1504 2004 2504 3004 3504 punt; 2 foute – geen Weight (in grams)/Gewig (in gram) punte) (4) 2.2.2 It will not deviate./it will remain the same.  answer/antwoord Dit sal nie afwyk nie./dit sal dieselfde bly. (1) 2.2.3 2 262,5 + 4  addition of 4/4 CA bygetel from = 2 266,5 grams/gram /van  answer/antwoord 2.1.2 (2) [11] 4 ### Page 5 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 3 y B(4 ; 8) A(− 3; k ) P   E 1 O C(5 ; 0) x D(− 2 ; − 4) • F 3.1 8−0  correct substitution mBC = 4−5 into gradient formula (swop x and y around 0/2)/korrekte vervanging in gradiënt formule (ruil x en y om 0/2/ mBC = −8  answer/antwoord (2) 3.2 AB = 65 = (− 3 − 4)2 + (k − 8)2  substitute A and B into distance formula/ 65 = 49 + k 2 − 16k + 64 vervang A en B in die k 2 − 16k + 48 = 0 afstandsformule ( k − 4 )( k − 12 ) = 0  standard form/ k = 4 or/of k = 12 standaardvorm  factors/faktore k = 4  k =4 OR/OF  substitute A and B into AB = 65 = (− 3 − 4) + (k − 8) 2 2 distance formula/ vervang A en B in die 65 = 49 + (k − 8) 2 afstandsformule (k − 8)2 = 16  isolate square/isoleer k − 8 = 4 kwadraat  square root both k = 4 or/of k = 12 sides/vierkantswortel k = 4 weerskante  k =4 (4) 5 ### Page 6 Downloaded from hlayiso.com 10612/22 3.3 8 − (− 4) mBD = =2 4 − (− 2) 4−0 1  mBD = 2 mAC = =− 1 −3−5 2  mAC = − 1 2  mBD  mAC = 2  − = −1  m BD  mAC = −1 2  BP ⊥ AC (must show the multiplication/ moet die OR/OF vermenigvuldiging aandui Coordinates of P = Midpoint of AC = Midpoint of BD/ Koördinate van P=Middelpunt van AC = Middelpunt van BD −3+5 4+0  − 2+ 4 − 4+8 = ;  = ;   2 2   2 2  = (1 ; 2) 8−2  mBP = 2 mBP = =2 4 −1 1 4−0 1  mAC = − mAC = =− 2 −3−5 2 1  mBP  mAC = −1  mBP  mAC = 2  − = −1 2 (must show the  BP ⊥ AC multiplication/ moet die vermenigvuldiging aandui (3) 3.4 Midpoint of AF = Midpoint of DC/ Middelpunt van AF = Middelpunt van DC x + ( −3) −2 + 5 y + 4 −4 + 0 = and/en =  x − coordinate/ 2 2 2 2 x = 6 and/en y = −8 x − koördinaat y – coordinate/  F(6 ; − 8) y − koördinaat Answer Only: Full Marks/Antwoord alleenlik: Volpunte OR/OF Method: Translation/Metode: Translasie A → D:  x − coordinate/ ( x ; y ) → ( x + 1; y − 8) x − koördinaat  by symmetry/as simmetrie :C → F:  y – coordinate/ y − koördinaat C ( 5;0 ) → F ( 5 + 1;0 − 8 )  F ( 6; −8 ) Answer Only: Full Marks/Antwoord alleenlik: Volpunte (2) 6 ### Page 7 Downloaded from hlayiso.com 10612/22 3.5 ONLY QUESTION (3.5) IN PAPER WHERE LEARNER WILL BE PENALISED FOR INCORRECT ROUNDING ENIGSTE VRAAG (3.5) IN VRAESTEL WAAR LEERLING GEPENALISEER WORD VIR VERKEERDE AFRONDING m AD = m BC = − 8  mAD = −8 tan  = m AD tan  = − 8  tan  = −8  = 180  − tan −1 (8)  = tan −1 (− 8) + 180 = 180  − 82,87 OR/OF = −82,87 + 180 = 97,13 = 97,13   = 97,13 tan  = mBD = 2   = 63,43  = 63, 43 ˆ ̂ ==63, O 11 43 ((vert.opp. s/oorst 63,43 . e ) s/regoorst.e) ˆ =  −  ( ext.  of Δ/buite  van  ) EDO = 97,13 − 63, 43  −  = 33, 7  answer/antwoord OR/OF  AP̂D = 90 In APD , AP̂D = 90  AP = 2 5 AP = 2 5  PD = 3 5 PD = 3 5 ˆ = AP tan EDO  trig ratio/ PD trig verhouding ˆ =2 5 tan EDO  substitution/ 3 5 vervanging ˆ = 33,69  33,7 EDO ( rounded off to ONE decimal place/rond af tot EEN desimale plek )  answer/antwoord OR/OF In APD , AP̂D = 90  AP̂D = 90 AP = 2 5 & AD = 65  AP = 2 5  AD = 65  trig ratio/ ˆ = AP sin EDO trig verhouding AD ˆ =2 5  substitution/ sin EDO vervanging 65 ˆ = 33,69  33,7 EDO  answer/antwoord ( rounded off to ONE decimal place/rond af tot EEN desimale plek ) OR/OF 7 ### Page 8 Downloaded from hlayiso.com 10612/22 In APD AP̂D = 90  AP̂D = 90 PD = 3 5  PD = 3 5 AD = 65  AD = 65 PD cos ˆ EDO =  trig ratio/trig AD verhouding ˆ =3 5 cos EDO  substitution/ 65 vervanging ˆ EDO = 33,69  33,7  answer/antwoord ( rounded off to ONE decimal place/rond af tot EEN desimale plek ) OR/OF In ABD  BD = 6 5 BD = 6 5  AD = 65 AD = 65  cosine rule/ ˆ = AD + BD − AB 2 2 2 cos EDO 2 ( AD )( BD ) kosinusreël ˆ =( 65 ) + ( 6 5 ) − ( 65 ) 2 2 2  substitution/ 3 cos EDO = vervanging 2 ( 65 )( 6 5 ) 13 3  ˆ = 33,69  33,7 EDO 13  answer/antwoord ( rounded off to ONE decimal place/rond af tot EEN desimale plek ) (6) 8 ### Page 9 Downloaded from hlayiso.com 10612/22  length of AC/ 3.6 ( 5 − ( −3) ) + ( 0 − 4 ) = 80 = 4 5 2 AC = 2 lengte van AC  length of DP/ = ((−−22−−)11)+ (+ −4( − −42 )− = 2) 245= = 3455 = 3 5 2 2 2 DP = DP lengte van DP 1  correct Area of/van ΔADC = AC  DP 2 substitution into ( ) ( ) 1 formula/ korrekte  Area of/van ΔADC = 4 5  3 5 vervanging in 2 formule = 30 = square units/vierkante 30 square eenhede units/vierkante eehede  answer/antwoord OR/OF ABCD is a rhombus, because the diagonals bisect perpendicularly and all sides are equal. ABCD is ’n ruit, want die hoeklyne halveer mekaar reghoekig en al die sye is ewe lank.  CD̂O = ED̂O (diags of rhombus bisect s of rhombus/diag van  CD̂O = ED̂O ruit halveer die e van die ruit)  AD̂C = 2  33,7 = 67,4  AD̂C = 67,4 1 ˆ Area of/van ΔADC =  AD  DC  sinADC  substitution into 2 formula/ 1 ( ) (  Area of/van ΔADC =  65  65  sin ( 67,4 ) 2 ) vervanging in formule = 30 square units/vierkante eenhede  answer/antwoord (4) [21] 9 ### Page 10 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 4 y P(5 ;4) Q N x O T 4.1 x 2 + y 2 − 4 x − 2 y − 13 = 0 completing the square/vierkants- x 2 − 4 x + 4 + y 2 − 2 y + 1 = 13 + 4 + 1 voltooiing (x − 2)2 + ( y − 1)2 = 18 Answer only: Full marks/  ( x − 2 ) + ( y − 1) 2 2 Antwoord alleenlik: Volpunte RHS/RK (3) 4.2 N (2 ;1) both x and y correct/ albei x en y korrek CA from/van 4.1 NT = 18 = 3 2 OR/OF 2,24 length of NT/lengte van NT (2) 4.3 Midpoint of TNP/Middelpunt van TNP, N (2 ;1) xT + 5 y +4 = 2 and/en T =1 2 2  x − value/waarde T (−1; − 2)  y − value/waarde OR/OF Method: Translation/Metode: Translasie P → N: (x ; y ) → (x − 3 ; y − 3) N → T : N (2;1) → T (2 − 3 ;1 − 3)  x − value/waarde  y − value/waarde  T (− 1;−2) 10 ### Page 11 Downloaded from hlayiso.com 10612/22 1 − (− 2) 3  m NT = 1 mNT = = =1 2 − (− 1) 3 radius ⊥ tangent  mQT = −1 mNT m QT = −1 radius ⊥ raaklyn mQT = −1 substitute/vervang m y = −x + c and/en T(− 1; − 2) y − (−2) = −1x − (− 1) CA from/van 4.3 − 2 = −1(−1) + c y + 2 = −1(x + 1) OR/OF c = −3 y = −x − 3 y = −x − 3  y = −x − 3 (6) 4.4 NS = 3NT NS2 = 9NT 2 ( a − 2 ) + ( b − 1) = 9 ( 18 ) = 162 ……………... (1) 2 2 2  equation/vergelyking NT ⊥ QT and/en ST ⊥ QT  mST = mNT = 1  mST = mNT = 1 b+2 b+2 =1  =1 a +1 a +1 b + 2 = a +1 …………………………………….. (2)  b = a −1 b = a −1 Substitute ((2) into Substitute (1 / vervang(2) into (1)/vervang (2)inin(1) (1) ( a − 2 ) + ( a − 1 − 1) = 162 substitution/ 2 2 vervanging ( a − 2 ) + ( a − 2 ) = 162 2 2 2 ( a − 2 ) = 162 2 ( a − 2 ) = 81 2 a−2=9 or/of a − 2 = −9 a = 11 or/of a = −7 But/maar a  0  a = −7  a = −7 b = ( −7 ) − 1 = −8  b = −8 S ( −7 ; − 8 ) OR/OF 11 ### Page 12 Downloaded from hlayiso.com 10612/22 NS = 3NT NS2 = 9NT 2 ( a − 2 ) + ( b − 1) = 9 ( 18 ) = 162 ………………….. (1) 2 2 2 equation/ (1) vergelyking (1) NT ⊥ QT and/en ST ⊥ QT  mST = mNT = 1  mST = mNT = 1 b+2 b+2 =1  =1 a +1 a +1 b + 2 = a +1  b = a −1 b = a − 1 ………………………………………….... (2) Substitute Substitute (2) into ((1)/vervang (2) into 1) / vervang(2) (2) in in (1) (1) substitution/ ( a − 2 ) + ( a − 1 − 1) = 162 2 2 vervanging ( a − 2 ) + ( a − 2 ) = 162 2 2 a 2 − 4a + 4 + a 2 − 4a + 4 = 162 2a 2 − 8a − 154 = 0 a 2 − 4a − 77 = 0 ( a − 11)( a + 7 ) = 0 a = 11 or/of a = −7 But/ maar a <00 But/Maar  a = −7  a = −7 b = ( −7 ) − 1 = −8  b = −8 S ( −7 ; − 8 ) OR/OF NTS is a straight line/NTS is ’n reguitlyn mSN = 1  mSN = 1 b −1 b −1 =1  a−2 a−2 b −1 = a − 2 b = a −1 ....................(1) equation/ (1) ST = 2NT = 2 18 vergelyking (1) ST 2 = 4(18) = 72 12 ### Page 13 Downloaded from hlayiso.com 10612/22 ( a + 1) + ( b + 2 ) = 72....... ( 2 ) equation/ (2) 2 2 vergelyking (2) Substitute (1) into ( 2 ) / 2(a + 1) = 72 2 vervang (1) in (2) (a + 1)2 = 36 ( a + 1) + ( a − 1 + 2 ) = 72 2 2 aa ++11==−–6 a + 1a =+ 61 = 6 6 oror/of substitution/ 2a + 4a − 70 = 0 2 vervanging OR/OF aa == −–7 a =a 5= 5 7 or/of a 2 + 2a − 35 = 0 a = −7 ( a + 7 ) ( a − 5) = 0 b = −7 − 1 = −8 aa==−–7 or/ofa a= = 7 or 55 S (− 7 ; − 8) aa==−7–7 and and/enb b = (=−7 ) − 1–1= =−8–8 Answer only: 2/7/ (–7) coordinates/ S ( −7 ; − 8 ) Antwoord alleenlik: 2/7 koördinate OR/OF N(2 ;1)  diagram 3 2 3  (2 ; − 2) T(–1 ;–2) (2 ;–2) 3  (2 ; − 8) 6 2 6 6 P(5 ;4)  S (− 7; − 8) S(a ;b) 6 (–1 ;–8) 3 (2 ;–8) OR/OF dividing of a line xT − x N y − yN 1 segment into a = T = given ratio/verdeel xS − x N yS − y N 3 lynsegment in −3 −3 1 gegewe verhouding = = a − 2 b −1 3 substitution/ vervanging a − 2 = −9 equation/ a = −7 vergelyking  a = −7 b − 1 = −9 b = −8 equation/ S (− 7 ; − 8) vergelyking  b = −8 (7) [18] 13 ### Page 14 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 5 5.1 −3 O −3 3 5.1.1 ( OP 2 = (− 3) + − 3 3 2 ) 2  Pythagoras OP = 6 OP = 6 −3 1 1 cos 2 = =−  cos 2 = − 6 2 2 (3) 5.1.2 cos 2 = 1 − 2 sin 2   double angle identity/ dubbelhoekidentiteit  2 sin 2  = 1 − cos 2 1 − cos 2  sin 2  = 2  1 correct substitution/ 1−  −  korrekte vervanging =  2 2 3 = 4 3 answer/antwoord  sin  = 2 OR/OF 1 − 1 = 1 − 2 sin 2  − = 1 − 2 sin 2  2 2 3  − = −2 sin 2  2 simplification/ 3  sin 2  = vereenvoudiging 4 3  sin  =  answer/antwoord 2 (3) 14 ### Page 15 Downloaded from hlayiso.com 10612/22 5.2 cos2 (180  + x ) + cos(− x ). tan x. cos(90 + x )  cos2 x = cos2 x + cos x. tan x.(− sin x )  cos x  − sin x .(− sin x ) sin x = cos2 x + cos x.  sin x cos x cos x = cos2 x − sin 2 x  cos2 x − sin 2 x = cos 2 x  cos 2 x (6) 5.3.1 5 tan  − 6 cos = 0  sin   sin   tan  = 5  − 6 cos = 0 cos  cos  5 sin  − 6 cos2  = 0 multiplying by cos  / ( 5 sin  − 6 1 − sin  = 0 2 ) vermenigvuldiging met cos  5 sin  − 6 + 6 sin 2  = 0  cos2  = 1 − sin 2  6 sin 2  + 5 sin  − 6 = 0 (3) 5.3.2 6sin 2  + 5sin  − 6 = 0 factors (must ( 3sin  − 2 ) ( 2sin  + 3) = 0 indicate)/faktore (moet 2 3 aandui) sin  = or/of sin  = − 3 2 both values of sin / no no solution/ geen oplossing albei waardes van sin  no solution/geen oplossing  = 41,81 + k .360; k   or/of  = 138,19 + k .360; k     = 41,81 or/of 138,19°  + k.360 ; k   (5) 5.4 This question has been removed from the question paper. Do not mark this question. Hierdie vraag is uit die vraestel verwyder. Moenie hierdie vraag merk nie. (0) 5.5 1 sin ( 3 −  ) = ref/verwys  = 45 2 3 −  = 180 − 45 3 −  = 135........................ (1)  3 −  = 135 1 tan ( 2 +  ) = ref/verwys  = 30 3 2 +  = 180 + 30 2 +  = 210........................ ( 2 )  2 +  = 210 5 = 345 (1) + ( 2 )  = 69   = 69  = 72   = 72 (4) [24] 15 ### Page 16 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 6 6.1 See the graph above/ asymptotes/asimptote Sien bostaande grafiek  x = 0 ; x =120 shape/vorm  (− 90 ; − 1) or/of (90 ; 1) (4) 6.2 180  method/metode Period of/Periode van g = = 120  answer/antwoord 1,5 Answer only: Full marks/ Antwoord alleenlik: Volpunte (2) 6.3 x  (0 ;180) OR/OF 0  x  180  values/waardes If values are incorrect 0/2  notation/notasie Indien waardes verkeerd is 0/2 (2) 6.4 x  (−60 ; − 30] OR/OF − 60  x  −30 both values/albei waardes CA or or −60 ; − 30 from x  (60 ; 90] OR/OF 60  x  90 notation/notasie graph/ both values/albei waardes vanaf 60 ; 90 grafiek If values are incorrect 0/4 notation/notasie Indien waardes verkeerd is 0/4 (4) 6.5 h ( x ) = g ( x − 30 ) 3  = − tan  ( x − 30 )  2   3 (x − 30) 2 3  = − tan  x − 45  2  (2) [14] 16 ### Page 17 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 7 A C B D h In ΔABC:sin = AC h  AC = AC in terms of h and  / sin  AC in terme van h en  In ΔACD: AC = AD ˆ = 180 − 2  s opp. = sides; s of Δ = 180 ˆ = ACD ADC / 2 [e teenoor = sye; som vd  van  =180]  ADCˆ = 90 −   AD̂C = 90 −  CD AC = sin 2 sin ( 90 −  ) correct subst. into sine rule/ AC.sin 2 CD = korrekte vervanging in sin ( 90 −  ) sinusreël h 2sin  .cos  CD as subject/CD as =  onderwerp sin  cos   sin 2 = 2 sin  cos  sin (90 −  ) = cos = 2h answer/antwoord OR/OF 17 ### Page 18 Downloaded from hlayiso.com 10612/22 h In ΔABC: sin = AC h AC in terms of h and  /  AC = AC in terme van h en  sin  In ΔACD: AC = AD AC = CD CD = AC + AD − 2 AC. AD cos 2 2 2 2 2 2 2 correct subst. into cosine  h   h   h  CD =   +  − 2  cos 2 2 rule/  sin    sin    sin   korrekte vervanging in kosinusreël CD 2 = h2 + sin  sin  2 h2 2 − 2 h2 sin  2 ( 1 − 2sin 2  )  cos 2 = 1 − 2 sin 2  2 2 h h multiplication/ CD 2 = 2 2 − 2 2 + 4h 2 sin  sin  vermenigvuldiging CD = 4h 2 2 simplification/ vereenvoudiging CD = 2h answer/antwoord (7) [7] 18 ### Page 19 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 8 S U 3 R T 2 1 5 4 P 3 2 1 O1 V 106º Q W 8.1.1 PŜR = 106 ext.  of cyclic quad/buite  van kvh. S R (2) 8.1.2 P̂4 = R̂ 3  s opposite = sides/  e teenoor = sye S/R OR/OF equal chords; equal  s /gelyke 180  − 106  koorde; gelyke  e R̂ 3 = sum of  s in a triangle/som vd  e v driehoek 2 S R̂ 3 = 37 S (3) 8.1.3 P̂5 = R̂ 3 = 37 tan chord theorem/raaklyn koordstelling S R (2) 8.1.4 Ô1 = 212  centre = 2   at circumference/ R middelpunts  = 2  omtreks  S (2) 19 ### Page 20 Downloaded from hlayiso.com 10612/22 8.1.5 P̂3 + P̂4 + P̂5 = 90 radius ⊥ tangent/radius ⊥ raaklyn R P̂3 = 90 − 37 − 37 = 16  P̂3 = 16 OR/OF PÔR = 148  ’s around a point/  ’e om ‘n punt R P̂3 = R̂ 2 = 16  ’s opposite radii/  ’e teenoor radiusse  P̂3 = 16 (2) 20 ### Page 21 Downloaded from hlayiso.com 10612/22 8.2 A 1 2 = B O E 1 F 2 3 1 = 4= C = 2 1= D 8.2.1  2 = Ê = x  s in the same segment/  e in dies. S/R segment ˆ =A Ê = A ˆ = x = chords subtend =  s /gelyke koorde S/R 1 2 OR/OF onderspan gelyke  e ˆ =A C ˆ = x  s opposite radii/  ’e teenoor radiusse S/R 2 2 Learner MUST first determine  2 in order to get the mark for Ĉ 2 Leerling MOET eers  2 bepaal, om die punt vir Ĉ2 te verdien (2) 8.2.2 Ĉ 2 + Ĉ 3 + Ĉ 4 = 90  s in a semi circle/  e in semi sirkel S/R D̂1 = 90 − x sum of  s in a triangle/som vd  e v S driehoek AB̂C = 90 + x opposite  s of cyclic quad/oorst  e v kvh S/R OR/OF Ô1 = 2 x  centre = 2   at circumference/ S/R middelpunts  = 2  omtreks  D̂1 = 90 − x sum of  s in a triangle/som vd  e v S driehoek AB̂C = 90 + x opposite  s of cyclic quad/oorst  e v kvh S/R (3) 8.2.3 Ĉ 3 + Ĉ 4 = 90 − x  s opposite radii/  ’e teenoor radiusse S/R  2 = Ĉ 2 = x  s opposite radii/  ’e teenoor radiusse S/R  Â1 = Ĉ 2 = x  AB || CO alternate  s =/verwis.  e gelyk R OR/OF (3) 21 ### Page 22 Downloaded from hlayiso.com 10612/22 Ĉ 3 + Ĉ 4 = 90 − x  s opposite radii/  ’e teenoor radiusse S/R Â1 +  2 = Ô1 = 2 x  centre = 2   at circ./ S/R middelpunt  = 2  omtrek   AB || CO corresponding  s =/ooreenkomst.  e gelyk R [19] 22 ### Page 23 Downloaded from hlayiso.com 10612/22 QUESTION/VRAAG 9 E D 2 = 1 F 21 2 A 2 1 1 ● O B 9.1.1 F̂2 = 90 line from centre to midpoint of chord/ S/R lyn v middelpunt na middelpunt v koord B̂1 + B̂ 2 = 90 radius ⊥ tangent/radius ⊥ raaklyn S/R  F̂2 + B̂ = 180 OBEF is a cyclic quad/ opp.  s are supplementary/ R OBEF is ʼn converse of opp.  s of a cyclic quad/ koordevierhoek oorst.  e is supplementêr/omgekeerde v oorst  e v kvh OR/OF F̂1 = 90 line from centre to midpoint of chord/ S/R lyn v middelpunt na middelpunt v koord B̂1 + B̂ 2 = 90 radius ⊥ tangent/radius ⊥ raaklyn ext.  = interior opposite  / S/R  F̂1 = B̂ converse of exterior.  of a cyclic quad/  = oorst binne  e /omgekeerde van buite R  van kvh (3) 23 ### Page 24 Downloaded from hlayiso.com 10612/22 9.1.2 D̂1 = 90  in a semi circle/  in semisirkel S/R D̂ 2 = 90 Adj. sup.  s /aangrensde suppl.  e In ΔADB and/en ΔBDE  = B̂2 tan chord theorem/raaklyn koordstelling S D̂1 = D̂ 2 proved/reeds bewys B̂1 = Ê sum of  s in  /som vd  e v  ΔADB ||| ΔBDE  R OR/OF D̂1 = 90  in a semi circle/  in semisirkel S/R D̂ 2 = 90 Adj. sup.  s /aangrensde suppl.  e In ΔADB and/en ΔBDE  = B̂2 tan chord theorem/raaklyn koordstelling S D̂1 = D̂ 2 proved/reeds bewys ΔADB ||| ΔBDE  R (3) 9.1.3 B̂1 = Ê 3rd  in  , ΔADB ||| ΔBDE S OB is a tangent/ converse of tan chord theorem/ R (2) OB is ’n raaklyn omgekeerde raaklyn koordstelling 9.2 AD BD = S BD DE ΔADB ||| ΔBDE BD 2 = AD  DE but/maar BD = 2 OF midpoint theorem/middelpuntstelling S/R BD = 4 OF 2 2 S 4 OF2 = AD  DE AD  DE OF2 = 4 (3) [11] 24 Text truncated here. Open the document page to read the rest. ## Related documents - [Gr12 Mathematics P1 (Afrikaans) SEP 2022 Possible Answers](https://www.hlayiso.com/library/gr12-mathematics-p1-afrikaans-sep-2022-possible-answers/38fb7466-deff-4faa-9c6c-825fcf932aae) — Mathematics · Memo · 2022 · Afr - [MATHS P2 QP GR12 JUNE 2022 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-june-2022-afrikaans/49feb536-3d0c-4e63-b857-c88def50a8c1) — Mathematics · QP · 2022 · Afr - [Gr 12 Gauteng June 2017 P2 Memo](https://www.hlayiso.com/library/gr-12-gauteng-june-2017-p2-memo/4aabd1d5-7a04-4e17-bff5-d1f3e3fa545a) — Mathematics · Memo · 2017 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — Mathematics · QP · 2025 · Afr - [MATHS P2 QP GR12 SEPT 2024 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-sept-2024-afrikaans/8572c34f-22e2-48bc-9d8a-f78c7e29781d) — Mathematics · QP · 2024 · Afr - [MATHS P2 GR12 ANSWER BOOK SEPT 2022 Afr Engl](https://www.hlayiso.com/library/maths-p2-gr12-answer-book-sept-2022-afr-engl/4744df95-ec05-48bc-b20c-0ae44310bc55) — Mathematics · Addendum · 2022 · Afr ## More from Grade 12 Mathematics - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [FS Gr 12 Sept 2024 Afr en Eng P1](https://www.hlayiso.com/library/fs-gr-12-sept-2024-afr-en-eng-p1/4d7be3dc-5f53-4da3-a833-8fa9c4d73be1) — QP · 2024 · Eng - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — QP · 2024 - [Gr 12 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-12-mathematics-p1-afrikaans-june-2024-question-paper/141815e5-df0c-4d8f-a354-4ef5b9b2a2e2) — QP · 2024 · Afr ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.