# Technical Mathematics Grade 12 P2 Gauteng Mock 2023 Memo > Grade 12 Technical Mathematics Paper 2 memorandum from the Gauteng Mock Exam 2023 (Afrikaans), 26 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-technical-mathematics-2023-gauteng-mock-exam-memorandum-paper-2-afrikaans/ad53fbdf-3326-4add-8f7e-fed7f23f3f43 - Updated: 2026-09-28 - Subject: Technical Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2023 - 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) › [Technical Mathematics](https://www.hlayiso.com/library/grade-12/technical-mathematics) › [Technical Mathematics Grade 12 P2 Gauteng Mock 2023 Memo](https://www.hlayiso.com/library/grade-12-technical-mathematics-2023-gauteng-mock-exam-memorandum-paper-2-afrikaans/ad53fbdf-3326-4add-8f7e-fed7f23f3f43) ## About this document Technical Mathematics · Grade 12 · Gauteng Mock Exam · 2023 · Afrikaans. Memorandum, 26 pages. 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Layout, diagrams and tables may not survive extraction. ### Page 1 PREPARATORY EXAMINATIONS VOORBEREIDENDE EKSAMEN 2023 MARKING GUIDELINES/ NASIENRIGLYNE TECHNICAL MATHEMATICS/TEGNIESE WISKUNDE (PAPER/VRAESTEL 2) 26 pages/bladsye Marking Codes/Nasienkodes A Accuracy/Akkuraatheid CA Consistent accuracy/Volgehoue akkuraatheid M Method/Metode R Rounding/Afronding NPR No penalty for rounding/Geen penalisering vir afronding nie NPU No penalty for units omitted/Geen penalisering indien eenhede weggelaat nie SF Simplification/Vereenvoudiging F Substitution in correct formula/Vervanging in korrekte formule ST Statement/Bewering ST/RE Statement and Reason/Bewering en Rede AO Answer Only / Slegs antwoord ### Page 2 11092/23 NOTES: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and did not redo it, mark the crossed-out version. • Consistent accuracy applies in all aspects of the marking guidelines. NOTAS: • Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • As ʼn kandidaat ʼn poging van ʼn vraag deurgehaal het en dit nie oorgedoen het nie, sien die doodgetrekte weergawe na. • Volgehoue akkuraatheid is deurgaans op alle aspekte van die nasienriglyne van toepassing. QUESTION/VRAAG 1 1.1 𝐴𝐵 = √(𝑥 − 𝑥 )2 + (𝑦 − 𝑦 )2 𝐵 𝐴 𝐵 𝐴 ✓ SF A 𝐴𝐵 = √(1 − (−3))2 + (3 − (−4))² 𝐴𝐵 = √16 + 49 𝐴𝐵 = √65 ✓ √65 CA 𝐴𝐵 = 8, 06 units/eenhede ✓ 8, 06 CA AO: Full marks (3) AO: Volpunte NPR 𝑥𝐴 +𝑥𝐶 𝑦𝐴 +𝑦𝐶 1.2 M ( ; ) 2 2 −3+3 −4−3 ✓𝑥=0 A = M( ; ) 2 2 1 5 ✓ 𝑦 = −2,5 A = M (0; −2 2) or (0; − 2) or/of (0; −2,5) (2) 2 ### Page 3 11092/23 1.3 𝑚𝐴𝐵 = 𝑦𝐵 − 𝑦𝐴 𝑥 −𝑥𝐵 𝐴 −4 − 3 𝑚𝐴𝐵 = −3 − 1 7 7 ✓ 𝑚𝐴𝐵 = 4 A 𝑚𝐴𝐵 = 4 7 𝑦 − 3 = 4 (𝑥 − 1) 7 7 𝑦 = 4𝑥 − 4+ 3 7 5 A 7 5 ✓ 𝑦 = 4𝑥 + 4 𝑦= 𝑥+ 4 4 for x – intercept, let y = 0 vir x–afsnit, stel y = 0 7 5 ✓M A 0= 𝑥+ 4 4 7 5 ∴ 𝑥=− 4 4 5 CA 5 ✓ 𝑥 = −7 𝑥=− 7 5 𝐷 (− ; 0) 7 OR/OF OR/OF 𝑦𝐵 − 𝑦𝐴 𝑚𝐴𝐵 = 𝑥𝐵 − 𝑥𝐴 −4 − 3 𝑚𝐴𝐵 = −3 − 1 7 7 ✓ 𝑚𝐴𝐵 = 4 A 𝑚𝐴𝐵 = 4 7 3−0 7 3−0 ✓ 4 = 1−𝑥 A = 4 1−𝑥 7 − 7𝑥 = 12 ✓S CA −7𝑥 = 5 5 CA 5 ✓ 𝑥 = −7 𝑥=− 7 5 (4) 𝐷 (− ; 0) 7 3 ### Page 4 11092/23 1.4 𝑚𝐵𝐶 = −2 ∴ 𝑡𝑎𝑛𝜃 = −2 ✓ SF A 𝑅𝑒𝑓. ∠ = 63,43° 𝜃 = 180° − 63,43° = 116,57° ✓ 116,57° CA (2) 1.5 𝑦 = −2𝑥 + 5….  𝑥2 + 𝑦2 = 𝑟2 𝑥 2 + 𝑦 2 = (√5)2 𝑥 2 + 𝑦 2 = 5…  ✓ 𝑥2 + 𝑦2 = 5 A Subst.  in  or/of Verv.  in  𝑥 2 + (−2𝑥 + 5)2 = 5 ✓ SF CA 𝑥 2 + 4𝑥 2 − 20𝑥 + 25 = 5 5𝑥 2 − 20𝑥 + 20 = 0 𝑥 2 − 4𝑥 + 4 = 0 ✓S CA (𝑥 − 2)(𝑥 − 2) = 0 𝑥=2 ✓𝑥=2 CA OR/OF OR/OF 4 ### Page 5 11092/23 𝑆(𝑥; −2𝑥 + 5) ✓ (𝑥; −2𝑥 + 5) A 𝑥 2 + 𝑦 2 = (𝑂𝑆)2 𝑥 2 + (−2𝑥 + 5)2 = (√5)2 ✓ SF A 𝑥 2 + 4𝑥 2 − 20𝑥 + 25 = 5 5𝑥 2 − 20𝑥 + 20 = 0 𝑥 2 − 4𝑥 + 4 = 0 ✓S CA (x – 2 (x – 2) = 0 𝑥=2 ✓𝑥=2 CA OR/OF OR/OF √5 = √(𝑥 − 0)2 + (𝑦 − 0)2 ✓M A 5 = 𝑥2 + 𝑦2 5 = 𝑥 2 + (−2𝑥 + 5)2 ✓ SF CA 5 = 𝑥 2 + 4𝑥 2 − 20𝑥 + 25 0 = 5𝑥 2 − 20𝑥 + 20 ✓S CA −(−20) ± √(−20)2 − 4(5)(20) 𝑥= 2(5) 20 ± √0 𝑥= 10 𝑥=2 ✓𝑥=2 CA (4) [15] 5 ### Page 6 11092/23 QUESTION/VRAAG 2 2.1 2.1.1 𝑟2 = 𝑥2 + 𝑦2 𝑟 2 = (−3)2 + (−4)2 ✓ SF A 𝑟 2 = 25 𝑥 2 + 𝑦 2 = 25 ✓ 𝑥 2 + 𝑦 2 = 25 CA (2) 6 ### Page 7 11092/23 𝑦 −𝑦 2.1.2 𝑚𝑂𝑃 = 𝑥𝑃 −𝑥 𝑂 𝑃 𝑂 −4−0 = −3−0 4 4 CA =3 ✓3 3 3 CA ∴ mtangent/raaklyn = − 4 ✓ mtang/raak = − 4 3 ✓ SF (−3; −4) CA 𝑦 − (−4) = − (𝑥 − (−3)) 4 3 9 𝑦=− 𝑥− −4 4 4 4𝑦 = −3𝑥 − 25 3𝑥 + 4𝑦 = −25 OR/OF OR/OF 𝑥. 𝑥1 + 𝑦. 𝑦1 = 𝑟 2 ✓F A 𝑥(−3) + 𝑦(−4) = 25 ✓ SF A −4𝑦 = 3𝑥 + 25 ✓ SF 25 CA 3𝑥 + 4𝑦 = −25 (3) 2.1.3 3𝑥 + 4𝑦 = −25…  4𝑥 − 3𝑦 = 25…  × 3 9𝑥 + 12𝑦 = −75…  ✓ 9𝑥 + 12𝑦 = −75 A × 4 16𝑥 − 12𝑦 = 100…  ✓ 16𝑥 − 12𝑦 = 100 A + 25𝑥 = 25 ✓ 25𝑥 = 25 CA 𝑥=1 ✓𝑥=1 CA 𝑦 = −7 ✓ 𝑦 = −7 CA ∴ R(1; −7) OR/OF OR/OF 3 25 𝑦=− 𝑥− 4 4 4𝑥 − 3𝑦 = 25 3 25 ✓ SF A 4𝑥 − 3 (− 𝑥 − ) = 25 4 4 9 75 ✓S CA 4𝑥 + 𝑥 + = 25 4 4 25 25 25 25 CA 𝑥= ✓ 4𝑥= 4 4 4 𝑥=1 ✓𝑥=1 CA 𝑦 = −7 ✓ 𝑦 = −7 CA ∴ R(1; −7) (5) 7 ### Page 8 11092/23 2.2 𝑥 2 𝑦 2 ✓ Both/Beide 𝑥 A + =1 ✓ Both/Beide 𝑦 A 72 52 ✓ Shape/Vorm CA 𝑥 = ±√49 = ±7 𝑦 = ±√25 = ±5 (3) [13] 8 ### Page 9 11092/23 QUESTION/VRAAG 3 3.1 3.1.1 cos (A + C) = cos (40° + 50°) = cos (90°) =0 ✓0 A AO: Full marks AO: Volpunte (1) 3.1.2 cos (90° + 2𝐴 − 𝐶) sin (180° − 3𝐶) cos (90° + 80° − 50°) ✓ SF A = sin (180° − 150°) cos 120° = sin 30° cos (180° − 60°) = 𝑠𝑖𝑛30° −cos 60° ✓ S – cos60° CA = sin 30° 1 ✓ S sin 30° CA −2 = 1 2 = −1 ✓ –1 CA 𝟏 AO: 𝟒 (4) 9 ### Page 10 11092/23 3.2 3.2.1 𝑥 cos 36° = 4 𝑥 = 4𝑐𝑜𝑠36° ✓ 4𝑐𝑜𝑠36° A 𝑥 = 3,24 ✓ 3,24 CA (2) 3.2.2 𝑦 ✓ 𝑠𝑖𝑛36° = 4 𝑦 A 𝑠𝑖𝑛36° = 4 𝑦 = 4𝑠𝑖𝑛36° 𝑦 = 2,35 ✓ 2,35 CA OR/OF OR/OF 𝑥2 + 𝑦2 = 𝑟2 (4,24)2 + 𝑦 2 = (4)2 ✓ SF A 𝑦 2 = 5,5024 𝑦 = 2,35 ✓ 2,35 CA (2) 10 ### Page 11 11092/23 𝜋 3.3 𝑠𝑖𝑛 ( ) 3 𝜋 1800 𝜋 ✓3× 1800 A = 𝑠𝑖𝑛 ( × ) 𝜋 3 𝜋 = 𝑠𝑖𝑛600 √3 √3 ✓ 2 A = 2 𝟏 P: if/as 0,87 𝟐 (2) [11] QUESTION/VRAAG 4 4.1 𝑠𝑖𝑛2 𝛽 ✓ 𝑠𝑖𝑛2 𝛽 A 4.2 tan(1800 − 𝛽) . sin(1800 + 𝛽) + 𝑐𝑜𝑠 2 (3600 − β) 𝑠𝑒𝑐𝛽 −𝑡𝑎𝑛𝛽. (−sinβ) ✓ −𝑡𝑎𝑛𝛽 A = + cos 2 𝛽 1 cos𝛽 sin𝛽 ✓ −𝑠𝑖𝑛𝛽 A = (− ) (cos𝛽)(−sin𝛽) + cos2 𝛽 cos𝛽 1 = sin2 𝛽 + cos2 𝛽 ✓ 𝑐𝑜𝑠𝛽 A =1 𝑠𝑖𝑛𝛽 A ✓ 𝑐𝑜𝑠𝛽 ✓ cos 2 𝛽 A ✓ 𝑠𝑖𝑛2 𝛽 + 𝑐𝑜𝑠 2 𝛽 CA ✓1 CA (7) 11 ### Page 12 11092/23 4.3 LHS/LK = 𝑐𝑜𝑠𝜃(1 + 𝑡𝑎𝑛2 𝜃) 𝑠𝑖𝑛2 𝜃 𝑠𝑖𝑛2 𝜃 A LHS/LK= 𝑐𝑜𝑠𝜃 (1 + 𝑐𝑜𝑠2 𝜃) ✓ I 𝑐𝑜𝑠2 𝜃 LHS/LK= 𝑐𝑜𝑠𝜃 ( 𝑐𝑜𝑠2 𝜃+𝑠𝑖𝑛2 𝜃 ) ✓S A 𝑐𝑜𝑠2 𝜃 1 LHS/LK= 𝑐𝑜𝑠𝜃 (𝑐𝑜𝑠2 𝜃) 1 LHS/LK= 𝑐𝑜𝑠𝜃 LHS/LK= 𝑠𝑒𝑐𝜃 ✓S A OR/OF OR/OF LHS/LK = 𝑐𝑜𝑠𝜃(1 + 𝑡𝑎𝑛2 𝜃) LHS/LK = 𝑐𝑜𝑠𝜃(𝑠𝑒𝑐 2 𝜃) ✓ I 𝑠𝑒𝑐 2 𝜃 A 1 1 LHS/LK = 𝑐𝑜𝑠𝜃 (𝑐𝑜𝑠2 𝜃) ✓ I 𝑐𝑜𝑠2 𝜃 A 1 LHS/LK = 𝑐𝑜𝑠𝜃 LHS/LK = 𝑠𝑒𝑐𝜃 ✓S A (3) 4.4 2 tan(𝑥 + 100 ) = 3,464 tan(𝑥 + 100 ) = 1, 732 ✓S A Ref. ∠/Verw.∠ = 60° ✓ 60° A 𝑥 = 600 − 100 or/of 𝑥 = 1800 + 60𝑜 − 100 ✓ 500 CA 𝑥 = 500 𝑥 = 230𝑜 ✓ 230𝑜 CA OR/OF OR/OF 2 tan(𝑥 + 100 ) = 3,464 tan(𝑥 + 100 ) = 1, 732 ✓S A 𝑥 + 10° = 60° ✓ 60° A 𝑥 = 50° ✓ 500 CA or/of 𝑥 + 10° = 180° + 60° CA 𝑥 = 230° ✓ 230𝑜 (4) [15] 12 ### Page 13 11092/23 QUESTION/VRAAG 5 5.1 𝑓:✓ Shape/Vorm A ✓ 𝑥-int./afs. A ✓ TP/DP A 𝑔:✓ Shape/Vorm A ✓ TP/DP A ✓ 𝑥-int./afs. A (6) 5.2 180° ✓ 180° A (1) 5.3 𝑠𝑖𝑛2𝑥 + 2𝑐𝑜𝑠𝑥 = 0 𝑠𝑖𝑛2𝑥 = −2𝑐𝑜𝑠𝑥 ✓ 900 CA ∴ 𝑥 = 900 or/of 𝑥 = 2700 ✓ 2700 CA (2) ✓ Endpoints/ 5.4 −2 ≤ 𝑦 ≤ 2 Eindpunte A ✓ Notation/Notasie A OR/OF OR/OF ✓ Endpoints/ 𝑦 ∈ [−2; 2] A Eindpunte A ✓ Notation/Notasie (2) [11] 13 ### Page 14 11092/23 QUESTION/VRAAG 6 AB2 +AC2 −BC2 6.1 𝑐𝑜𝑠A = 2.AB.AC ✓F A 62 +72 −82 𝑐𝑜𝑠A = 2.6.7 ✓ 𝑐𝑜𝑠A = A 1 𝑐𝑜𝑠A = ✓ SF A 4 ̂ A = 75, 50 ✓S A OR/OF OR/OF ̂ BC2 = AB2 + AC2 − 2AB. AC. cosA ✓F A ̂ 82 = 62 + 72 − 2(6)(7)cosA ✓ SF A ̂ −21 = −84cosA ✓S A 1 ̂=1 A ̂= cosA ✓ cosA 4 4 ̂ = 75, 50 A (4) ̂ C = 2BA 6.2 BO ̂C ∠ at cent. = 2 x ∠ at circ. ✓ RE A mdpt. ∠ = 2 x omtr.∠ ̂ C = 2(75 ,50 ) BO ̂ C = 1510 CA BO ✓ 1510 (2) 14 ### Page 15 11092/23 6.3 1800 − 1510 ̂C = OB 2 ̂ C = 14, 50 OB ✓ 14, 50 CA OC BC 0 = ✓ SF CA sin14,5 sin1510 sin14,50 × 8 OC = sin1510 OC = 4, 13 ✓ 4, 13 CA ∴ 𝑑 = 2 × OC = 2(4,13) = 8, 26 units/eenhede ✓ 8, 26 CA OR/OF OR/OF ̂C BC2 = OB2 + OC2 − 2OB. OC. cosBO ✓F A 82 = OB2 + OB2 − 2. OB. OB. cos151° ✓ SF CA 64 = 2. OB2 + 1,75. OB 2 3,75. 𝑂𝐵 2 = 64 OB = 4,13 ✓ 4, 13 CA ∴ d = 2(4,13) = 8,26 units/eenhede ✓ 8, 26 CA NPR/NPU (4) 15 ### Page 16 11092/23 1 ̂C 6.4 Area of/Opp van ∆BOC = 2 × OB × OC × 𝑠𝑖𝑛BO 1 Area of/Opp van ∆BOC = 2 × 4,13 × 4,13 × 𝑠𝑖𝑛1510 ✓ SF CA Area of/Opp van ∆BOC = 4,1 unit2/eenheid2 ✓ 4,1 CA OR/OF OR/OF P/Omtrek ∆OBC = OB + OC + BC P/Omtrek ∆OBC = 16,26 P 16,26 Semi – Perimeter/Omtrek = 2 = = 8,13 2 𝑃 𝑃 𝑃 𝑃 A/Opp ∆OBC = √ 2 ( 2 − 𝑂𝐵) (2 − 𝑂𝐶) ( 2 − 𝐵𝐶) A/Opp ∆OBC = √8,13(4)(4)(0,13) ✓ SF CA A/Opp ∆OBC = 4,1 units2/eenhede2 ✓ 4,1 CA NPR/NPU (2) [12] 16 ### Page 17 11092/23 QUESTION/VRAAG 7 7.1 the angle that the chord subtends in the alternate segment. ✓ RE A die hoek in die teenoorstaande sirkel segment. (1) 7.2 7.2.1 ∠𝑠 opp. = sides/ ✓ RE A ∠𝑒 teenoor = sye (1) 7.2.2 Ĉ1 = E ̂2 = 35° ∠ tan chord ✓ ST/RE A ∠ raaklyn koord ̂ A = Ĉ1 = 35° ∠ tan chord ✓ ST A ∠ raaklyn koord Ĉ4 = E ̂2 = 35° ∠𝑠 opp = sides ✓ ST A ∠𝑒 teenoor = sye OR/OF OR/OF Ĉ1 = E ̂2 = 35° ∠ tan chord ✓ ST/RE A ∠ raaklyn koord 𝐴̂ = 𝐸̂2 = 35° ∠𝑠 in same segment ✓ ST/RE A ∠𝑒 in dies. segment Ĉ4 = E ̂2 = 35° ∠𝑠 opp equal sides ✓ ST A ∠𝑒 teenoor = sye (3) 17 ### Page 18 11092/23 7.2.3 EĈB = 90° ∠ in semi-circle ✓ ST/RE A ∠ in halwe sirkel ̂ B3 = 180° − 90° − 35° Int. ∠∆/ ̂ 3 = 55° ✓ ST CA B binne ∠𝑒 ∆ CA Ĉ5 = ̂ B3 = 55° ∠ tan chord/∠ raaklyn koord ✓ ST (3) 7.2.4 ̂ C = 2 × CB ̂O EO ∠ at cent. = 2 x ∠ at circ. ✓ ST A A mdpt.∠ = 2x Omtr. ∠ ✓ RE (2) [10] QUESTION/VRAAG 8 8.1 corresponding/ooreenstemmende ✓ ST A (1) 8.2 8.2.1 BF AE = CE prop theorem/eweredigheid; EF ∥ AB ✓ ST A CF 𝑟 12 = ✓ RE A 4 8 CA 𝑟 = 6 ✓6 (3) 8.2.2 Parallelogram. Both pairs of opposite sides are parallel ✓ ST A ✓ RE A Twee pare teenoorst. sye ewewydig (2) 18 ### Page 19 11092/23 8.2.3 AD AE DB = EC prop theorem/eweredigheid; DE ∥ BC ✓ RE A 𝑡 12 = ✓ ST A 15 8 45 𝑡= 2 45 1 𝑡 = 22,5 or/of 2 or/of 22 2 ✓ 22,5 CA NPR (3) 8.2.4 AD 22,5 3 = = ✓ ST CA EF 15 2 AE 12 3 = = ✓ ST CA EC 8 2 DE 6 3 = = ✓ ST CA FC 4 2 AD AE DE ∴ = = EF EC FC ∴ ∆ADE||| ∆EFC sides in prop./sye in verhouding ✓ RE CA OR / OF OR / OF ̂=E A ̂1 Corr. ∠′𝑠 =; AB//EF ✓ ST/RE CA Ooreenst. ∠𝑒 =; AB//EF ̂3 = Ĉ E Corr. ∠′𝑠 =; DE//BC ✓ ST CA Ooreenst. ∠𝑒 =; DE//BC D1 = F̂2 ̂ 3𝑟𝑑 ∠ ∆ ✓ ST CA 3𝑑𝑒 ∠ ∆ ∴ ∆ADE||| ∆EFC [∠, ∠, ∠] ✓ RE CA (4) [13] 19 ### Page 20 11092/23 QUESTION/VRAAG 9 9.1 9.1.1 Ŝ = Ŝ Common ∠/Gemeensk. ∠ ✓ ST/RE A ̂ = XY Q ̂S Both/Beide 90° (Given / Gegee) ✓ ST/RE A ̂ ̂S R = YX 3𝑟𝑑 ∠ ∆/3e ∠ ∆ ✓ ST/RE A ∆SYX|||∆SQR [∠, ∠, ∠] ✓ RE A (4) 9.1.2 SY SQ ✓ SR A = SX SR (1) 9.1.3 SY SQ = SX SR SY 30 ✓ SF A = 25 SY + 4√5 SY(SY + 4√5) = 750 SY 2 + 4√5SY − 750 = 0 ✓S CA 2 −4√5±√(4√5) −4(1)(−750) ✓ SF CA SY = 2(1) SY = 23,28 or/of SY = −32,22 ✓ 23,28 CA N. A./𝑁. 𝑣. 𝑇. NPR/NPU (4) 20 ### Page 21 11092/23 9.2 9.2.1 𝐴𝑟𝑒𝑎 ∆SYX 1 ×(𝑌𝑆)(𝑋𝑌) ✓ SF = 12 CA 𝐴𝑟𝑒𝑎 ∆SQR ×(𝑄𝑆)(𝑄𝑅) 2 𝐴𝑟𝑒𝑎 ∆SYX 1 2 ×(23,28)(5√5) ✓ SF = 1 A 𝐴𝑟𝑒𝑎 ∆SQR (30)(15) 2 𝐴𝑟𝑒𝑎 ∆SYX = 0,60 units2/eenhede2 ✓ 0,60 CA 𝐴𝑟𝑒𝑎 ∆SQR NPR/NPU (3) 9.2.2 AQRYS = A∆SQR − A∆SYX ✓M A 1 1 AQRYS = 2 (30)(15) − 2 (23,28)(5√5) ✓ SF CA AQRYS = 225 − 130,39 … AQRYS = 94,86 units2/eenhede2 ✓ 94,86 CA NPR/NPU (3) [15] 21 ### Page 22 11092/23 QUESTION/VRAAG 10 10.1 11000 𝑛 = 60 550 𝑛 = 3 rev/s 𝑛 = 183,33 rev/s ✓ 183,33 A NPR/NPU (1) 10.2 𝑣 = πDn ✓F A 11000 𝑣 = 𝜋(0,32) ( 60 ) ✓ SF A 𝑣 = 58,6667𝜋 m/s 𝑣 = 184,30 m/s ✓ CA NPR/NPU (3) 10.3 𝜔 = 2𝜋𝑛 ✓F A 11000 𝜔 = 2𝜋 ( ) ✓ SF CA 60 𝜔 = 366,67𝜋 rad/s ✓ 𝜔 = 1151,90 rad/s CA 1151,90 NPR/NPU (3) 22 ### Page 23 11092/23 10.4 In ∆ODC: OD OD OD = 𝑠𝑖𝑛30° OR/OF = cos60° ✓ 16 A 16 16 ✓ 𝑠𝑖𝑛30° OD = 16sin30° OR/OF OD = 16cos60° OR/OF A cos60° OD = 8 cm ℎ = 16 − OD = 16 − 8 ℎ = 8 𝑐𝑚 ✓8 CA (3) 10.5 4ℎ2 − 4𝑑ℎ + 𝑥 2 = 0 ✓F A 4(8)2 − 4(2 × 16)(8) + 𝑥 2 = 0 ✓ SF A 256 − 1024 + 𝑥 2 = 0 𝑥 2 = 768 𝑥 = 16√3 𝑥 = 27,71 cm ✓ 27,71 CA OR/OF OR/OF AC2 = OA2 + OC2 − 2. OA. OC. cos120° ✓F A AC2 = 162 + 162 − 2(16)(16). cos120° ✓ SF A AC = 16√3 = 27,71 cm ✓ 27,71 CA NPR/NPU (3) [13] 23 ### Page 24 11092/23 QUESTION/VRAAG 11 𝜋 11.1 80° × 180° 4 = 𝜋 9 = 1,4 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 ✓ 1,4 A NPR/NPU (1) 11.2 𝑠 = 𝑟𝜃 ✓F A 4 𝑠 = (7) ( 𝜋) ✓ SF CA 9 𝑠 = 9,77 cm ✓ 9,77 CA NPR/NPU (3) 𝑠𝑟 11.3 𝐴𝐵𝑖𝑔𝑔𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 2 (9,77)(7) 𝐴𝐵𝑖𝑔𝑔𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 2 𝐴𝐵𝑖𝑔𝑔𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 34,20 cm2 ✓ 34,20 CA 𝑟 2𝜃 𝐴𝑆𝑚𝑎𝑙𝑙𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 2 4 (3)2 ( 𝜋) 𝐴𝑆𝑚𝑎𝑙𝑙𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 9 2 𝐴𝑆𝑚𝑎𝑙𝑙𝑒𝑟 𝑆𝑒𝑐𝑡𝑜𝑟 = 6,26 cm2 ✓ 6,26 CA 𝐴𝑇 = 34,20 − 6,28 ✓M A 𝐴𝑇 = 27,92 cm2 ✓ 27,92 CA NPR/NPU (4) [8] 24 ### Page 25 11092/23 QUESTION/VRAAG 12 12.1 12.1.1 152 = 92 + (8 + 𝑦)2 ✓M A 144 = (8 + 𝑦)2 ✓S CA 12 = 8 + 𝑦 𝑦 = 4 𝑐𝑚 ✓4 CA (3) 12.1.2 V = Vcone/keël – Vcut-off/afsny 1 1 𝑉 = 𝜋𝑟 2 𝐻 − 𝜋𝑟 2 𝑦 ✓M A 3 3 1 1 𝑉 = (3,14)(9)2 (4 + 8) − (3,14)(3)2 (4) ✓ SF A 3 3 𝑉 = 1017,36 − 37,64 𝑉 = 979,72 𝑐𝑚3 ✓ 979,72 CA PR (3) 25 ### Page 26 11092/23 12.1.3 TSA = SAcone/keël – SAcut-off/afsny + SABASE/BASIS SAcone/keël = (3,14)(9)(15) SAcone/keël = 423,9 … 𝑐𝑚2 ✓ 423,9 CA 𝑙 cut-off/afsny = √42 + 32 = 5𝑐𝑚 ✓5 A SAcut-off/afsny = (3,14)(3)(5) SAcut-off/afsny = 47,1 … 𝑐𝑚2 ✓ 47,1 CA SABASE/BASIS = (3,14)(3)2 = 28,26. . 𝑐𝑚2 ✓ 28,26 CA TSA = 423,9 − 47,1 + 28,26 TSA = 405,06 𝑐𝑚2 405,06 Number of tins/Aantal blikkies = 90 Number of tins/Aantal blikkies = 4,5 … = 5 tins/blikkies ✓5 CA (5) 12.2 𝑂1 + 𝑂𝑛 𝐴𝑇 = 𝑎 ( + 𝑂2 + 𝑂3 + . . . +𝑂𝑛−1 ) ✓F A 2 7,4 + 8,1 𝐴𝑇 = 5 ( + 6,9 + 5,6 + 7,8 + 6,7 + 7,4) ✓ SF A 2 𝐴𝑇 = 5(42,15) 𝐴𝑇 = 210,75 m2 ✓ 210,75 CA (3) [14] TOTAL/TOTAAL: 150 26 ## Related documents - [TECH MATHS P2 MEMO JUNE 2023 Afrikaans+English](https://www.hlayiso.com/library/tech-maths-p2-memo-june-2023-afrikaans-english/8fb4241f-2e03-413c-8367-8e8e5cecc687) — Technical Mathematics · Memo · 2023 · Afr - [Gr12 Technical Mathematics P2 (English and Afrikaans) 2021 Preparatory Possible Answers](https://www.hlayiso.com/library/gr12-technical-mathematics-p2-english-and-afrikaans-2021-preparatory-possible-answers/615318ee-d80c-4db5-a0ba-3926b9f96288) — Technical Mathematics · Memo · 2021 · Afr - [TECH MATHS P2 ANSWER BOOK JUNE 2023 Afrikaans+English](https://www.hlayiso.com/library/tech-maths-p2-answer-book-june-2023-afrikaans-english/120c2552-197e-40ac-8eb4-460d2eaab01e) — Technical Mathematics · Addendum · 2023 · Afr - [Gr 12 Technical Mathematics P2 (Afrikaans) 2025 Question Paper X05](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-gauteng-mock-exam-question-paper-paper-2-afrikaans/673174da-3296-4dd9-b0ed-c262fb1d1c63) — Technical Mathematics · QP · 2025 · Afr - [Gr12 Technical Mathematics P2 (Afrikaans) 2021 Preparatory Question Paper](https://www.hlayiso.com/library/gr12-technical-mathematics-p2-afrikaans-2021-preparatory-question-paper/062749a9-c572-415d-8d76-eab76e1a2e74) — Technical Mathematics · QP · 2021 · Afr - [Tech MATH P2 (English and Afrikaans) X10 2019 Preparatory Examination Answer book](https://www.hlayiso.com/library/tech-math-p2-english-and-afrikaans-x10-2019-preparatory-examination-answer-book/35283687-efdb-40c5-a5e3-383bee3f78ba) — Technical Mathematics · Addendum · 2019 · Afr ## More from Grade 12 Technical Mathematics - [Technical Mathematics P1 Nov 2025 Afr](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-nsc-november-exam-question-paper-paper-1/00c4af24-67df-4872-a5e8-a2928fb0fe99) — QP · 2025 · Afr - [TECH MATHS P2 AB AFR ENG JUNE 2025](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-nw-june-exam-addendum-paper-2-afrikaans/7615a50d-6554-42bc-a0f1-879ea3150d06) — Addendum · 2025 · Afr - [TECH MATHS P2 QP AFR JUNE 2025](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-nw-june-exam-question-paper-paper-2-afrikaans/4a488928-ed3b-48aa-a9c1-a53a6ffc4f7a) — QP · 2025 · Afr - [TECH MATHS P2 ANSWER BOOK JUNE 2024 Afr+Engl](https://www.hlayiso.com/library/tech-maths-p2-answer-book-june-2024-afr-engl/b5a5c28f-caaa-41b0-b1a1-136115668877) — Addendum · 2024 · Afr - [TECH MATHS P2 ANSWER BOOK GR12 JUNE 2024 (Backup)](https://www.hlayiso.com/library/tech-maths-p2-answer-book-gr12-june-2024-backup/5931090d-affa-49bb-b628-82a8b751ff35) — Addendum · 2024 - [Tech Maths P2 Erratum June Common 2024](https://www.hlayiso.com/library/tech-maths-p2-erratum-june-common-2024/5751ef8a-8c68-4bef-bf01-27b139b3c084) — Errata · 2024 ## 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. 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