# Mathematics Grade 10 Paper 2 Gauteng June 2019 Memo

> Grade 10 Mathematics Paper 2 memorandum from the Gauteng June 2019, 6 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/gr10-math-p2-afr-eng-june-2019-possible-answers/ce7f376e-26a0-4d19-a715-9814e814452a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Memorandum
- Year: 2019
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 10 · Gauteng June · 2019. Memorandum, 6 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                GAUTENG DEPARTMENT OF EDUCATION /
               GAUTENGSE DEPARTEMENT VAN ONDERWYS

      PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN
                         JUNE / JUNIE 2019
                           GRADE / GRAAD 10

                        MATHEMATICS / WISKUNDE

                          PAPER / VRAESTEL 2

                            MEMORANDUM

6 pages / bladsye

### Page 2

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                                                            MATHEMATICS / WISKUNDE
                                              MEMORANDUM
                                                                    GRADE 10 / GRAAD 10

                             GAUTENG DEPARTMENT OF EDUCATION /
                            GAUTENGSE DEPARTEMENT VAN ONDERWYS
                                        PREPARATORY EXAMINATION /
                                         VOORBEREIDENDE EKSAMEN

                                                           MATHEMATICS / WISKUNDE
                                                           (Paper / Vraestel 2)

 QUESTION 1 /
 VRAAG 1
1.1.1 cos 
            b                                                     b
                                                             
            c                                                     c
                                                                                    (1)
1.1.2    sin (90   )
             b                                                      b
                                                              
             c                                                      c
                                                                                    (1)
1.2.1    r 5
         1 2
           sin 
         4
                                                                  4
           1  4 
                      2                                       
                                                               5
           4 5 
            4                                                     4
                                                             
           25                                                     25
                                                                                    (2)
1.2.2   LHS//LK
              LK           RHS//RK
                                 RK                             5
                                                                        2

                                                                 1
        sec  1
             2
                            tan 
                               2
                                                                3
                            4 
                 2                  2
                                                                 4 
                                                                            2
          5
           1                                            
          3               3                                  3 
          16                 16                                 16
                                                            
           9                  9                                  9
                                                                                    (3)
                                                                                    [7]

                                                    2

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                                                             MATHEMATICS / WISKUNDE
                                             MEMORANDUM
                                                                     GRADE 10 / GRAAD 10

 QUESTION 2 /
 VRAAG 2
2.1  2,34                                                      answer / antwoord
                                                                                     (1)
2.2   cos (2 10)
                     0, 091
            2
      cos (2 10)  0,184
                                                                 79, 4
            2 10  79, 4                                     44, 7
                     44, 7                                                        (2)
      ONLY PLACE IN PAPER WHERE LEARNER IS
      PENALIZED FOR INCORRECT ROUNDING OFF /
      ENIGSTE PLEK IN VRAESTEL WAAR LEERDER
      GEPENALISEER WORD VIR VERKEERDE
      AFRONDING

2.3                                   1
      2sin 2 60  sin 45 . sec 45  tan10 . cot10
                                      4                             3
               2
                                                               
            3  2  2  1                                       2
       2        .    (1)                             2  2 
            2   2   2 4                                  
                                                                  2  .  2 
           3       1                                               
       2   1                                              1
           4       4
                                                                 3
        3                                                      
                                                                4
        4
                                                                                     (4)
                                                                                     [7]

                                                         3

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                                              MATHEMATICS / WISKUNDE
                                 MEMORANDUM
                                                      GRADE 10 / GRAAD 10

 QUESTION 3 /
 VRAAG 3
3.1

                                              shape     ) / vorm
                                              turning points     )/
                                               draaipunte

                                              shape      ) / vorm
                                              turning points     )/
                                               draaipunte

                                                                                (4)
3.2.1   y [2 ;  2]                          answer / antwoord

                                              OR/OF
        OR / OF

        2  y  2                            answer / antwoord
                                                                                 (1)
3.2.2   90 x  270                         correct critical values / korrekte
                                              kritieke waardes
                                              correct inequality / korrekte
                                              ongelykheid
                                                                                 (2)
                                                                                 [7]

                                     4

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                                                                  MATHEMATICS / WISKUNDE
                                                     MEMORANDUM
                                                                          GRADE 10 / GRAAD 10

QUESTION 4 /
VRAAG 4
4.1 Midpoint RS/ Middelpunt RS = (2;1)                            2
                                                                  1
                                                                                                   (2)
4.2   2  x             1 y                                       2  x              1 y
             2               1                                           2 and / en      1
        2                 2                                            2                 2
           x6              y 1                                   x6
                                                                   y 1
                                                                                                   (3)
      T (6;1)
4.3   RU  (2  2)2  (5 1) 2                                     (2  2)2  (5 1)2
           32                                                     32
                                                                                                   (2)
4.4              1 5                   y 3
      M RU                    M SP         1
                2  2                  x2                        M RU = 1
           1                           y  x 5                   y  x 5

      SP  ( x  2) 2  ( y  3) 2  2 32                          ( x  2)2  ( y  3)2
        x 2  4 x  4  y 2  6 y  9  2 32                       128
                                                                  substitute / vervanging
      x 2  4 x  13  ( x  5) 2  6( x  5) 128
                            2 x 2  8 x  90  0
                             x 2  4 x  45  0                    x 2  4 x  45  0
                            ( x  9) ( x  5)  0
                                                                   x 9 ; x   5
            x 9         or / of     x5
           y 95                     y  55
          y4                         y   10                     y  4 ; y   10
      P(9; 4)                        P( 5;  10)                                                  (8)

                                                                                                  [15]

                                                         5

### Page 6

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                                                                   MATHEMATICS / WISKUNDE
                                                MEMORANDUM
                                                                           GRADE 10 / GRAAD 10

QUESSTION 5 /
VRAAG 5
5.1 parallelogram                                                  answer / antwoord
                                                                                               (1)
5.2.1 E1 = x                           <’s opp equal sides /       S/R
                                       <’e teenoor gelyke sye
      E3 = 2x
                                       <’s opp equal sides /       S
      x + 90° + 2x = 180°              <’e teenoor gelyke sye
                                       <’s straight line /         S
                     x = 30°
                                       <’e op reguityn             S
                                                                                               (4)
5.2.2 A = 120°                         sum <’s  / som vd <’e     S

      C1 = 60°                         sum <’s  / som vd <’e 
                                       alt <’s AD||BC/
      C2 = 60°                         verwis <’e AD||BC           R
                                                                   S
      A=C                             opp <’s equal / oorst <’e
      ABCD is a parm                   gelyk                       R
                                                                                               (4)
                                                                                               [9]

QUESTION 6
VRAAG 6
    AO = OC  2 xy                     given / gegee                AO = OC  2xy
      BO = OD  x 2  y 2              given / gegee                  BO = OD  x2  y 2
      BO 2 + OC 2
       ( x 2  y 2 ) 2  (2 xy ) 2                                 ( x2  y 2 )2  (2 xy)2
       x4  2x2 y 2  y 4  4x2 y 2
       x4  2x2 y 2  y 4
       ( x 2  y 2 )2
                                                                    ( x 2  y 2 )2

      BC2
       ( x 2  y 2 )2
      BC2  BO 2 + OC 2
                                       pythagoras                  S/R
      BOC =90 
             90°
                                       diagonals bisect at 90 /   R
      ABCD is a rhombus / is ŉ         diagonale halveer 90                                   (5)
      ruit
                                                                                               [5]
                                         TOTAL / TOTAAL: 50

                                                             6

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