Mathematics P2 Feb March 2017 Memo Afr Eng_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL/NASIONALE SENIOR CERTIFICATE/SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 FEBRUARY/MARCH/FEBRUARIE/MAART 2017 MEMORANDUM MARKS / PUNTE: 150 This memorandum consists of 21 pages. Hierdie memorandum bestaan uit 21 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 2 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum NOTE:  If a candidate answered a question TWICE, mark only the FIRST attempt.  If a candidate has crossed out an attempt to answer a question and did not redo it, 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. LET WEL:  Indien 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.  As 'n kandidaat 'n poging om 'n vraag te beantwoord, doodgetrek en nie oorgedoen het nie, sien die doodgetrekte poging na.  Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Staak nasien by die tweede berekeningsfout.  Om antwoorde/waardes om 'n probleem op te los, te veronderstel, word NIE toegelaat NIE. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 3 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 1 Ogive/Ogief 70 Cumulative frequency/Kumulatiewe frekwensie (60, 65) 60 (50, 61) 50 (40, 45) 40 30 (30, 25) 20 (20, 12) 10 0 (10, 0) 0 10 20 30 40 50 60 70 Money spent/Geld gespandeer (R) Amount of money/ Bedrag geld 10  x  20 20  x  30 30  x  40 40  x  50 50  x  60 (in R) Frequency a 13 20 b 4 Frekwensie 1.1 65 learners/leerders  answer (1) 1.2 Modal class/Modale klas: 30  x  40  answer (1) 1.3 a = 12  answer b = 61 – 45 = 16  answer (2) 1.4 No. of learners/Aantal leerders = 65 – 54 OR/OF 65 – 55  54 or 55 = 11 = 10  11 or 10 (2) Answer only: full marks [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 4 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 2 2.1 Class/Klas A Class/Klas B 45 51 65 72 20 30 40 50 60 70 80 90 100 2.1.1 IQR of Class B/IKV van Klas B = Q 3  Q1 = 72 – 51 72 and 51 = 21 marks/punte  21 only (2) 2.1.2 Although the boxes contain the same number of data points, the   Class A is marks for Class A are more widely spread./Alhoewel die monde more widely dieselfde aantal datapunte bevat, is die punte van Klas A meer spread verspreid. (2) OR/OF Although the boxes contain the same number of data points, the   Class B is marks for Class B are more clustered./Alhoewel die monde dieselfde more clustered aantal datapunte bevat, is die punte van Klas B nader aan mekaar. (2) 2.1.3 Medians are the same/Mediane is dieselfde   any TWO Ranges are the same OR Maximum and minimum values are the of the 3 same/Variasiewydtes is dieselfde OF die maksimum en minimum reasons waarde is dieselfde mentioned 75% of both classes obtained 51 and above/75% van albei klasse behaal 51 en meer. (2) 2.2 COUPLE/PAAR 1 2 3 4 5 6 7 8 JUDGE 1/ 18 4 6 8 5 12 10 14 BEOORDELAAR 1 JUDGE 2/ 15 6 3 5 5 14 8 15 BEOORDELAAR 2 2.2.1 a = –0,03  value a b = 0,93  value b ŷ = –0,03 + 0,93x  equation (3) 2.2.2 ŷ = –0,03 + 0,93(15)  substitution = 13,92 OR/OF 13,85  14  answer (2) 2.2.3 Yes OR they are consistent, because r = 0,9. (r = 0,89567…)/Ja OF  statement hulle is konsekwent, want r = 0,9. (r = 0,89567…)  r = 0,9 (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 5 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 3 y S R(10 ; 7) T(0 ; 4) M(5 ; 2)  β x O Q(3 ; 0) 3.1 40 mTQ  03 4 =   answer 3 (1) 3.2 d  ( x 2  x1 ) 2  ( y 2  y1 ) 2 RQ  (10  3) 2  (7  0) 2 substitution/substitusie RQ  98  7 2  answer in surd form (2) 3.3 mFQ  mTQ  equating gradients/stel gradient 8 4 gelyk  8 k 3 3 OR/OF  mFQ  k 3 4k  12  24  simplification/vereenvoudig k 9  answer OR/OF (4) 4 Equation of TQ: y  x4  gradient 3  equation of TQ/vgl van TQ 4 8   k + 4  substitution of (k ; - 8) 3 /substitusie van (k ; - 8) k=9  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 6 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 3.4   x–value/waarde Using transformation/Gebruik transformasie:   y–value/waarde S(7 ; 11) (4) OR/OF Midpoint of TR = midpoint of SQ [diag ||m/hkle||m] 11 Midpoint of TR = (5 ; 2 )  x–value/waarde of/van T xS  3 yS  0 11  y–value/waarde of/van T  5 and  2 2 2  xS  7 and yS  11  S(7 ; 11)  x–value/waarde of/van S  y–value/waarde of/van S OR/OF (4)  72  Equation of TS: y   x  4  x  4  equations of TS and RS/vgls  10  5  van TS en RS 4 Equation of RS: y  7   ( x  10) 3 4 61 y  x 3 3 4 61 x4 x  equating / gelykstel 3 3 7 x  49 x7  y = 11  x–value/waarde S(7 ; 11)  y–value/waarde (4) 3.5 TŜR  TQ̂R [opp s of ||m/teenoorst e ||m] ˆ R    TQ ˆ R     TQ 4 tan   mTQ    tan   mTQ 3  α = 180° – 53,13° = 126,87° α 7 tan   mRQ   1  tan   mRQ 7 β = 45°  TQ̂R  126,87  45 = 81,87° TŜR  81,87  answer (6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 7 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum TQ = SR = 5  length of TQ OR SR TR = 100  9  109  length of TR RQ = TS = 49  49  98  length of RQ OR TS TQ2  RQ 2  TR 2 cosRQ̂T  cosTŜR  2.TQ.RQ 25  98  109  correct subst into cosine rule  2(5)( 98 )  0,141...  simplification  answer RQ̂T  TŜR  81,87 (6) 3.6.1 MQ  (5  3) 2  (2  0) 2 substitution/substitusie MQ  8  MQ  8  2 2 MQ 8 Answer only: full marks  RQ 98 2  or 0 ,29  answer 7 (3) 3.6.2 1 .QM .  h area of ΔTQM 2  [h same/dieselfde] area of TQR 1 .QR.  h 2 area of ΔTQM 2  QM 2    QR 7 area of TQR 7 area of ΔTQM area of ΔTQM   area of parm RQTS 2  area of  TQR area parm RQTS  2area TQR 1 2 1     answer 27 7 (3) OR/OF area of ΔTQM QM  area of TQR QR 2 area of ΔTQM 2 =   7 area of TQR 7 area of ΔTQM area of ΔTQM   area of parm RQTS 2area of TQR area parm RQTS  2area TQR 1 2 1 =   =  answer 27 7 (3) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 8 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 1 1 area of ΔTQM QM.  h QM.  h  2  2 area of parm RQTS RQ.  h RQ.  h 1 2    27 1 2 1     27 7  answer (3) OR/OF 1 QT.QM. sin(   ) area of ΔTQM   2 area of parm RQTS 2area of ΔQTR area parm RQTS  2area TQR 1 QT.QM. sin(   ) 1 QT.QM. sin(   )  2 2 1  2[ .QT.QR. sin(   )] 1 2 2[ .QT.QR. sin(   )] 2 1 2    27 1  7  answer (3) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 9 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 4 y P S(–3 ; 8) N T(0 ; 5) M O x R 4.1 line from centre to midpt of chord /  answer lyn vanaf midpt na midpt van koord (1) 4.2 85  subst (–3 ; 8) and mST  30 (0 ; 5) into  1 gradient formula  mST mST  mNP  1 [TS  NP]  m NP  m NP  1 y = x  c 8 = –3 + c y  y1  1( x  x1 )  subst (–3 ; 8) into c = 11 OR/OF y  8  1( x  3) equation of a line y = x  11 y  x  11  equation (5) 4.3 P(0 ; 11) [y-intercept of chord NP]  coordinates of P/  radius is 6 units koördinate v P R(0 ; –1)  coordinates of R Equations of the tangents to the circle parallel to the x-axis/ koördinate van R Vgls van die raaklyne aan die sirkel || aan die x-as: y  11 and y  1  answers (4) 4.4 M(–11 ; 0) [x–intercept of/x-afsnit van NP]  coordinates of M MT  (0  11) 2  (5  0) 2  substitution MT  146 = 12,08  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 10 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 4.5 MT = diameter/middellyn [conv in 12 circle/omgek  in 12 sirkel] 146  radius of circle radius = units 2 Centre of circle/Middelpunt v sirkel = Midpoint MT /Middelpunt MT  x value of M   11 5   y value of M = ;   2 2 2 2 Equation of circle through S, T and M:  x     y    11 5 146  LHS of equation  2  2 4  RHS of equation 2 2 OR/OF  x  5    y  2   1 1 73  6,04  2  2 2 (5) [19] QUESTION/VRAAG 5 5.1 a  1  answer b2  answer (2) 5.2 f(3x)= – sin 3x 360 360 Period of f(3x) =  3 3 = 120° Answer only: Full marks  answer (2) 5.3 x  [90° ; 135°)  {180°}  90° and 135° in interval form  180° as single value  correct brackets OR/OF (3) 90° ≤ x < 135° or x = 180°  90° and 135° in interval form  180° as single value  correct inequalities (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 11 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 6 6.1.1 sin (360° – 36°) = – sin 36  answer (1) 6.1.2 cos 72  cos(2  36)  double angle/dubbelhoek  1  2 sin 2 36 Answer only: Full marks answer (2) 6.2 tan  2 R.T.P.: 1   cos 2  1  tan  2 1  tan 2   tan 2  LHS   writing as a single fraction/skryf 1  tan 2  as enkelbreuk 1  sin 2   quotient identity/ 1 cos 2  kwosiëntidentiteit 1  cos   sin 2  2  denominator as a single fraction / Noemer as enkelbreuk cos 2  1  1  square identity/vierkantidentiteit cos 2  (4)  cos 2  = RHS OR/OF 1  tan 2   tan 2  LHS   writing as a single fraction/skryf 1  tan 2  as enkelbreuk 1  quotient identity /  sin 2  kwosiëntidentiteit 1 cos 2  1 cos 2  cos 2    × sin 2  cos 2  cos 2  1 cos 2  cos 2   cos 2   sin 2  cos 2    square identity/vierkantidentiteit 1  cos 2  = RHS (4) OR/OF  quotient identity/ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 12 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum  sin 2   sin 2    kwosiëntidentiteit LHS  1     1  cos 2   cos 2       writing as a single fraction/ skryf  sin 2  cos 2    1    as enkelbreuk  cos 2  cos 2   sin 2      square identity/vierkantidentiteit  sin 2  cos 2    1     cos 2  1   simplification/vereenvoudiging    1  sin 2   cos 2  (4)  RHS 6.3 1 1 1 1 cos 2 x   cos 2 x 2 4 2 4 1 1 1 cos x  or  2 2 2 1 1  60° and 300° x  60  k .360 or x  300  k.360 or 2 2 1 1 120° and 240° x  120  k .360 or x  240  k .360 2 2  write at least one general x  120  k.720 or x  600  k.720 or x  240  k.720 or x  480  k.720; k  Z 1 solution as x    k.360 2  write at least one general solution as x    k.720 ; kZ OR/OF (6) 1 1 1 1 cos 2 x  cos 2 x 2 4 2 4 1 1 1 cos x  or  2 2 2 1 1 x  60  k .360 or x  120  k.360  ±60° ±120 2 2 x  120  k.720 or x  240  k.720; k  Z  write at least one general 1 solution as x    k.360 2  write at least one general solution as x    k.720 kZ (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 13 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 6.4.1 sin(A  B) = cos[90  (A  B)]  co–ratio/ko-verhouding  cos[(90  A)  (B)]  writing as a difference of A & B/ skryf as verskil van A & B  cos(90  A)cos(B)  sin(90  A)sin(B)  expansion/uitbreiding  sin AcosB  cos A(sinB)  all reductions/alle reduksies  sin AcosB  cos AsinB (4) OR/OF sin(A  B) = cos[90  (A  B)]  co–ratio/ko-verhouding  cos[(90  B)  A]  writing as a difference of A & B/  cos(90  B)cosA  sin(90  B)sinA skryf as verskil van A & B   sin BcosA  cos BsinA  expansion/uitbreiding  sin AcosB  cos AsinB  all reductions/alle reduksies (4) 6.4.2 sin( x  64) cos( x  379)  sin( x  19) cos( x  244)  sin( x  64) cos( x  19)  sin( x  19)[ cos( x  64)]  cos( x  379)  cos( x  19)  sin( x  64) cos( x  19)  cos( x  64) sin( x  19)  cos( x  244)   cos( x  64)  sin[ x  64  ( x  19)] compound formula identity/  sin 45 saamgestelde identiteit 1  sin 45  2 (6) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 14 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 7 A C 8,6 D B 10 40° 70° 27° E 7.1 CD sin 27   substitution in correct trig ratio / 8,6 substitusie in korrekte trig verh CD  8,6 sin 27  answer CD  3,90 m (2) 7.2 10  substitution in correct trig ratio / cos 40  substitusie in korrekte trig verh AE 10 AE  cos 40 AE  13,05 m  answer (2) 7.3 AC2  CE 2  AE 2  2 CE.AE(cos AÊC)  correct use of cosine rule in ACE/ 2 2  (8,6)  (13,05)  2(8,6)(13,05)(cos 70) korrekte gebruik van reel in ACE  167,49  correct subst into cosine rule  AC2 AC  12,94 m  answer (4) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 15 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 8 Q V 3 2 1 1 R P 2 108° 2 42° 1 T S 8.1 Q̂  72 [opp s of cyclic quad/teenoorst e koordevh] SR (2) 8.2 R̂ 2  P̂1 [s opp equal sides/e teenoor gelyke sye]  S/R 180 - 72 R̂ 2  2 [sum of s in ∆/som v e in ]  answer  54 (2) 8.3 P̂2  42 [tan chord theorem/raakl-koordst] SR (2) 8.4 R̂ 3  P̂1  P̂2 [ext  of cyclic quad/buite van koordevh] R = 54° + 42° S = 96° (2) OR/OF  R̂1  30 R̂1  180  108  42  30 [sum of/som vans/e in ∆] R̂ 3  180  R̂ 1  R̂ 2 [s on str line/e op reguitlyn] = 180° – 30° – 54° [sum of/som vans/e in ∆] = 96° S (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 16 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 9 P Q 2 1 W T 2 1 S V R 9.1.1 ST SW = [prop theorem/eweredighst; TW || QP] S TQ WP 2 = S 3 (2) 9.1.2 SV SW = [prop theorem/eweredighst; VW || RP] VR WP 2 =  answer 3 (1) 9.2 ST SV WS = [both equal/beide gelyk ] TQ VR PW S  TV || QR [line divides 2 sides of ∆ in prop/lyn verdeel 2 sye  S  R van  in dies verh]  T̂1  Q̂1 [corresp/ooreenkomst s/e; TV || QR] R (4) 9.3 VWS ||| RPS  RPS (any order) (1) 9.4 WV SW  [ VWS ||| RPS]  ratio PR SP 2  answer  (2) 5 OR/OF [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 17 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 10 10.1 P 12 O 1 2 A B Constr/Konst : Draw line PO and extend /Trek lyn PO en verleng  construction Proof/Bewys : OP  OA [radii]  P̂1  Â [s opp/teenoor  sides/sye]  S/R but Ô1  P̂1  Â [ ext  of ]  S/R  Ô1  2 P̂1 S Similarly/ Netso , Ô 2  2 P̂2  Ô1  Ô 2  2(P̂1  P̂2 ) S i.e. AÔB  2AP̂B (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 18 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 10.2 R y O 1 2 P 2 A 1 S 3 3 2 4 1 Q T 10.2.1 s in the same segment/ e in dieselfde sirkelsegment R (1) 10.2.2 P̂2  Ŝ1  y [s opp equal sides/e teenoor = sye] SR SR Ŝ1  P̂3  y [tan chord theorem/raakl-koordst]  P̂2  P̂3  PQ bisects TP̂S (4) 10.2.3 PÔQ  2Ŝ1  2 y [at centre =2×at circ/midpts= 2 omtreks]  S  R (2) 10.2.4 TP̂A  P̂2  P̂3  2 y [proved/bewys in 11.2.2]  TP̂A  PÔQ  TP̂A  PÔQ [proved/bewys in 11.2.3]  PT = tangent [converse tan chord theorem/omgek raakl-koordst]  R (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 19 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 10.2.5 OP̂Q  OQ̂P  180  2 y [sum of/sum v s/e in ∆] S  OQ̂P = 90° – y [s opp equal sides/e to = sye; OP = OQ] S R In PAQ: OQ̂P  P̂2  QÂP  180 90 - y  y  QÂP  180 [sum of/sum v s/e in ∆] S S QÂP  90  OÂP  90 [s/e on straight line/op reguitlyn] (5) OR/OF OP̂T  90 [radius  tangent/raaklyn] SR S  P̂1  90  2 y P̂1  Ô  OÂP  180 [sum of/sum v s/e in ∆] (90  2 y)  2 y  OÂP  180 S S  OÂP = 90° (5) OR/OF POSQ is a kite/'n vlieër  S  OQ  PS [diag of a kite/hoeklyne v vlieër]  R  OÂP = 90° (5) OR/OF In OAP and OAS OP = OS (radii) S OA is common S P ˆOA  2 y  2 ˆP 2 ˆS  QO S OAP  OAS (SAS) R ˆ P  OA OA ˆ S (s) ˆ P  OA OA ˆ S  90 (s on str line) S (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 20 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 11 L S x  P 1 2 N 1 3 2 T K R 11.1 N̂ 2 = 90° [ in semi-circle/halfsirkel] SR  TPLN is a cyclic quad/ ‘n koordevh [opp s of quad is suppl/ R teenoore v vh is suppl] (3) OR N̂ 2 = 90° [ in semi-circle/halfsirkel] SR  TPLN is a cyclic quad [ext  = int opp /buite = to binne ] R (3) 11.2 T̂2 = PL̂N  x [ext  of cyclic quad/buite van koordevh] R K̂  90  x [sum of/som v s/e in ∆] N̂1  K̂  90  x [tan chord theorem/raakl-koordst]  S R (3) OR/OF K̂  90  x [sum of/som v s/e in ∆] R N̂1  K̂  90  x [tan chord theorem/raakl-koordst]  S R (3) OR/OF N̂3  x [tan chord theorem/raakl-koordst] R N̂ 2  90 [in semi circle/ halfsirkel] S N̂1  90  x [straight line/reguitlyn] S (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/ Wiskunde V2 21 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 11.3.1 In KTP and KLN: PK̂T  LK̂N [common/gemeen] S KP̂T  KN̂L  90 [given/gegee] S   KTP | | |  KLN [] R (3) OR/OF In KTP and KLN: PK̂T  LK̂N [common/gemeen] S KP̂T  KN̂L  90 [given/gegee] S T̂2  PL̂N  x [proved in 11.2 OR sum of s in ∆] S   KTP | | |  KLN (3) 11.3.2 KT KP  [||| s]  S/R KL KN KT . KN = KP . KL S But KL = 2KP [radii: PK = LP] S KT . KN = KP . 2KP = 2KP2 S = 2(KT2 – TP2 ) [Theorem of Pythagoras] S  2KT2  2TP2 (5) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou