# Mathematics Grade 10 P2 Gauteng June 2018 Memo (Afrikaans)

> Grade 10 Mathematics Paper 2 memorandum from the Gauteng June 2018 (Afrikaans), 6 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr10-math-p2-afrikaans-june-2018-possible-answer/0fd19b16-8994-4859-b474-e323fc17e7d3
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Language: Afrikaans
- Document type: Memorandum
- Year: 2018
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 10 · Gauteng June · 2018 · Afrikaans. Memorandum, 6 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Matching paper and memo

- [Gr10 Math P2 (Afrikaans) June 2018 Question Paper](https://www.hlayiso.com/library/gr10-math-p2-afrikaans-june-2018-question-paper/8eca90bd-adf7-477d-8f55-fbd705a35b52) — QP · 2018 · Afr

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

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            GAUTENGSE DEPARTEMENT VAN ONDERWYS
                    PROVINSIALE EKSAMEN
                        JUNIE 2018
                         GRAAD 10

                         WISKUNDE
                        VRAESTEL 2

                       NASIENRIGLYNE

6 bladsye

### Page 2

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                                                     WISKUNDE
                                    MEMORANDUM (Vraestel 2)                      GRAAD 10

                        GAUTENGSE DEPARTEMENT VAN ONDERWYS
                                    PROVINSIALE EKSAMEN

                                                         WISKUNDE
                                                         (Vraestel 2)

                                                         NASIENRIGLYNE

VRAAG 1

1.1   1.1.1

                                                                        Pythagoras

              √         √

                                                                        

                                                                                            (2)

      1.1.2                                                             

                                                                        

                                                                                            (2)

      1.1.3   LK: sin  . cot  . sec           RK: 1

                     12 5 13                                            vervang korrekte
                                                                     waardes
                     13 12 5
                    1
                                                                        1

                                                                                            (2)

                                                                                            [6]

                                             2

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                                                WISKUNDE
                                   MEMORANDUM
                                                (Vraestel 2)           GRAAD 10

VRAAG 2

2.1                4sin120
              tan 200  cos 70

              =157,82                                          antwoord

                                                                               (2)

2.2   2.2.1

                                                               vereenvoudig

                                                               

                                                                               (2)

      2.2.2     (         )

                                                               
                                                               

                                                                               (2)

                                                                               [6]

                                        3

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                                                            WISKUNDE
                                             MEMORANDUM
                                                            (Vraestel 2)                    GRAAD 10

VRAAG 3

3.1                           √                                                1
                         2
                3             1                                                √
                         √                                                
                                   
                 2            2
                                                                               
                                                                                   √

                                                                               

                                                                                                    (4)

3.2                   cos 50 . cos 30 . sec50                                3
      x . tan 60 
                                tan 45
                                                                                        3
                                                                               
                                        3                                              2
                (cos 50 . sec50 ).
      x. 3                            2
                              1                                                 cos50 . sec50  1
                3
      x  1.       3                                                          1
               2
           1                                                                       1
      x                                                                       
           2                                                                       2

                                                                                                    (5)

3.3   3.3.1
                                             𝑦                                 f ( x)  2 tan x

                                                                               vorm

                                                                               asimptote
                                                   2                           (           )

                                                   0                       𝑥   g ( x)  cos x  1

                                                                               vorm
                                                   -2
                                                                                -afsnit

                                                                                -afsnit

                                                                                                    (6)

                                                        4

### Page 5

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                                                          WISKUNDE
                                      MEMORANDUM
                                                          (Vraestel 2)            GRAAD 10

       3.3.2    1                                                        1

                                                                                             (1)

       3.3.3    180                                                     1

                                                                                             (1)

       3.3.4    x  90 ; 90                                       kritieke waardes

                                                                         ongelykheid

                                                                                             (2)

                                                                                          [19]

VRAAG 4

4.1                                     AB//DC Ko-binnehoeke             B
          =     (          )                                             R
          =
                                                                           +       =
          +   =
                                                                         R
       AD//BC                           Ko-binnehoeke aanvullend
                                                                                             (4)

4.2   ABCD is ŉ parallelogram.          Oorstaande sye parallel          parallelogram

                                        Oorstaande hoeke gelyk           korrekte rede

                                                                                             (2)

                                                                                             [6]

                                              5

### Page 6

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                                                  WISKUNDE
                               MEMORANDUM
                                                  (Vraestel 2)                  GRAAD 10

VRAAG 5

5.1   Cˆ1  Dˆ1                  Verwisselende hoeke BD // CR    R

      Cˆ  Cˆ
        1      2
                                                   ˆ
                                 Gegee CR halveer DCE             Cˆ1  Cˆ 2
      Cˆ 2  Bˆ 2                Ooreenkomstige hoeke BD // CR
                                                                 B/R
      Dˆ  Bˆ
        1      2
                                 Sye teenoor gelyke hoeke
      BC  CD                                                    B

                                                                 R
                                                                                     (5)

5.2   CD = BC                    Reeds bewys

      CD = AB                    Oorstaande sye parallelogram    B/R

      AD = BC                    Oorstaande sye parallelogram    B/R

      ABCD is ŉ rombus           Al 4 sye gelyk                  R

                                                                                     (3)

5.3   BO = OD = 12               Diagonale halveer mekaar        B/R

      Oˆ  90                   Diagonale halveer reghoekig     B/R

                                 Pythagoras
      AB 2  AO 2  BO 2
      (13) 2  AO 2  (12) 2                                     korrekte vervang
      AO 2  169  144
      AO 2  25
                                                                 AO = 5
      AO  5

      AO  OC
                                 Diagonale halveer mekaar        AO = OC
      AC  10cm
                                                                 AC = 10

                                                                                     (5)

                                                                                    [13]

                                                                    TOTAAL:          50

                                       6

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