# Mathematics Grade 11 Paper 2 Gauteng June 2019 Memo

> Grade 11 Mathematics Paper 2 memorandum from the Gauteng June Exam 2019, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr11-math-p2-afr-eng-june-2019-possible-answers/641902e1-464c-4127-a4d6-a5395f3d42d8
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2019
- Exam period: Gauteng June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 Paper 2 Gauteng June 2019 Memo](https://www.hlayiso.com/library/gr11-math-p2-afr-eng-june-2019-possible-answers/641902e1-464c-4127-a4d6-a5395f3d42d8)

## About this document

Mathematics · Grade 11 · Gauteng June Exam · 2019. Memorandum, 10 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

                                             testpapers.co.za

                  GAUTENG DEPARTMENT OF EDUCATION /
                 GAUTENGSE DEPARTMENT VAN ONDERWYS
                        PROVINCIAL EXAMINATION /
                          PROVINSIALE EKSAMEN
                               JUNE 2019 /
                                JUNIE 2019
                               GRADE 11 /
                                GRAAD 11

                        MATHEMATICS / WISKUNDE

                           PAPER 2 / VRAESTEL 2

                MARKING GUIDLINE / NASIENRIGLYNE

10 Pages / 10 bladsye

                                    1

### Page 2

Downloaded from hlayiso.com
                                                                         MATHEMATICS / WISKUNDE
                                                  MEMORANDUM                 Paper 2 / Vraestel 2
                                                                              Grade / Graad 11

                            GAUTENG DEPARTMENT OF EDUCATION /
                           GAUTENGSE DEPARTEMENT VAN ONDERWYS
                                    PROVINCIAL EXAMINATION /
                                      PROVINSIALE EKSAMEN

                                                     MATHEMATICS Paper 2 / WISKUNDE Vraestel 2

                                                     MARKING GUIDLINE / NASIENRIGLYNE

INSTRUCTIONS AND INFORMATION
       Consistent accuracy applies as a general rule.
       If a candidate answers a question twice and does not delete either, mark the FIRST attempt.
       If a candidate answers a question, crosses it out and does not re-do it, mark the deleted attempt.

        Marks are awarded as per the guideline, and the following symbols are used:
A         –     Accuracy
C.A       –    Continued Accuracy
S         –    Statement
R         –    Reason
S. R      –    Statement and Reason

INSTRUKSIES EN INLIGTING
       Deurlopende akkuraatheid geld as ŉ algemene reël.
       Indien ŉ kandidaat ŉ vraag twee keer beantwoord en nie een van die antwoorde kanselleer nie,
        merk die EERSTE poging.
       Indien ŉ kandidaat ŉ vraag beantwoord en dit doodtrek maar nie oordoen nie, merk die
        doodgetrekte poging.

        Punte word toegeken volgens die riglyn en die volgende simbole word gebruik:
A        – Akkuraatheid
D.A.     – Deurlopende Akkuraatheid
S        – Stelling
R        – Rede
S. R     – Stelling en Rede

                                                      2

### Page 3

Downloaded from hlayiso.com
                                                            MATHEMATICS / WISKUNDE
                                               MEMORANDUM       Paper 2 / Vraestel 2
                                                                 Grade / Graad 11

        QUESTION 1 / VRAAG 1
1.1
                                                                         correct formula /
1.1.1            =√                      –                                korrekte form.
                                                                         correct subst. /
                  (2  (2))2  (1  (2))2
                                                                          korrekte verv.
                  25                                                    answer /
                 5                                                       antwoord
                                                                                           (3)
1.1.2     x1  x2   y y
                  ; 1 2
             2        2                                                 correct formula /
                                                                         korrekte form.
        2=                    ; 0=                                      correct subst. /
         4=1+                 ; 0=2+                                     korrekte verv.
         3=                   ;   –2 =

        D (3 ; –2)                                                      answer /
                                                                         antwoord

                                                                                             (3)
1.1.3   mAC  mCE

          –                       –                                      equate m. /
                    =                                                     gelykstelling m
          –                   –
         – –                  –
                      =                                                  correct subst. /
         – –                      –
                                                                          korrekte verv.
        2     1
          
        4 p2                                                             simplification /
        2p 4  4                                                          vereenvoudiging
        2p  8                                                             (2p – 4 = 4)
        p4
                                                                        
                                                                                             (4)

1.1.4   quadrilateral ABDE is a kite when:           .                  
        vierhoek ABDE is ŉ vlieër wanneer:
             –                    –
                      .
           –                      –
         – –                  –                                          correct subst. /
                      .                                                   korrekte verv.
         – –                  –
          . (             )                                              –1
        –1
        since the product of the gradients = –1 /
        produk van van gradiënte = –1                                    explanation /
           AC BC                                                          verduideliking
           ABDE is a kite (diagonals intersect 90 /
           ABDE is ŉ vlieër (hoeklyne deursny 90                                             (4)
                                                 3

### Page 4

Downloaded from hlayiso.com
                                                               MATHEMATICS / WISKUNDE
                                          MEMORANDUM               Paper 2 / Vraestel 2
                                                                    Grade / Graad 11

1.2
1.2.1     P(0 ; 4)                                                              answer / antwoord
                                                                                                (1)
1.2.2     R(4 ; 0)                                                              answer /
                                                                                 antwoord
                                                                                                (1)
1.2.3          = tan 45
                  =1                                                                   =1
          y = mx + c
          4 = 1(0) + c
          4=c
            g(x) = x + 4                                                        answer /
                                                                                 antwoord
                                                                                                (2)
1.2.4.1          = –1                                                                  = –1
              = tan (–1)
              = 180 – 45                                                         answer /
              = 135                                                               antwoord
                                                                                                (2)
1.2.4.2   PR̂Q  45        (supplementary angles) (suppl. hoeke)                S.R
               = 180 – 90   (sum of angles of a / (som van hoeke in driehoek)
                                                                                 answer /
               = 90
                                                                                  antwoord
                                                                                                (2)
                                                                                               [22]

                                              4

### Page 5

Downloaded from hlayiso.com
                                                                     MATHEMATICS / WISKUNDE
                                                  MEMORANDUM             Paper 2 / Vraestel 2
                                                                          Grade / Graad 11

        QUESTION 2 / VRAAG 2
2.1
2.1.1                            y

                                                                                correct sketch /
                                                                                 korrekte skets
                       2    5

            x                θ

        x2  y 2  r 2
                                                                                application Pyth. /
        x 2  (5)2  (2)2
                                                                                 Pythagoras
        x 2  16
         x  4             ……..restriction / beperking (90;270)              choose / kies
         x  4                                                                  x  4

                 x 4                                                           answer /
        cos      
                 r   5                                                           antwoord
                                                                                                    (4)
2.2
2.2.1   cos2 45  sin330.tan 2 210
         cos2 45  sin(360  30).tan 2 (180  30)                         correct expansion /
                                                                                 korrekte
         cos2 45  sin30.tan 2 30                                            uitbreiding.
                   2
            2 1  1 
                                     2
                                                                                correct reduction /
              .                                                          korrekte reduksie
            2  2  3                                                         substitution /
           1
                                                                                     vervanging
           3                                                                    answer / antwoord
                                                                                                  (6)
2.2.2

                                                                                  cos x
          cos x.( tan x)( cos x)                                                tan x
        
                 1.( sin x)                                                      sin x
                 sin x                                                          sin x
          cos x                                                                
                         (cos x)                                                 cos x
                cos x 
                   sin x
                                                                                simplification /
          sin x.cos x
                                                                                vereenvoudiging
             sin x                                                             answer / antwoord
          cos x                                                                                 (6)
                                                     5

### Page 6

Downloaded from hlayiso.com
                                          MATHEMATICS / WISKUNDE
                             MEMORANDUM       Paper 2 / Vraestel 2
                                               Grade / Graad 11

2.3
      √ –          –     –

                                                     –tan
      √ – –
                                                      tan
      √                                               identity / identiteit
                                                     simplification /
      √                                               vereenvoudiging
                                                     answer / antwoord
      sec                                                                (5)

2.4   x  0                                         answer / antwoord
      x  90                                        answer / antwoord
                                                     answer / antwoord
      x  180                                                       (3)
                                                                    [24]

                               6

### Page 7

Downloaded from hlayiso.com
                                                             MATHEMATICS / WISKUNDE
                                      MEMORANDUM                 Paper 2 / Vraestel 2
                                                                  Grade / Graad 11

        QUESTION 3 / VRAAG 3
3.1
3.1.1                                                                     answer / antwoord
                                                                          answer / antwoord
                                                                          answer / antwoord
                                                                                          (3)
3.2
3.2.1                                                                    answer / antwoord
        x = 325                                                          answer / antwoord
                                                                                             (2)
3.2.2   90    x   325                                                    corr. critical value /
                                                                          korr. kritieke
                                                                           waarde
                                                                         corr. inequality
                                                                          korr. ongelykhede
                                                                                           (2)
3.3      4                                                                asymptotes /
                                                                           asimptote
         3
                                                                          shape / vorm
         2                                                               x-intercepts /
                                                                          x-afsnitte
         1                                                               y-intercept /
                                                                          y-afsnit
         0
                        45°   90°   135°       180°   225°        270°
         -1

         -2

         -3

         -4

                                                                                            (4)
                                                                                           [11]

                                           7

### Page 8

Downloaded from hlayiso.com
                                                                     MATHEMATICS / WISKUNDE
                                              MEMORANDUM                 Paper 2 / Vraestel 2
                                                                          Grade / Graad 11

      QUESTION 4 / VRAAG 4
4.1   ... twice angle at circumference / twee keer die hoek op die omtrek.       answer /
                                                                                  antwoord
                                                                                               (1)

4.2   Construction / konstruksie: Join CO and produce it./ lyn CO verleng        construction /
                                                                                  konstruksie
      Proof / bewys: In ACO:
      Oˆ 2  Aˆ  Cˆ 2 (exterior angle of a triangle is
                                                                                  SR
      equal to sum of int. opp. angles / buite hoek van )
      but / maar Aˆ  Cˆ 2 (OA  OC )       (radii / radius)
       Ô2 = Ĉ2+ Ĉ2
      Ô2 = 2 Ĉ2
      Similarly we can prove that / kan ook bewys word dat:        Ô1= 2Ĉ1      S
      Ô2+ Ô1= 2 Ĉ2+ 2 Ĉ1                                                         equating /
       AÔB = 2 AĈB                                                               gelykstelling
                                                                                 substitution /
                                                                                  vervanging
                                                                                                (6)
                                                                                                [7]

                                                  8

### Page 9

Downloaded from hlayiso.com
                                                                       MATHEMATICS / WISKUNDE
                                                 MEMORANDUM                Paper 2 / Vraestel 2
                                                                            Grade / Graad 11

        QUESTION 5 / VRAAG 5
                                                                                           answer /
 5.1    ... supplementary / suppl.                                                          antwoord
                                                                                                            (1)
 5.2
5.2.1   Ŝ  Q̂  x (subt. by same arc / omtrekshoeke onderspan deur dieselfde boog)       S.R
                                                                                           S.R
        QR̂S  Q̂  x          (alternate angles/ verwisselende binnehoeke: PQ || RS)
                                                                                           S.R
            Ŝ  QP̂S  x      (alternate angles: verwisselende binnehoeke PQ || RS)
                                                                                                            (6)
5.2.2   PT̂R  x  x      (ext. angle of    sum of int. opp. angles / buitehoek van     )   S.R
                2x                                                                        answer /
                                                                                            antwoord
                                                                                                            (3)
5.2.3   PT̂R  2 x (ext. angle of       sum of int. opp. angles / buitehoek van )
                                                                                           SR
        PÔR  2 x (angle at centre th. / middelpunt stelling)                             SR
        Line PR subtends equal angles at PT̂R and PÔR                                    conclusion /
        gelyke omtrekshoeke onderspan deur dieselfde koord                                 gevolgtrekking
          quadrilateral PTOR is concyclic /                                                             (3)
          PTOR is ŉ koordevierhoek
                                                                                                        [13]
        QUESTION 6 / VRAAG 6
6.1
6.1.                                                                                      SR
             =              (midpoint Th. / middelpunt stelling)

                                                                                          substitution /
             =                                                                             vervanging

           =
                                                                                          conclusion /
          AE = EC                                                                          gevolgtrekking
                                                                                                         (4)
6.2     Eˆ 1  90    (line from centre of a circle bisecting the chord is to chord)      S
                                                                                           R
                      (lyn van middelpunt van koord na middelpunt van sirkel loodreg)
                                                                                                         (3)
6.3     OE 2  102  82     (Theorem of Pythagoras / Pythagoras Stelling)                 S R
         OE  6cm                                                                         OE = 6cm
         ED  4cm                                                                         answer /
                                                                                           antwoord
                                                                                                         (4)
                                                                                                        [11]

                                                     9

### Page 10

Downloaded from hlayiso.com
                                                                      MATHEMATICS / WISKUNDE
                                               MEMORANDUM                 Paper 2 / Vraestel 2
                                                                           Grade / Graad 11

        QUESTION 7 / VRAAG 7

7.1.    Let:      = x
              = = x (tan / chord / raaklyn / koord)                                     S R
             =    = x (alternate angles / verwisellende hoeke, DR║ PS)                  S R
         but / maar     =      (given / gegee)
             = = x (v.o.a / teenoorstande hoeke)                                        S R
        hence / dus    =     =x
           TS is a tangent (angle between tang. and chord is to angle in alter. seg.)   R
           TS is ŉ raaklyn(tussen lyn en koord)
                                                                                                  (6)

7.1.2   T̂2  D̂ 2 (angles subt. by the same chord / omtrekshoeke onderspan              S R
                   deur dieselfde koord)
                                                                                         S R
        but / maar D2  PŜT (tan / chord / raaklyn / koord)                             S R
         PŜT  T2              (alternate angles / verwisselende binnehoeke )
        TW || PS                                                                                   (6)
                                                                                                 [12]
                                                                TOTAL / TOTAAL:                   100

                                                   10

## Related documents

- [Gr11 Math P2 (Afrikaans) November 2019 Possible Answers](https://www.hlayiso.com/library/gr11-math-p2-afrikaans-november-2019-possible-answers/d5138457-8292-415c-8ae5-52ff1920ee97) — Mathematics · Memo · 2019 · Afr
- [Gr 11 Math P2 (English) June 2023 Possible Answers](https://www.hlayiso.com/library/gr-11-math-p2-english-june-2023-possible-answers/7d433fdb-38a7-4d1e-b09f-472448486aa3) — Mathematics · Memo · 2023 · Eng
- [Gr11 Math P1 (Afrikaans) June 2019 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-june-2019-question-paper/8a92fc2a-58f8-4c91-9d87-bc71603b08cd) — Mathematics · QP · 2019 · Afr
- [Gr11 Math P1 (English) June 2019 Question Paper](https://www.hlayiso.com/library/gr11-math-p1-english-june-2019-question-paper/9e010e98-da6e-4074-a4f9-ae653c922762) — Mathematics · QP · 2019 · Eng
- [G11 Math P2 (Afrikaans) November 2019 Question Paper](https://www.hlayiso.com/library/g11-math-p2-afrikaans-november-2019-question-paper/ef5d8a14-92a5-4e9c-ba98-30f5140b04de) — Mathematics · QP · 2019 · Afr
- [Mathematics GR 11 Paper 2](https://www.hlayiso.com/library/mathematics-gr-11-paper-2/d067d79e-bc2e-41a9-af62-97830243ae5c) — Mathematics · QP · 2019

## More from Grade 11 Mathematics

- [FS Maths Grade 11 June 2025 P1 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p1-only/82015d8f-bcc0-478e-ad47-d3b295dabb01) — QP · 2025
- [FS Maths Grade 11 June 2025 P2 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p2-only/5d0e7dd8-cb37-419c-8fe2-0aee27fbdc84) — QP · 2025
- [KZN Maths Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p2-and-memo/a2aac931-a969-41c8-bde2-85d23035d042) — QP + Memo · 2025
- [MATHS P2 ANSWER BOOK GR11 2024 ENG AFR D watermark](https://www.hlayiso.com/library/maths-p2-answer-book-gr11-2024-eng-afr-d-watermark/91fd4f6b-a252-4d30-b6b1-f8f546015336) — Addendum · 2024 · Afr
- [MATHS P1 QP GR11 NOV2024 ENG D watermark](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov2024-eng-d-watermark/739068f1-b085-4977-8a45-18f9bb927644) — QP · 2024 · Eng
- [Gr 11 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-question-paper/1a0f4a9d-f18a-4aca-af77-5a81acfb8e0c) — QP · 2024 · Afr

## Collection

- [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
