# Technical Mathematics Grade 12 P2 Gauteng Mock 2018 Memo

> Grade 12 Technical Mathematics Paper 2 memorandum from the Gauteng Mock Exam 2018 (Afrikaans), 19 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-nsc-technical-mathematics-p2-afrikaans-prepartory-possible-answer/cff5d010-4cbb-45ab-845a-6c8e4354fb8a
- Updated: 2026-09-27
- Subject: Technical Mathematics
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2018
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 19

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## About this document

Technical Mathematics · Grade 12 · Gauteng Mock Exam · 2018 · Afrikaans. Memorandum, 19 pages. Read online or download the PDF.

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## Document text

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### Page 1

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      VOORBEREIDENDE EKSAMEN
                         2018
                NASIENRIGLYNE

             TEGNIESE WISKUNDE (VRAESTEL 2)

19 bladsye

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                     GAUTENGSE DEPARTEMENT VAN ONDERWYS
                              VOORBEREIDENDE EKSAMEN

                                                    TEGNIESE WISKUNDE
                                                    (Vraestel 2)

                                                    NASIENRIGLYNE

VRAAG 1

1.1   Middelpunt van BC = (             )              x = – 1
                                                       y = – 1
                                                                                           (2)
                     =M(–1 ; – 1)
1.2         =           OF          =

                –                           –
            =                       =                  substitusie in die regte formule

            =                       =
                                                       antwoord
            =                       =                                                      (2)
1.3   y = mx + c

      7 =       +c                                     substitusie (3 ; 7) of (– 3 ; 4)
                                                       substitusie van        =
      c=                                               c =
                                                                                           (3)
        y= x+

                                                2

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        OF
                                                 substitusie (3 ; 7) of (– 3 ; 4)
                                                 substitusie van        =
                                                c =

         y= x+                                                                       (3)
1.4.1

                                                 substitusie

                                                 Verwysingshoek = 68,20
        Verwysingshoek = 68,20

                                                                                     (2)
1.4.2

                                                

                                                                            (2)
1.5     KC2 = (x 2 – x 1 )2 + (y2 – y1 )2       gebruik van afstandsformule

        17 = (–3 – e)2 +(4 – 2e -1)2             substitusie van die regte
                                                  waardes in die regte formule
        17 = 9 + 6e + e2 + 9 – 12e + 4e2
                                                standaardvorm
                  5e2 – 6e + 1 = 0

                (5e – 1)(e – 1) = 0
                                                beide e – waardes
                          or e = 1              beide f – waardes

                f = 2(1) +1 = 3
                                                                                     (5)
                f = 2( ) +1 =
                                                                                  [16]

                                            3

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VRAAG 2
2.1

2.1.1   d=5                                             d = 5
                                                                                   (1)
2.1.2 Radius                                            R
                                                                                   (1)
2.1.3            =                                              =

                 =       Produk van loodregte lyne se           =
                         gradiënte = -1

        y = mx + c

        –4=          –   +c                             substitusie van gradiënt
                                                        substitusie van B(– 3 ; – 4)
             –
        c=                                                                         (4)

        Verlangde vergelyking is: y =       –

        3x + 4y + 25 = 0
                                                        OF
        OF

                                                4

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        Gebruik die formule: x.x 1 + y.y1 = r2          regte formule

        x 1 = – 3 en y1 = – 4

        (–3)x + (–4)y = 25                              substitusie van B(– 3 ; – 4)
        –3x – 4y = 25                                   substitusie van    = 25
        3x + 4y + 25 = 0.                               vereenvoudiging           (4)
2.1.4   CA OA Raaklyn             Radius

        xc = 5                                          identifiseer dat x = 5 by C

        Substitusie van x = 5 in 3x + 4y + 25 = 0

        m.a.w.: 3(5) + 4y + 25 = 0                      substitusie van x c = 5 in
                                                         BC
                 4y + 40 = 0

                 y = –10
                                                        y = –10
        C(5 ; – 10)
                                                                                      (3)
2.2                =1                                   vergelyking in
                                                         standaardvorm
        x-afsnitte: (– 3 ; 0) ; (3 ; 0)                  x –afsnitte
        y-afsnitte: (0 ; – 2) ; ( 0 ; 2)                 y-afsnitte
                                                                                      (3)
                                                y

                                           2

                                                                                            x
                                               0
        -3                                                                3

                                           -2

                                                                                  [12]

                                                    5

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VRAAG 3
 3.1.1 √ tan            =–1

                    –                                     –
        tan     =                           tan     =
                    √                                     √

                                            diagram met
                                            terminale been in die
                                            regte kwadrant

        Gebruik Pythagoras: r = 2           r = 2

        sin    =                            sin     =            (4)

3.1.2                       √               substitusie
                        (       )

                                            antwoord
        cos2 =

        OF                                  OF

        cos2       = 1 – sin2               identiteit
                   = 1 – ( )2
                                            antwoord
                =                                                 (2)

3.1.3   Gebruik sin             =

                                             Verwysingshoek
        Verwysingshoek = 30o
                                              = 30o
               = 150o
                                                  = 150o
                                                                  (2)
        OF
                                    –
                                            OF
        Gebruik tan             =
                                    √

        Verwysingshoek = 30o                 Verwysingshoek
                                              = 30o
               = 150o
                                                  = 150o
                                                                  (2)

                                        6

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 3.2       cot (x – y) = cot (            –   )           substitusie

                                                          
                               = cot( )

                               =
                                                          antwoord
                               = 0,27                                        (3)
                                                                                   [11]
VRAAG 4
 4.1

                                                          sin x
       =               –                                  cos x
                                                          – cos x
       =                                                  
              –

                                                          Teller
       =                                                  Noemer

       =1                                                                           (6)
4.2    RK      =

                   =
                           –                              
                                                                 –
                                                  –       
                   =

                                                          
                   =                                                 –
                                                  –

                                                          
                   =       ( –     )–                                   –

                  =        –

                  = LK

           LK = RK

       : –
                                                                                    (5)

                                                      7

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4.3   2 cos 2x = √

      cos 2x = √                             cos 2x = √

      Verwysingshoek: 2x = 30o               2x = 30o

      Kwadrant I:            Kwadrant IV:

                                              x = 15o
      x = 15o                                                       (4)
                                                                           [15]
VRAAG 5

5.1   Ko-Binne hoeke; QV RW                  R
                                                                            (1)
5.2       ̂ = 90 – 30
                o        o

             = 60o                           S

      In QVR:

      ̂                          Binne

                                             S/R                          (2)
5.3   In PQR:

                                             

                                                        √
                     √

                                         8

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      In QRV:
                                              gebruik van die sin-reël
                                              korrekte substitusie in die
                                               sin-reël

                 √

                                                                              (4)
                   √
5.4   Oppervlakte van       QRV =

                     =          √     √              √           √

                     = 739,382 m2             739,382
                                              
      Oppervlakte van       RVW =                             √
                                                          √
                                              VW =
                                √                    √
                        =             √                      √

                        = 270,63 m2           270,63

       Oppervlakte van QVWR = 1010,01 m2      1010,01                        (7)

                                                                                    [14]

                                          9

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VRAAG 6
6.1 Sien die onderstaande skets:
    Vir f(x)
    x –afsnitte 0o ; 180o ; 360o
    vorm
    assimptote by 90o & 270o

     Vir g(x)
     y –afsnit (0 ; – 2)
     Draaipunte by (90o ; – 1) en (270o ; – 3)
     vorm
                                                                                              (6)

            f

                                                  f

                                                                                  f

                                                      g

6.2 1                                                     1                                 (1)
6.3 180o                                                  180o
                                                                                              (1)
       o           o                                              o         o
6.4 90 < x < 270                                           90 < x < 270                     (2)
                                                                                             [10]

                                             10

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VRAAG 7

7.1   ….. is loodreg op die koord.                                  loodreg op die
                                                                       koord          (2)
7.2   In die onderstaande diagram is O die middelpunt van die sirkel waarvan die radius
      10 cm is. Die lengte van die koord BC is 16 cm en OD BC. M is die middelpunt
      van koord CE.

7.2.1 BD = 8 cm               Loodlyn uit midpt.    na koord       SR               (2)
                       2        2      2
7.2.2 In BOD:       OB = OD + BD             Pythagoras            S/R

                    (10)2 = OD2 + (8)2                             substitusie

                    OD = 6 cm                                      OD = 6 cm          (3)
7.2.3 OM    EC                Midpt. ʘ ; Midpt. koord              SR

      m.a.w.:    ̂ C = 90o

       ̂ C = 90o              Loodlyn uit midpt.    na koord

       ̂C +       ̂C = 180o                                        S

                                                                   R
       MODC is ʼn koordevierhoek            teenoorst. e van
                                                                                       (4)
                                            vierhoek is suppl.
                                                                                     [11]

                                              11

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VRAAG 8

8.1   Raaklyn            Radius                                    R
                                                                                (1)
8.2       ̂ S = 40   o
                                                                   S
                                                                                (1)
8.3   ̂ = 90o                                         in ½       SR
                                                                                (2)
8.4   In TWR:

      50o +         ̂R +        ̂ R = 180o   Binne                 S/R

           ̂R = 90o                          e op reguit lyn       S/R

      50o + 90o +             ̂ R = 180o

          ̂ R = 40o                                                             (2)
8.5   ̂ =       ̂ S = 40o                    e in dies. ʘ segm.   SR
                                                                                (2)
8.6   ̂         ̂          ̂                 Buite                 R

           = 90o + ̂                                               S

                                                                   S
      Maar: ̂             ̂                  e in dies. ʘ segm.

            ̂        ̂
                                                                   S
                = 90o + ̂                                                       (4)
                                                                              [12]

                                                     12

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VRAAG 9
 9.1
                                          P

                                                   ( x + 2) cm
                            x cm

                      M                                          S

                                                                     3 cm
               2 cm

           Q                                                                   R

     In        : MS // QR

                               lyn || een sy van                        S/R

                                                                        substitusie
       =

     3x = 2x + 4                                                        
                                                                        vereenvoudiging

                                                                        antwoord
                                                                                        (4)

                                         13

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 9.2                                                A
                                           1
                                       B
                                   3                       F
                               C

                               1
                         D
                                                               G

                  E                                                      M

9.2.1a        =                    lyn || een sy van               S

                                                                   antwoord
                                                                                (2)
              = 4:1
9.2.1b                             lyn || een sy van               S

                                                                   antwoord
              = 3: 1                                                           (2)

9.2.2    In    AEM: DG // EM

                                       lyn || een sy van           S

         DE =                                                      antwoord
                                                                                (2)
                                                                               [10]

                                               14

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VRAAG 10

                                                     10         L

                                      K

                                                                    15
                                                         20
                            30

                                                                         M
                                                    40
                 N

10.1   In            en           :                                      

                                                                         

                                                                         

                      |||             Sye van                                             (3)
                                      eweredig
10.2   Omdat                |||               van 10.1 hierbo

        ̂            ̂                                                   S

            LK    MN                      Verwissellende                 R
                                                                                          (2)
10.3                                          |||                        S

                                                                         Substitusie
                                                                         MO = 20 – KO

       10(20 – KO) = 40 (KO)

       200 – 10.KO = 40.KO

       200 = 50 KO

            KO = 4 eenhede                                               antwoord
                                                                         (4)
                                                                                          [9]

                                                          15

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VRAAG 11

                                                     C

                                                         6 cm

                                  A        15 cm                15 cm   B
                                                     D
                                                          O

                                                         E
11.1.1   4h2 – 4dh + x 2 = 0

         4(6)2 – 4d(6) + (30)2 = 0                                          substitusie in die
                                                                             regte formule
         144 – 24d + 900 = 0

         24d = 1044

         d = 43,5cm                                                         d = 43,5

         Maar: d = DOE + 6

         DOE = 37,5 cm                                                      DOE = 37,5 cm
                                                                                                (3)
11.1.2   In   AOB:

         AB2 = AO2 + OB2 – 2(AO)(BO)cos        ̂B                           substitusie in die
                                                                             cos-reël
         (30)2 = (21,75)2 + (21,75)2 – 2(21,75)(21,75). cos      ̂B

              ̂      0,04875…..                                             vereenvoudiging
                                                                            antwoord       (3)
            ̂ B = 87,21o
11.1.3   Middelpuntshoek = 87,21o = 1,52 radiale                            87,21o = 1,52
                                                                                      radiale
         Booglengte = s = r
                                                                             substitusie in die
           s = (21,75).(1,52)                                                regte formule
         s = 33,06 cm
                                                                            antwoord           (3)

                                                16

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11.2.1   Hoeksnelheid =              2

                                 =                                  substitusie in die
                                                                    regte formule
                                 =       radiale/sekonde
                                                                   antwoord            (2)
11.2.2   Deursnee = 0,24 m                                         d = 0,24 m

         Omtreksnelheid = v =

                                         =                          substitusie in die
                                                                    regte formule
                                         = 5,04 m/s
                                                                   antwoord            (3)
                                                                                          [14]
QUESTION 12
12.1.1 l2 = r2 + h2                  Pythagoras                 S

         l=√                                                    antwoord
                                                                                           (2)
12.1.2        =            ………..(1)

         Substitusie van l = √                  in (1) hierbo

              √            ) =                                  substitusie van
                                                                 l=√
         √            =
                                                                √            =
         r2 + h 2 = 4h 2
          2       2
                                                                 r2 + h 2 = 4h 2
         r = 3h                                                                         (3)
12.1.3   Volume van ʼn kegel =                                   substitusie van r2 = 3h 2

                                         =

                                                                antwoord
         Volume van ʼn kegel
                                                                                           (2)

                                                      17

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12.1.4a   Volume silinder – Volume kegel = 54

          π      –             = 54                                          opstel van die
                                                                              vergelyking
                     )h –         = 54

          2h3 = 54

          h3 = 27                                                           vereenvoudiging
          h= 3
                                                                                                   (2)
          2           2
12.1.4b   r = 3(3)                                                          substitusie

          r= √                                                              antwoord
                                                                                                   (2)
 12.2
           y1         y2         y3   y4   y5   y6   y7   y8   y9                     y10   y11
           430        793        1167 1475 1409 1431 1365 1205 870                    529   347

                          y1     y2   y3    y4    y5    y6   y7   y8   y9       y10   y11

12.2.1    AT = a[                + y2 + y3 + y4 + y5 +….+y10 ]               substitusie in die
                                                                             regte formule
             = 110[388,5 +
          793+1167+1475+1409+1431+1365+1205+870+529]                        

              = 110[388,5 + 10244]
                                                                            antwoord
              = 116 9575 km2

          OF                                                                OF

                                                       18

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         AT = a[m1 + m2 + m3 +…..+ m10 ], waar
         a = 110 ; en

         m1 =                     ;
         m2 =         = 980;                                  3-5 mid-ordinate reg
         m3 =             = 1321;
                                                              OF
         m4 =                                                 al 10 mid-ordinate reg
         m5 =             = 1420
         m6 =            = 1398
         m7 =            = 1285
         m8 =         = 1037,5
         m9 =         = 699,5
         m10 =           = 438

         AT = 110[10 632,5] = 1 169 575 km 2

                                                              antwoord
                                                                                      (3)
12.2.2   Oppervlakte van die ingekleurde gebied =

         (1475             ) – 1 169 575                      (1475            )

         = 452 925 km2                                        antwoord
                                                                                     (2)
                                                                                    [16]
                                                    TOTAAL:                         150

                                           19

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