Mathematics Grade 12 Paper 2 NSC November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 23 pages. Hierdie memorandum bestaan uit 23 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 DBE/November 2014 NSC/NSS – Memorandum 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. • Assuming answers/values in order to solve a problem is NOT acceptable. NOTA: • As 'n kandidaat 'n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging. • As 'n kandidaat 'n poging om die vraag te beantwoord, doodgetrek het en nie dit oorgedoen het nie, merk die doodgetrekte poging. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas. • Aanvaarding van antwoorde/waardes om 'n probleem op te los, is ONaanvaarbaar. QUESTION/VRAAG 1 1.1 816 816 x= = 68  12 12  68 (2) 1.2 σ = 18,42  answer/antw (1) 1.3 (68 – 18,42 ; 68 + 18,42) = (49,58 ; 86,42)  interval ∴ 6 candidates had a mark within one standard deviation of the  answer/antw mean/6 kandidate het 'n punt binne een standaardafwyking vanaf (3) die gemiddelde. 1.4 a = 22,828... = 22,83  value of a/ waarde van a b = 0,66429... = 0,66  value of b/ waarde van b ∴ yˆ = 0,66 x + 22,83 OR/OF yˆ = 22,83 + 0,66 x  equation/vgl (3) 1.5 yˆ = 0,66 x + 22,83 y = 0,66(60) + 22,83  subs of 60 into 62,43...% ≈ 62% equation  answer/antw OR/OF (2) 62,69% ≈ 63%  answer/antw (2) 1.6 (82 ; 62)  answer/antw (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 2 2.1 50 < x ≤ 60 OR/OF 50 ≤ x < 60 OR/OF  answer/antw between 50 and 60/tussen 50 en 60 (1) 2.2.1 Class Frequency Cumulative frequency Klas Frekwensie Kumulatiewe frekwensie 20 < x ≤ 30 1 1 30 < x ≤ 40 7 8 8 40 < x ≤ 50 13 21 50 < x ≤ 60 17 38 60 < x ≤ 70 9 47 70 < x ≤ 80 5 52 80 < x ≤ 90 2 54 90 < x ≤ 100 1 55  55 (2) 2.2.2 60  grounding at (20 ; 0)/ anker by (20 ; 0)  plotting at 50 upper limits/ plot by boonste limiete 40  smooth shape of curve/gladde kurwe Cumulative Frequency/ 30 20 Kumulatiewe frekwensie 10 0 0 10 20 30 40 50 60 70 80 90 100 Speed in km per hour/ Spoed in km per uur (3) 2.3 55 – 44 (accept/aanvaar 43 – 45)  44 ≈ 11 motorists/motoriste  11 (accept/aanvaar 10 – 12 motorists/motoriste) (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 3 y S K L N M(5 ; 4) x O A B P 3.1 r = MN = 5  answer/antw (1) 3.2 ( x − 5) 2 + ( y − 4) 2 = 25 equation/vgl (1) 3.3 A(x ; 0)  substitute into eq/ (x – 5)2 + (0 – 4)2 = 25 (x – 5)2 + (0 – 4)2 = 25 vervang in vgl x2 – 10x + 25 + 16 = 25 (x – 5)2 + 16 = 25 y=0 2 x – 10x + 16 = 0 OR/OF (x – 5)2 = 9  standard form/ (x – 8)(x – 2) = 0 (x – 5) = ± 3 standaardvorm or ∴x = 8 or/of x = 2 ∴x = 8 or/of x = 2 perfect square ∴ A(2 ; 0) ∴ A(2 ; 0) form/kwadr vorm  answer/antw (3) 3.4.1 4−0  subst M and B mMB = 5−8 into form/vervang 4 M and B in form =− 4 3  mMB = − 3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 DBE/November 2014 NSC/NSS – Memorandum 3.4.2 mMB × mPB = −1 (tangent ⊥ radius/ rkl ⊥ radius)  3 mMB × mPB = −1 mPB = 3 4  mPB = 3 3 4 y= x+c OR/OF y – y1 = (x – x 1 ) 4 4 3 3 0= (8) + c y–0 = (x – 8) 4 4 3 3 y= x–6 y = x–6 4 4  equation/vgl (3) 3.5 yK = yM + r = 4 + 5  9 y = 9  equation/vgl (2) 3.6 At/By L: 3 x−6 =9  equating 4 simultaneously 3 x − 24 = 36  simplification 3 x = 60 x = 20 ∴ L(20 ; 9) (2) 3.7 L(20 ; 9)  correct subst ML = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 OR/OF ML = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 into distance formula/ = (20 − 5) 2 + (9 − 4) 2 = (15) 2 + (5) 2 korrekte subst in afstand- = 225 + 25 = (5) 2 (9 + 1) formule = 250 or / of 5 10 = 250 or / of 5 10  answer in surd form/antw in wortelvorm (2) 3.8 MK ⊥ KL OR/OF MK ˆ L = 90° (radius ⊥ tangent/radius ⊥ rkl) S ∴ML is a diameter as it subtends a right angle/ML is middellyn  value ML 250 125 r= = = or 7,91 of/waarde 2 2 2 van r Centre of circle = midpoint of ML/Midpt van sirkel = midpt v ML 5 + 20 25 4 + 9 13  x = 12,5 x= = = 12,5 y= = = 6,5 2 2 2 2  y = 6,5 Centre/midpt: (12,5 ; 6,5) Equation of the circle KLM /Vgl van sirkel KLM:  answer in 250 125 correct ∴ ( x − 12,5) 2 + ( y − 6,5) 2 = = = 62,5 form/ antw in 4 2 korrekte vorm OR/OF (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 DBE/November 2014 NSC/NSS – Memorandum MK ⊥ KL OR/OF MK ˆ L = 90° (radius ⊥ tangent/radius ⊥ rkl) S ∴ML is a diameter as it subtends a right angle/ML is middellyn Centre of circle = midpoint of ML/Midpt van sirkel = midpt v ML 5 + 20 25 4 + 9 13 x= = = 12,5 y= = = 6,5  x = 12,5 2 2 2 2  y = 6,5 Centre/midpt: (12,5 ; 6,5) Equation of the circle KLM /Vgl van sirkel KLM: ( x − 12,5) 2 + ( y − 6,5) 2 = r 2  value subst (5 ; 4): (5 − 12,5) 2 + (4 − 6,5) 2 = r 2 of/waarde 62,5 = r 2 van r2 250 125 ∴ ( x − 12,5) 2 + ( y − 6,5) 2 = = = 62,5  answer in 4 2 correct form/antw in OR/OF korrekte vorm (5) By symmetry about LM/deur simmetrie om LM: MK ⊥ KL OR/OF MK ˆ L = 90° (radius ⊥ tangent/radius ⊥ rkl) S ∴ML is a diameter as it subtends a right angle/ML is middellyn ML is a diameter /ML is 'n middellyn ML 250 125  value r= = = or /of 7,91 of/waarde 2 2 2 van r Centre of circle = midpoint of ML/Midpt van sirkel = midpt v ML 5 + 20 25 4 + 9 13 x= = = 12,5 y= = = 6,5  x = 12,5 2 2 2 2  y = 6,5 Centre/midpt: (12,5 ; 6,5) Equation of the circle KLM /Vgl van sirkel KLM:  answer in 250 125 ∴ ( x − 12,5) 2 + ( y − 6,5) 2 = = = 62,5 correct 4 2 form/antw in korrekte vorm (5) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 4 y F B(1 ; 5) D 45° M A x E O 4.1 y = 0: 3x + 8 = 0  y-value/waarde 8 x= − 3  x-value/waarde  2   8  ∴ E  − 2 ; 0  OR/OF E  − ; 0  (2)  3   3  4.2 tan DÊO = mDE = 3  tan DÊO = 3 ∴ DÊO = 71,565... = 71,57°  71,565…° DÂE = 71,565...° − 45° = 26,57°  26,57° (3) 4.3 mAB = tan 26,57°  mAB = tan 26,57° 1 1 =  mAB = 2 2 1 1 y= x + c OR/OF y – y1 = (x – x 1 )  subst of m and 2 2 1 1 (1 ; 5)into formula/ 5 = (1) + c y–5 = (x – 1) subst m en (1 ; 5) in 2 2 formule 1 1 1 9  equation/vgl y= x+4 y = x+ 2 2 2 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 DBE/November 2014 NSC/NSS – Memorandum 4.4 Solve x – 2y + 9 = 0 and y = 3x + 8 simultaneously: x – 2(3x+8) + 9 = 0  subst/vervang x – 6x –16 + 9 = 0 – 5x = 7 2 x = −1  x-value/waarde 5 2 2 ∴ y = 3( − 1 ) + 8 OR/OF − 1 − 2 y + 9 = 0  subst/vervang 5 5 4 4 y=3 y=3  y-value/waarde 5 5 (4) 2 4 ∴ D( − 1 ; 3 ) 5 5 OR/OF x = 2y – 9 y = 3(2 y − 9) + 8  subst/vervang y = 6 y − 27 + 8 4 ∴y=3  y value/waarde 5 4 x = 2(3 ) − 9 OR/OF  subst/vervang 5 2 x = −1 5  x-value/waarde 2 4 ∴ D( − 1 ; 3 ) 5 5 OR/OF (4) 1 1 3x + 8 = x+4  equating/gelyk stel 2 2 6x + 16 = x + 9 5x = – 7 2 ∴ x = −1  x value/waarde 5 2 1 2 1  subst/vervang ∴ y = 3( − 1 ) + 8 OR/OF y= (−1 ) + 4 5 2 5 2 4 4 y=3 y=3 5 5  y-value/waarde 2 4 ∴ D( − 1 ; 3 ) 5 5 (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 DBE/November 2014 NSC/NSS – Memorandum x – 2y = –9 ……(1) –6x + 2y = 16 …...(2) (1) + (2): –5x =7 2  adding/optelling ∴ x = –1 5 2 2  x-value/waarde ∴ –1 – 2y = –9 OR/OF y = 3(–1 )+8 5 5 4 4 y=3 y=3 5 5  subst/vervang 2 4 ∴ D( − 1 ; 3 ) 5 5  y-value/waarde OR/OF (4) y = 3x + 8 ……….(1) 6y = 3x + 27 …….(2) (1) – (2): –5y = –19  4 subtracting/aftrekking ∴ y=3 5  y-value/waarde 4 4 3 = 3x + 8 x = 2(3 ) − 9 5 5 OR/OF  subst/vervang 2 2 x = −1 x = −1 5 5  x-value/waarde 2 4 (4) ∴ D( − 1 ; 3 ) 5 5 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 DBE/November 2014 NSC/NSS – Memorandum F 8 3 B(1 ; 5) D( ; ) 45° M(0 ; 4 ) 135° G 4 26,57° C A(– 9 ; 0) x E( ; 0) O 4.5 area DMOE = area ∆AMO – area ∆ADE  correct method/ x A = 2(0) – 9 ∴ A(–9 ; 0) korrekte metode  xA = – 9 area ∆AMO area ∆ADE 1 1 = . AO. OM = . AE . y D 1 1 2 2  (9)(4 ) 1 1 1 2 2 = (9)(4 ) = . (AO – EO). y D 2 1 2 2 2  AE = 9 − 2 = 6 1 2  4  3 3 = 20,25 =  9 − 2  3  4 2 3  5   yD = 3 = 12,03 5 OR/OF OR/OF area ∆ADE 1 = AD.AE.sin DÂE 19 5 2  AD = 5 1 19 5 1 = ( ).6 .sin 26,57° 1 2 5 3  AE = 6 3 = 12,03 ∴ area DMOE = 8,22 square units/vk eenh  answer/antw (6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 DBE/November 2014 NSC/NSS – Memorandum area DMOE = area rectangle DCOG + area ∆DMG + area ∆DEC  correct method/ 2 4 1  2 7 1 4 19 korrekte metode = (1 × 3 ) + 1 ( ) + (3 )( ) 4 5 5 2  5  10 2 5 15  3 = 8,22 square units/vk eenh 5 2  1  0,7 5 19  15  answer (6) OR/OF area DMOE = area ∆EDO + area ∆ODM  correct method/ = (EO × y D ) + (OM × − x D ) 1 1 korrekte metode 2 2 19 4  yD = or 3 1  8 19   9 7  5 5 =  ×  +  ×  8 2  3 5   2 5   EO = 3 1  304 + 189  =   7 2 30   –x D = 5 493 13 9 1 = or/of 8 or/of 8,22 square units/vk eenh  OM = or 4 60 60 2 2  answer/antw (6) OR/OF area DMOE = area ∆EOF – area ∆DMF  correct method/ = (EO × OF) − (OF − OM )(− x D ) 1 1 korrekte metode 2 2  yF = 8 1  8   7 7  8 =  × 8  +  ×   EO = 2  3   2 5  3 1  640 − 147  7  –x D = =   5 2 30  1 493 13  FM = 3 = or 8 or 8,22 square units/vk eenh 2 60 60  answer/antw (6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 DBE/November 2014 NSC/NSS – Memorandum area ∆EOM = 1 (EO × OM ) 2 18 9 =  ×  23 2 = 6 sq units/vk eenh  area ∆EOM 2 2 2 2  7 8   19   7   9 19  ED = − +  +   and DM =   + −   5 3  5  5 2 5   ED = 19 10 19 10 7 5 15 = or 4,005... = or 1,565.. 7 5 15 10  DM = 1 ( area ∆EDM = ED × DM × sin ED̂M 2 ) 10 1  19 10   7 5  =  sin 135° 2  15   10  133 = or 2,216...  area ∆EDM 60 ∴area DMOE = area ∆EOM + area ∆EDM  correct method/ korrekte metode = 6 + 2,216... 493 13 = or/of 8 or/of 8,22 square units/eenh2 60 60  answer/antw (6) [19] F 8 3 B(1 ; 5) D( ; ) 45° M(0 ; 4 ) 135° G 4 26,57° C A(– 9 ; 0) x E( ; 0) O Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 5 C 4 x 8 A P y 4 D 5.1 CP sin CÂP =  correct sine ratio/ AP korrekte sin-verh 4 3 3 sin x = = 3 8 2  2 x = 60° (2) OR/OF sin 90° sin x  correct sine ratio/ = 8 4 3 korrekte sin-verh 4 3 3 3 sin x = =  8 2 2 x = 60° (2) 5.2 CP̂A = DP̂A = 30° ( AP bisects DP̂C)  DP̂A = 30° AD 2 = AP 2 + DP 2 − 2.AP.DP.cos AP̂D  correct subst into = 8 2 + 4 2 − 2(8)(4) cos 30° cosine rule/ 3 korrekte subst in = 8 2 + 4 2 − 2(8)(4)( ) cos-reël 2 = 24,57...  24,57... AD = 4,96  4,96 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 DBE/November 2014 NSC/NSS – Memorandum 5.3 sin DÂP sin AP̂D = DP AD  correct subst into sin y sin 30° = sine rule/ 4 4,96 korrekte subst in 4 sin 30° sin-reël sin y =  sin y subject 4,96 = 0,403...  23,78° y = 23,78° (3) OR/OF AD 2 = AP 2 + DP 2 − 2.AP.DP.cosDÂP 4 2 = 8 2 + (4,96) 2 − 2(8)(4,96). cos y  correct subst into cosine rule/ korrekte subst in cos-reël 8 2 + (4,96) 2 − 4 2 cos y =  cos y subject 2(8)(4,96) cos y = 0,9148... y = 23,82°  23,82° (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 6 6.1 cos 2 (180° + x) + tan( x − 180°) sin(720° − x) cos x = (− cos x) 2 + [−(− tan x)](− sin x)(cos x)  ( − cos x )2 or cos2x  tan x or –(–tan x)  sin x   –sin x = cos 2 x +  (− sin x)(cos x)  cos x  sin x  tan x = = cos 2 x − sin 2 x cos x = cos 2 x  cos x − sin 2 x 2 (5) 6.2 sin(α − β ) rewrite as/herskryf = cos[90° − (α − β )] cos[(90° − α ) + β ] = cos[(90° − α ) + β ]  expansion/ = cos(90° − α ) cos β − sin(90° − α ) sin β uitbreiding = sin α cos β − cos α sin β simpl/vereenv (3) OR/OF sin(α − β )  rewrite as/herskryf = cos[90° − (α − β )] cos[(90° + β ) + (−α )] = cos[(90° + β ) + (−α )]  expansion/ = cos(90° + β ) cos(−α ) − sin(90° + β ) sin( −α ) uitbreiding = (− sin β ) cos α − cos β (− sin α ) simpl/vereenv = sin α cos β − cos α sin β (3) 6.3 x2 – y2 = sin 2 76° – cos276° = –( cos276° – sin 2 76°) –( cos276° – sin 2 76°) = – cos 2(76°)  recognition of cos = – cos 152° double angle = – (–cos 28°) OR/OF = – cos (90° + 62°)  – cos 152° = cos 28° = – (–sin 62°) = cos (90° – 62°) = sin 62°  cos 28° = sin 62° OR/OF (4) 2 2 x –y = sin 2 76° – cos276°  cos 14° = sin 76° sin 76° – cos 76° cos 76°  sin 14° = sin 76° cos 14° – cos 76° sin 14°  recognition of sine = sin (76° – 14°) compound angle = sin 62°  sin(76° – 14°) OR/OF (4) x2 – y2 = sin 2 76° – cos276°  cos2 14° = cos2 14° – sin2 14°  sin2 14° = cos 2(14°)  recognition of cos = cos 28° double angle = sin 62°  cos 28° (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 7 7.1 0 ≤ y ≤ 2 or y ∈ [0 ; 2]  critical values/ kritieke waardes  notation/notasie (2) 7.2 sin x + 1 = cos 2x sin x + 1 = 1 – 2sin2 x  1 – 2sin2x 2 2sin x + sin x = 0  st form/st vorm sin x(2sin x + 1) = 0 (2) 7.3 sin x(2sin x + 1) = 0  sin x = 0 or 1 1 sin x = 0 or sin x = – sin x = – 2 2 x = 0° + k .360° or x = 210° + k .360° or  0° ; 180° OR/OF x = k.180° x = 180° + k.360° x = 330° + k .360°, k∈Z  210° ; 330° OR/OF  k.360°, k ∈ Z x = k.180°, k∈Z (4) 7.4 2  y-intercept/afsnit f  x-intercepts/afsnitte 1  min/max points/ min/maks punte -90 -45 45 90 135 180 225 270 g -1 -2 (3) 7.5 f (x) = g(x) at/by: x = –30° ; 0° ; 180° ; 210°  –30° ; 0° ; 180° ; 210° ∴f (x + 30°) = g(x + 30°) at/by: x = –60° ; –30° ; 150° ; 180°  –60° ; –30° ; 150° ; 180° (3) 7.6 Series will converge if/Reeks sal konvergeer as: –1 < r < 1  –1 < r < 1 –1 < 2cos 2x < 1  r = 2cos 2x 1 1 1 1 – < cos 2x <  – < cos 2x < 2 2 2 2 ∴ 30° < x < 60° or x ∈ (30° ; 60°)  30° < x < 60° (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 DBE/November 2014 NSC/NSS – Memorandum QUESTION/VRAAG 8 8.1 A 48° O 1 x C 2 y 1 2 B 8.1.1 x = 96° (∠ at centre = 2∠ at circumference/  SR ∠ by midpt = 2∠ by omtrek) (2) 8.1.2 Ĉ 2 + B̂ 2 = 180° − 96° = 84° (sum of ∠s in ∆/ som v∠e in ∆ ) S y = B̂ 2 = 42° (∠s opp = sides/∠e teenoor = sye) S (2) 8.2 A B O 1 30° C F D 8.2.1 F̂1 = 90° (line from centre to midpt chord/  SR lyn vanaf midpt na midpt kd) (2) 8.2.2 AB̂C = 150° (opposite ∠s of cyclic quad/  SR tos ∠e v koordevh ) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 DBE/November 2014 NSC/NSS – Memorandum 8.3 A x E 13 C x+7 B 8.3.1 (a) tangent ⊥ radius/diameter / raaklyn ⊥ radius/middellyn R (1) 8.3.1 (b) tangents from common pt OR tangents from same pt / R raaklyne v gemeensk pt OF raaklyne vanaf dies pt (1) 8.3.2 AB² + BC² = AC²  AB² + BC² = AC² x + ( x + 7) 2 = 13 2 2 (Theorem of/Stelling vanPythagoras)  x + x + 14 x + 49 = 169 2 2 x 2 + ( x + 7) 2 = 13 2 2 x 2 + 14 x − 120 = 0  standard form x 2 + 7 x − 60 = 0 ( x − 5)( x + 12) = 0 x = 5 ( x ≠ −12 )  answer (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 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

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 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

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 21 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

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 22 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

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 23 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] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou