# Technical Mathematics Grade 12 P2 Gauteng Prelim 2024 Memo > Grade 12 Technical Mathematics Paper 2 memorandum from the Gauteng Prelim Exam 2024 (Afrikaans), 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-12-technical-mathematics-p2-english-and-afrikaans-2024-possible-answers/a57aa3c8-17e4-416b-bb0e-e0f3307b2346 - Updated: 2026-09-27 - Subject: Technical Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2024 - Exam period: Gauteng Prelim Exam - Paper: 2 - Curriculum: CAPS - Pages: 19 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 Prelim 2024 Memo](https://www.hlayiso.com/library/gr-12-technical-mathematics-p2-english-and-afrikaans-2024-possible-answers/a57aa3c8-17e4-416b-bb0e-e0f3307b2346) ## About this document Technical Mathematics · Grade 12 · Gauteng Prelim Exam · 2024 · Afrikaans. Memorandum, 19 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 AMENDMENT TO MARKING GUIDELINES PREPARATORY EXAMINATIONS – 2024 FOR ATTENTION: THE CHIEF INVIGILATOR SUBJECT / VAK TECHNICAL MATHEMATICS / TEGNIESE WISKUNDE PAPER / VRAESTEL 2 DATE OF EXAMINATION 3 SEPTEMBER 2024 The errata for the Marking Guidelines of TECHNICAL MATHEMATICS P2 has reference. There was an OMISSION OF FORMULAE in Question 11 on both the English and Afrikaans versions of the question paper. This matter was addressed at the Marking Standardisation Meeting. To ensure that candidates are not disadvantaged and prejudiced in any way, you are advised to please ask your Technical Mathematics educator(s) to ignore Questions 11.1.1 to 11.1.4 when marking. In other words, the paper must be marked out of a total of 141 instead of 150 and then the learners’ marks must be converted to a mark out of 150. E.g., Should a learner attain 85/141 then that mark is recalculated as 91/150. Use the a b formula: x 100 = b. Then, x 150 = c 141 100 c is the mark that is entered into SASAMS out of 150. DIRECTOR: EXAMINATIONS MANAGEMENT 3 September 2024 ### Page 2 Downloaded from hlayiso.com PREPARATORY EXAMINATIONS VOORBEREIDENDE EKSAMEN 2024 OFFICIAL MARKING GUIDELINES/ AMPTELIKE NASIENRIGLYNE TECHNICAL MATHEMATICS/TEGNIESE WISKUNDE (PAPER/VRAESTEL 2) 16 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 is nie S Simplification/Vereenvoudiging SF Substitution in correct formula/Vervanging in korrekte formule ST Statement/Bewering ST/RE Statement and Reason/Bewering en Rede 2 ### Page 3 Downloaded from hlayiso.com 11092/24 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. LET WEL: • 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 0 = −2𝑥 + 17 ✓ y= 0 A 2𝑥 = 17 17 17 𝑥= 2 ✓ 𝑥= 2 CA 17 1 K ( 2 ; 0) or/of K (8 2 ; 0) or/of K(8,5; 0) (2) 1.2 𝛽 = 𝑡𝑎𝑛−1 (2) ✓ SF A ° ✓ 63, 43 ∴ Ref/𝑉𝑒𝑟𝑤. ∠ = 63, 43° CA 𝛽 = 180° − 63, 43° ✓ 116, 57° ∴ 𝛽 = 116, 57° CA (3) ### Page 4 Downloaded from hlayiso.com 11092/24 1.3 Midpoint/Middelpunt 𝑥 +𝑥 Translation/Translasie: ✓M A 𝑇𝑥 = 𝑆 2 𝑃 7 + 𝑥𝑃 1= S(7; 3) → T(1; 1) 2 2 − 7 = 𝑥𝑃 OR/OF (𝑥 − 6; 𝑦 − 2) ∴ 𝑥𝑃 = −5 T(1; 1) → P(𝑥; 𝑦) ✓ 𝑥𝑃 = −5 A Midpoint/Middelpunt 𝑦 +𝑦 P(1 − 6; 1 − 2) 𝑇𝑦 = 𝑆 2 𝑃 3 + 𝑦𝑃 1= P(−5; −1) 2 2 − 3 = 𝑦𝑃 ∴ 𝑦𝑃 = −1 ✓ 𝑦𝑃 = −1 A 𝑃(−5; −1) AO: Full marks/ (3) Volpunte 𝑦 −𝑦 1.4 𝑚SP = 𝑥𝑃 −𝑥 𝑆 𝑃 𝑆 −1−3 𝑚SP = −5−7 ✓ SF CA −4 𝑚SP = −12 1 1 CA 𝑚SP = 3 ✓ 𝑚SP = 3 OR/OF OR/OF 𝑦 −𝑦 𝑚𝑆𝑃 = 𝑥𝑆 −𝑥 𝑇 𝑆 𝑇 3−1 𝑚SP = 7−1 ✓ SF CA 2 𝑚SP = 6 1 1 CA 𝑚SP = 3 ✓ 𝑚SP = 3 OR/OF OR/OF 𝑦 −𝑦 𝑚𝑆𝑃 = 𝑥𝑃 −𝑥 𝑇 𝑃 𝑇 −1−1 𝑚SP = −5−1 ✓ SF CA −2 𝑚SP = −6 1 1 CA 𝑚SP = 3 ✓ 𝑚SP = 3 (2) 4 ### Page 5 Downloaded from hlayiso.com 11092/24 1.5 Finding equation of QR/Vind die vergelyking van QR: CA 1.3 & 1.4 1 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) ; 𝑚 = 3 and/en 𝑝𝑡. (−1; 5) 1 𝑦 − 5 = 3 [𝑥 − (−1)] 1 16 𝑦 = 3𝑥 + 3 or/of 1 5 = 3 (−1) + 𝑐 16 𝑐= 3 1 16 1 16 CA 𝑦 = 3𝑥 + 3 ✓ 𝑦 = 3𝑥 + 3 1 16 Subst./Vervang 𝑦 = −2𝑥 + 17 in 𝑦 = 3 𝑥 + 3 1 16 M CA ∴ −2𝑥 + 17 = 3 𝑥 + 3 7 35 7 35 −3𝑥 = − 3 or/of 𝑥= 3 3 𝑥=5 ✓𝑥=5 CA 𝑦 = −2(5) + 17 ∴𝑦=7 ✓𝑦=7 CA ∴ 𝑅(5; 7) OR/OF OR/OF 1 𝑦−5 CA 1.3 & 1.4 = 𝑥+1 3 𝑥 + 1 = 3𝑦 − 15 𝑥 = 3𝑦 − 16  𝑥 = 3𝑦 − 16 CA Subst./Vervang 𝑥 = 3𝑦 − 16 in 𝑦 = −2𝑥 + 17 𝑦 = −2(3𝑦 − 16) + 17 M CA 𝑦 = −6𝑦 + 32 + 17 7𝑦 = 49 𝑦=7 ✓𝑦=7 CA 𝑥 = 3(7) − 16 𝑥=5 ✓𝑥=5 CA R(5; 7) (4) [14] 5 ### Page 6 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 2 2.1 2.1.1 𝑟 2 = 𝑥 2 + 𝑦 2 𝑟 2 = (0)2 + (10)2  SF A 𝑟 2 = 100 𝑥 2 + 𝑦 2 = 100  𝑥 2 + 𝑦 2 = 100 CA OR/OF OR/OF 2 2 2 𝑟 =𝑥 +𝑦 𝑟 2 = (−8)2 + (−6)2  SF A 𝑟 2 = 100 𝑥 2 + 𝑦 2 = 100  𝑥 2 + 𝑦 2 = 100 CA (2) 2.1.2 N(10; 0) ✓ N(10; 0) A ACCEPT/AANVAAR: 𝑥 = 10 and/en 10 (1) 2.1.3 E (𝑥M +𝑥N ; 𝑦M +𝑦N ) 2 2 E( −8+10 −6+0 ; ) ✓𝑥=1 CA 2 2 E(1; −3) ✓ 𝑦 = −3 A 𝐿𝐸 = √(0 − 1)2 + (10 − (−3))2 ✓ SF CA 𝐿𝐸 = √1 + 169 𝐿𝐸 = √170 𝐿𝐸 = 13,04 ✓ 13,04 CA PR (4) 6 ### Page 7 Downloaded from hlayiso.com 11092/24 2.1.4 𝑚𝑟 = −6−0 −8−0 3 3 A 𝑚𝑟 = 4  𝑚𝑟 = 4 4 4 CA ∴ 𝑚𝑡 = − 3  𝑚𝑡 = − 3 4 𝑦 − (−6) = − 3 [𝑥 − (−8)]  SF CA 4 32 𝑦 = −3𝑥 − 3 − 6 4 50 4 50 CA ∴ 𝑦 = −3𝑥 − 3  𝑦 = −3𝑥 − 3 OR/OF OR/OF 𝑥. 𝑥1 + 𝑦. 𝑦1 = 𝑟 2 F A 𝑥(−8) + 𝑦(−6) = 100  SF (−8; −6) A −6𝑦 = 8𝑥 + 100  SF 100 A 4 50 4 50 CA 𝑦 = −3𝑥 − 3  𝑦 = −3𝑥 − 3 (4) 2.2.1 𝑥2 𝑦2 + 81 = 1 36 𝑥2 𝑦2 + =1  62 and/en 92 A 62 92 (1) 2.2.2  Shape/Vorm A  All 4 intercepts/ CA Al 4 afsnitte (2) [14] 7 ### Page 8 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 3 P(8; 6) Q(t; 3) 2 2 2 3.1.1 𝑟 2 = 𝑥2 + 𝑦2 𝑟 = 6 +8 ✓ SF A ∴ 𝑟 = 10 ✓ 𝑟 = 10 CA 𝑠𝑖𝑛𝐴 6 = 10 3 6 3 =5 ✓ 10 or/of 5 or/of 0,6 CA (3) 3.1.2 𝑡𝑎𝑛𝐴. [cos(2𝜋 − 𝐴)] = 𝑡𝑎𝑛𝐴. (𝑐𝑜𝑠𝐴) ✓ 𝑐𝑜𝑠𝐴 A 6 = (8) (10) 8  SF CA 6 = 10 3 3 =5  5 or/of 0,6 CA OR/OF OR/OF = 𝑡𝑎𝑛𝐴. (𝑐𝑜𝑠𝐴) ✓ 𝑐𝑜𝑠𝐴 A 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝐴 = . 𝑐𝑜𝑠𝐴 CA 𝑐𝑜𝑠𝐴 = 𝑠𝑖𝑛𝐴 3 3 =  5 or/of 0,6 CA 5 (3) 8 ### Page 9 Downloaded from hlayiso.com 11092/24 3.1.3 𝑠𝑖𝑛𝐴 = 3  Ratio A 5 3 𝐴 = 𝑠𝑖𝑛−1 (5) ° ∴ 𝐴 = 36, 87  36, 87° CA OR/OF OR/OF 8  Ratio 𝑐𝑜𝑠𝐴 = 10 A 4 𝐴 = 𝑐𝑜𝑠 −1 (5) ° ∴ 𝐴 = 36, 87  36, 87° CA OR/OF OR/OF 6  Ratio 𝑡𝑎𝑛𝐴 = 8 A 3 𝐴 = 𝑡𝑎𝑛−1 (4) ∴ 𝐴 = 36, 87°  36, 87° CA (2) 3.1.4 𝑡𝑎𝑛𝐵 = −0, 3 3  Ratio 𝑡𝑎𝑛𝐵 = − A 𝑡 3  SF −0, 3 = − 𝑡 A 𝑡 = 10  𝑡 = 10 CA (3) 3.2 cot (𝑚 + 𝑛) 180° 𝑚 = 1,427 × = 81,76° 𝜋 180° 𝑛 = 3,324 × = 190,45° 𝜋 = cot (1, 427 + 3, 324) = cot (81,76° + 190,45°) ✓ SF A = cot (4, 751) = 𝑐𝑜𝑡272,21° 1 1 ✓I = tan (4,751) = 𝑡𝑎𝑛272,21 ° A = −0,04 = −0,04 ✓ −0,04 CA NPR (3) [14] 9 ### Page 10 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 4 4.1 𝑐𝑜𝑠𝑒𝑐 2 𝑥 − 1 = 𝑐𝑜𝑡 2 𝑥 ✓I A (1) 4.2 tan(180° +𝑥).sin(360° −𝑥).𝑐𝑜𝑠225° ✓ t𝑎𝑛𝑥 5𝜋 A 𝑠𝑖𝑛2 𝑥.𝑡𝑎𝑛 4 tan 𝑥.− sin 𝑥.cos (180° +45° ) ✓ −𝑠𝑖𝑛𝑥 = 𝜋 A 𝑠𝑖𝑛2 𝑥.tan (𝜋+ ) 4 𝑠𝑖𝑛𝑥 .− sin 𝑥.(−cos 45° ) ✓ −𝑐𝑜𝑠45° 𝑐𝑜𝑠𝑥 = 𝜋 A 𝑠𝑖𝑛2 𝑥.tan ( 4 ) −𝑠𝑖𝑛2 𝑥 √2 𝜋 𝑐𝑜𝑠𝑥 .(− ) 2 ✓ 𝑡𝑎𝑛 ( 4 )/𝑡𝑎𝑛45° = A 𝑠𝑖𝑛2 𝑥.(1) 𝑠𝑖𝑛𝑥 √2 √2 ✓ 𝑐𝑜𝑠𝑥 = or/of = 𝑠𝑒𝑐𝑥 A 2𝑐𝑜𝑠𝑥 2 −√2 ✓ A 2 S CA √2 ✓ 2 𝑠𝑒𝑐𝑥 CA (8) 4.3 √3 sin(𝜃 + 40° ) = 2 √3 Ref/𝑉𝑒𝑟𝑤𝑦𝑠 ∠ = sin−1 2 Ref/𝑉𝑒𝑟𝑤𝑦𝑠 ∠ = 60°  Ref/𝑉𝑒𝑟𝑤𝑦𝑠 A ∠ = 60° 𝜃 + 40° = 60° or/of 𝜃 + 40° = 180° − 60°  𝜃 = 20° CA 𝜃 = 20° or/of 𝜃 = 80°  𝜃 = 80° CA (3) [12] 10 ### Page 11 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 5 5.1 −1 ≤ 𝑦 ≤ 1  ≤ 𝑦 ≤ or/of [;] A OR/OF 𝑦 ∈ [−1; 1]  −1 & 1 A (2) 5.2 120°  120° A (1) 5.3 𝑓(360° ) − 𝑔(360° ) = 𝑠𝑖𝑛3𝑥 − cos (𝑥 − 30° ) = 𝑠𝑖𝑛3(360° ) − cos (360° − 30° )  SF A √3 √3 = − 2 or/of − 0,87 − 2 CA AO: Full (2) marks/Volpunte 5.4 𝑥 = 30°  30° A 𝑥 = 120°  120° A 𝑥 = 210°  210° A (3) 5.5 180° < 𝑥 < 240°  < 𝑥 < or/of (;) A OR/OF  180° & 240° A (180°; 240°) (2) [10] 11 ### Page 12 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 6 6.1.1 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐶  𝑠𝑖𝑛𝐶 = A 𝑎 𝑐 (1) 6.1.2 𝑏 2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐. 𝑐𝑜𝑠𝐵 𝑏 2 A (1) 6.2 ̂ 6.2.1 US 2 = UT 2 + TS 2 − 2. UT. TS. cosT (In ∆UTS) US = (120) + (200) − 2(120)(200)cos110° 2 2 2  SF A US 2 = 70 816, 967 S CA US = 266, 11 m  266, 11 CA NPU (3) ̂ 6.2.2 US 2 = UV 2 + VS 2 − 2. UV. VS. cosV (In ∆UVS) 2 2 2 (266,11) = (70) + (280) − 2(70)(280)cosV̂  SF CA ̂ −12 485,4679 = −39 200cosV S CA ̂ −1 −1 V = cos (0,31851)  cos A ̂ V = 71,43°  71, 43° CA OR/OF OR/OF UV2 +VS2 −US2  SF 𝑐𝑜𝑠V = CA 2UV.VS (70)2 +(280)2 −(266,11)2 S 𝑐𝑜𝑠V = CA 2(70)(280) ̂ = 𝑐𝑜𝑠 −1 (0,31851) V  cos −1 A ̂ V = 71,43°  71, 43° CA (4) 12 ### Page 13 Downloaded from hlayiso.com 11092/24 6.2.3 Area of/van ∆UTS = 1 . UT. TS. sinT ̂ 2 1 Area of/van ∆UTS = 2 × 120 × 200 × 𝑠𝑖𝑛110° Area of/van ∆UTS = 11 276, 31unit 2  11 276, 31 CA 1 ̂ Area of/van ∆UVS = 2 . UV. VS. sinV 1 Area of/van ∆UVS = 2 . 70 × 280 × 𝑠𝑖𝑛71, 43° Area of/van ∆UVS = 9289, 76unit 2  9 289, 76 CA The area of the plot/Die area van die grond = 11 276, 31 + 9289,76 M A The area of the plot/Die area van die grond = 20 566,08 𝑚2  20 566,08 CA (4) [13] QUESTION/VRAAG 7 7.1 perpendicular/loodreg ✓ ST A (1) 7.2 7.2.1 ̂ A2 = 30° tan-chord theorem ✓ST A ∠ raaklyn-koord ✓RE A (2) 7.2.2 In ∆ 𝐴𝐵𝐶 ̂C = 90° AT ∠ in semicircle  ST CA ∠ 𝑖𝑛 ℎ𝑎𝑙𝑤𝑒 𝑠𝑖𝑟𝑘𝑒𝑙  RE A 30° + 90° + TĈO = 180° sum of  ' of ∆ TĈO = 60° 𝑏𝑖𝑛𝑛𝑒 ∠𝑒 ∆  ST CA OR/OF OR/OF ̂3 = 90° − 30° T tan. ⊥ radius ̂3 = 60° T 𝑟𝑎𝑎𝑘𝑙𝑦𝑛 ⊥ 𝑟𝑎𝑑𝑖𝑢𝑠  ST/RE CA TO = CO Radii  ST/RE A Radiusse TĈO = T ̂3 ∠𝑠 opp. = sides TĈO = 60° ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 = 𝑠𝑦𝑒  ST CA (3) 13 ### Page 14 Downloaded from hlayiso.com 11092/24 7.2.3 In ∆ AOT T2 = ̂ ̂ A2 ∠𝑠 opp. = sides  ST CA ̂ T2 = 30° ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 = 𝑠𝑦𝑒  RE A OR/OF OR/OF In ∆ CAT ̂2 = 90° − 60° T ∠ in semi circle  ST CA ̂2 = 30° T ∠ 𝑖𝑛 ℎ𝑎𝑙𝑤𝑒 𝑠𝑖𝑟𝑘𝑒𝑙  RE A (2) 7.2.4 BĈT = ST̂C Alt. ∠s =; BC║TS  ST A ̂ BCT = 30° 𝑉𝑒𝑟𝑤. ∠𝑒 =; BC║TS  RE A ̂ 1 = BĈT A ∠𝑠 in same segment  ST CA ̂ 1 = 30° A ∠𝑒 𝑖𝑛 𝑑𝑖𝑒𝑠. 𝑠𝑒𝑔𝑚𝑒𝑛𝑡  RE A (4) 7.3 TĈO = 60° Proven ̂ ∴ BCA = 30° BĈA ≠ DA ̂C  ST CA AD not // to BC 𝐴𝑙𝑡. ∠𝑠 ≠ AD nie // aan BC 𝑉𝑒𝑟𝑤. ∠𝑒 ≠  RE A ABCD not a parallelogram. ABCD is nie 'n paralellogram nie. OR/OF OR/OF ̂ BAD = 110° Proven/Bewys CA BĈD = 70° Proven/Bewys ∴ BÂ D ≠ BĈD ABCD not a parallelogram.  ST CA ABCD is nie 'n paralellogram nie. Opp. angles not equal/Teenoorst. hoeke nie gelyk  RE A (2) [14] 14 ### Page 15 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 8 8.1.1 Double (twice)/Dubbel (twee keer)  ST A (1) 8.1.2 equal/gelyk  ST A (1) 8.1.3 supplementary/supplementêre  ST A (1) 8.2.1 ̂ B1 = 40° tan-chord  ST/RE A raaklyn-koord ̂2 = B D ̂1 = 40° ∠s opp. = sides  ST/RE A ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 = 𝑠𝑦𝑒 (2) 8.2.2 Ĉ = 180° − 40° − 40° s Int. ∠ ∆  ST A Ĉ = 100° 𝐵𝑖𝑛𝑛𝑒 ∠𝑒 ∆ (1) 8.2.3 ̂ A = 180° − 100° Opp. ∠s cyclic quad  ST CA ̂ A = 80° 𝑇𝑒𝑒𝑛𝑜𝑜𝑟𝑠𝑡. ∠𝑒 𝑘𝑑𝑣ℎ. (1) ̂ 1 = 2 × 80° 8.2.4 O ∠ 𝑐𝑒𝑛𝑡𝑟𝑒 = 2 × ∠ 𝑐𝑖𝑟𝑐.  ST CA ̂ 1 = 160° O 𝑚𝑑𝑝𝑡. ∠ = 2 × 𝑜𝑚𝑡𝑟. ∠  RE A (2) [9] QUESTION/VRAAG 9 9.1 proportion/eweredig  ST A (1) 9.2 9.2.1 QS 2 =5 ✓ Ratio A SP QS 2 =  SF 10 A 10 5 20 QS = 5 S CA QS = 4 cm  𝑄𝑆 = 4 CA (4) 15 ### Page 16 Downloaded from hlayiso.com 11092/24 9.2.2 SR PS = RS ∆PSR///∆RSQ  ST A SQ RS 2 = SQ × PS S CA RS = √4 × 10  SF CA RS = √40 S CA RS = 6,32 cm  6,32 (√40) CA (5) 9.2.3 In ∆PSR (PR)2 = (RS)2 + (PS)2 F A 2 (PR)2 = (√40) + (10)2  SF CA PR = √40 + 100 PR = √140  11,83 (√140) CA PR = 11,83 cm (3) 9.2.4 In ∆PRQ PT PS = QS Proportionality; ST║QR PR  ST A Eweredigheid; ST║QR PT 5 =7 PR 5 PT = 7 × PR 5 PT = 7 × √140  SF CA PT = 8,45 cm  8,45 ( 10√35 ) CA 7 (3) [16] 16 ### Page 17 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 10 10.1 10.1.1 𝑛 = 150 = 2,5𝑟/𝑠  Both/Beide C A 60 2×22,86 𝐷= = 0,4572𝑚 100 𝑣 = 𝜋𝐷𝑛 F A 𝑣 = 𝜋(0,4572)(2,5)  SF CA 𝑣 = 3,59 𝑚/𝑠  3,59 CA (4) 10.1.2 𝜔 = 2𝜋𝑛 F A 𝜔 = 2𝜋(2,5)  SF CA 𝜔 = 15,71 rad/s  15,71 CA (3) 10.2 4ℎ2 − 4𝑑ℎ + 𝑥 2 = 0 F A 4ℎ2 − 4(2 × 22,86)ℎ + (20)2 = 0  SF A 4ℎ2 − 182,88ℎ + 400 = 0 −(−182,88)±√(−182,88)2 −4(4)(400) M ℎ= CA 2(4) ℎ = 43,42 or/of ℎ = 2,30 ∴ ℎ = 2,30 cm  2,30 CA OR/OF OR/OF OM 2 = OR2 − MR2  SF A OM 2 = (22,86)2 − (10)2  20,55 CA OM 2 = 422,5 … OM = 20,55 … h (BM) = 22,86 − 20,55 … M A h (BM) = 2,30cm  2,30 CA (4) 17 ### Page 18 Downloaded from hlayiso.com 11092/24 10.3 ̂ B = 360° − 240° = 120° 10.3.1 AO  120° A (1) ̂ B = 120° × 𝜋 10.3.2 AO M A 180° 2 ̂B = 𝜋 AO 3 ̂ B = 6,58 rad AO  6,58 CA (2) 10.3.3 s = rθ F A 2  SF s = (30) ( π) CA 3 s = 20π s = 62,83 cm  62,83 CA (3) 10.3.4 Areasector/sektor = r . s 2 (30) .(20π)  SF Areasector/sektor = CA 2 Areasector/sektor = 300π = 942,48 cm2  942,48 CA 1 A∆ = (30)(30) sin 120° 2 A∆ = 389,71 cm2  389,71 CA A = 700 − (942,48 − 389,71) M A A = 147,23 cm2  147,23 CA (5) [22] 18 ### Page 19 Downloaded from hlayiso.com 11092/24 QUESTION/VRAAG 11 11.1.1 DO NOT MARK / MOET NIE MERK NIE. 11.1.2 DO NOT MARK / MOET NIE MERK NIE. 11.1.3 DO NOT MARK / MOET NIE MERK NIE. 11.1.4 DO NOT MARK / MOET NIE MERK NIE. 11.2 O1 + On Area = a ( + O2 + O3 + . . . +On−1 ) F A 2 26 + 78 Area = 80 ( + 52 + 143 + 290 + 182)  SF A 2 Area = 80(719) Area = 57 520 mm2  57 520 CA OR/OF OR/OF o +o o +o o +o o +o o +o A = a ( 1 2 2 + 2 2 3 + 3 2 4 + 4 2 5 + 5 2 6) F A 26+52 52+143 143+290 290+182 182+78 A = 80 ( + + + + )  SF A 2 2 2 2 2 A = 80(719) A = 57 520 mm2  57 520 CA (3) [3] TOTAL/TOTAAL: 141 19 ## Related documents - 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