# Mathematics Grade 12 Paper 2 NSC November 2022 Memo

> Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2022, 24 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-p2-nov-2022-mg-afr-eng/003f5383-067b-458f-bbb1-b45c4abceef5
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: NSC November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 24

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Mathematics · Grade 12 · NSC November Exam · 2022. Memorandum. Read online or download the PDF.

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### Page 1

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                                       NATIONAL
                                  SENIOR CERTIFICATE/
                                       NASIONALE
                                   SENIOR SERTIFIKAAT

                                        GRADE 12/GRAAD 12

                             MATHEMATICS P2/WISKUNDE V2

                                         NOVEMBER 2022

                       MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                            These marking guidelines consist of 24 pages.
                             Hierdie nasienriglyne bestaan uit 24 bladsye.

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Mathematics P2/Wiskunde V2                          2                                   DBE/November 2022
                              NSC/NSS – Marking Guidelines/Nasienriglyne

NOTE:
 If a candidate answers a question TWICE, only mark the FIRST attempt.
 If a candidate has crossed out an attempt of a question and not redone the question,
  mark the crossed out version.
 Consistent accuracy applies in ALL aspects of the marking memorandum. Stop
  marking at the second calculation error.
 Assuming answers/values in order to solve a problem is NOT acceptable.

NOTA:
 As 'n kandidaat 'n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging.
 As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, merk die
   doodgetrekte poging.
 Volgehoue akkuraatheid word in ALLE aspekte van die memorandum toegepas. Hou
   op nasien by die tweede berekeningsfout.
 Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie.

                                  GEOMETRY/MEETKUNDE

                A mark for a correct statement

                (A statement mark is independent of a reason)
         S
                'n Punt vir 'n korrekte bewering

                ('n Punt vir 'n bewering is onafhanklik van die rede)

                A mark for the correct reason

                (A reason mark may only be awarded if the statement is correct)
         R
                'n Punt vir 'n korrekte rede

                ('n Punt word slegs vir die rede toegeken as die bewering korrek is)

                Award a mark if statement AND reason are both correct
        S/R
                Ken 'n punt toe as die bewering EN rede beide korrek is

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Mathematics P2/Wiskunde V2                                            3                                    DBE/November 2022
                                                NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

                                       SCATTER PLOT / SPREIDIAGRAM
                                      25

                  Number of votes /
                                      20

                                      15

                   Getal stemme
                                      10

                                      5
                                           98            103          108         113          118
                                                            Intelligence quotient (IQ) /
                                                             Intelligensiekwosiënt (IK)

Popularity score (x)
                                                32     89      35     82     50     59       81      40     79       65
Gewildheidspunt (x)
Number of votes (y)
                                                9      22      10     21     11     15       20      12     19       16
Getal stemme (y)

1.1.1        155                                                                                     155
         y
              10
            15,5                                              ANSWER ONLY: Full marks               answer
                                                                                                                          (2)
1.1.2   SD  4,59                                                                                     answer
                                                                                                                          (1)
1.2      y  SD
        = 15,5 – 4,59
        = 10,91                                                                                       value of y  SD
         10 – 2 = 8 learners                                                                         answer
                                                                                                                     (2)
1.3     a = 1,7709…                                                                                   a
        b = 0,2243…                                                                                   b
        yˆ  1,77  0,22 x                                                                            equation
                                                                                                                      (3)
1.4      yˆ  1,77  0,2272                                                                         substitution
           = 17,61                                                                                    answer
            18 votes
                                                                                                                           (2)
        OR/OF
        yˆ  17,92  18 votes                                                                         answer
                                                                                                                           (2)
1.5.1   Points are all scattered therefore low correlation and unrealistic                            R
        prediction./Punte is versprei daarom 'n lae korrelasie en
        onrealistiese voorspelling.                                                                                        (1)
1.5.2   r = 0,98/correlation very strong/korrelasie baie sterk                                        S
         a reliable prediction/'n betroubare voorspelling                                                                 (1)
                                                                                                                          [12]

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Mathematics P2/Wiskunde V2                                                 4                                     DBE/November 2022
                                                     NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2

                                                                    OGIVE
                                                                     OGIVE/ /
                                                                           OGIEF
                                                                              OGIEF
                             70 70

                             60 60

                             50 50

             Cumulative
           Cumulative   frequency
                      frequency /
                                        40
                             40

                                        30
                             30

           Kumulatiewe Frekwensie
                                        20
                             20
                                        10
                             10
                                        0
                                    0            5             10         15         20            25          30          35
                                             5            10         15         20         25            30          35
                                                             Percentage of salary spent on fuel /
                                                                 Percentage
                                                         Persentasie          of op
                                                                     van salaris fuelbrandstof
                                                                                      spent spandeer

2.1    60 employees                                                                                answer (A)
                                                                                                                                          (1)
2.2     20  x  25                                                                                answer
                                                                                                                                          (1)
2.3    60 – 34                                                                                     34
       = 26 employees                                    ANSWER ONLY: Full marks                   answer
                                                                                                                                      (2)
2.4             100                                                                                method
        Salary      2400
                 7
       Salary = R34 285,71                               ANSWER ONLY: Full marks                   answer
                                                                                                                                      (2)
2.5     Ogive/Cumulative frequency graph will shift to the                                        answer
         right/will become steeper.
        Ogief/Kumulatiewe frekwensie grafiek sal na regs
         skuif/sal steiler wees.                                                                                                      (2)
                                                                                                                                      [8]

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Mathematics P2/Wiskunde V2                          5                                    DBE/November 2022
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3
                                            y
          D

                                                                     A(4 ; 2)

                                                                            α
                                                                                                            x
                  P                         O          F                     R

                                            S

                          C(–2 ; –3)
                                                           T
                                                                                      B(6 ; –4)

3.1.1             2   4                     42
         mAB                 OR mAB 
                    46                          64                        substitution
         mAB  3                      ANSWER ONLY: Full marks              answer
                                                                                                    (2)
3.1.2    tan  mAB  3                                                    tan  mAB  3
          108,43                                                         answer
                                       ANSWER ONLY: Full marks
                                                                                                      (2)
3.1.3     x x y y 
         T 1 2 ; 1 2 
           2       2 
            2 6 3 4
         T       ;     
           2        2 
                                                                                                 7
           7                                                             xT  2  yT 
         T 2 ;                                                                                  2
               2                                                                                    (2)
3.1.4    5(0)  6 y  8                                                     xS  0
                          4                                                         4
                   y                                                      yS 
                          3                                                          3
            4
         S  0; 
            3                                                                                       (2)
3.2      mCD  mAB  3                                                     gradient
          3  3 2  c OR           y   3  3x   2
                                                                            substitution of
         c = –9                         y  3x  9                          C(–2 ; –3)
         y  3x  9
                                                                            equation
                                                                                                      (3)

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Mathematics P2/Wiskunde V2                            6                                      DBE/November 2022
                                NSC/NSS – Marking Guidelines/Nasienriglyne

3.3.1   5x  6 y  8
             5    8
        y     x
             6    6
                        5
        tan  mAC                                                   tan  mAC 
                                                                                         5
                        6
                                                                                         6
          39,81                                                      39,81
        Â  108,43  39,81
            68,62                                                   Â  68,62
        DĈA  68,62                   [alt s ; DC||AB]             answer
                                                                                                       (4)
3.3.2   P 3; 0 and F(1,6 ; 0)                                      P 3; 0
        Area POSC = Area FPC – Area OFS                             method
                                                                        1
                    1             1        4                         4, 6  3
                    4, 6  3  1, 6                             2
                    2             2        3
                                                                        1      4
                 = 6,9 – 1,07                                         1, 6   
                                                                        2      3
                 = 5,83 units2
                                                                      answer
                                                                                                      (5)
        OR/OF
        P 3; 0                                                     P 3; 0
                            2
                   8
        FC    2     3  0 
                                  2  3 61
                   5                 5

        Area Δ PFC  PFFCsinOF̂S
                     1
                     2
                                                                         1  23   3 61 
                             1  23  3 61                                          sin 39,81
                                        sin 39,81                 2  5   5 
                             2  5  5 
                          6,90                                          1  8  4 
                        1  8  4                                         
           Area Δ OFS                                               2  5  3 
                        2  5  3 
                   = 1,07
                                                                      method
        Area POSC = 6,90 – 1,07
                                                                      answer
                   = 5,83 units2                                                                      (5)

        OR/OF

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Mathematics P2/Wiskunde V2                              7                                   DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

                 P            R                        O

                      C(–2 ; –3)

                                                                        P 3; 0

        P  3;0 
                                                                        method
        Area of POSC = Area of OSCR + Area of PRC
                                                                           14           1
                                                                              3   2  1 3
                            14         1
                             3   2  1 3                       
                                                                           23           2
                            23         2
                            35
                          
                             6
                           5,83units 2                                 answer
                                                                                                       (5)
        OR/
        OF
                 P            R                        O

                              W           2
                      C(–2 ; –3)

        P 3; 0                                                       P 3; 0

        Area POSC = Area ROSW + Area PRC + Area                        method
        WSC
                                                                          1
                       4        1         1      5
                         2   1 3   2                        1 3
                       3        2         2      3                    2
                                                                         4        1    5
                                                                           2   2  
                         35                                              3        2    3
                     
                          6
                                                                        answer
                      5,83units     2                                                                 (5)

        OR/OF

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Mathematics P2/Wiskunde V2                              8                                        DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

        P 3; 0
                                                                        P 3; 0
                      1             ˆ
        Area of PSC = (PC)(CS)sin DCA
                      2

                                             
                       
                            1
                                 10   361  sin 68, 62                                
                            2                                        
                                                                           1
                                                                           2
                                                                                10   361  sin 68, 62
                                                                                            
                       = 3,833..

                            1
        Area of POS =        (PO)(OS)
                            2

                            1 4
                             3  
                            2 3
                                                                              3  
                                                                           1        4
                                                                       
                                                                           2 3
                       =2

        Area POSC = 3,833… + 2
                                                                        method
                      5,83units2
                                                                        answer
                                                                                                              (5)

                                                                                                          [20]

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Mathematics P2/Wiskunde V2                            9                                        DBE/November 2022
                                   NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 4

                                           y
                                                     K

                                                                                                              x

                                                                     N(7 ; –2)

                                                         M(3 ; –5)

                                                                                 L
                                   P

4.1     Px; y  ; N7;2 ; M3;5
        x7                 y2
              3                5
          2                  2
          x  1            y8                                                             xP  1  yP  8
        P 1;  8                                                                                             (2)
        r 2  7  3   2   5          OR/OF r 2   1  3   8   5 
4.2.1                2                 2                             2                  2    substitution into
                                                                                              distance formula
        r 2  25
        ( x  3) 2  ( y  5) 2  25                                                         ( x  3)2  ( y  5)2
                                                                                             r 2  25
                                                                                                                        (3)
4.2.2              5   2 3                                                              substitution
        mradius             
                     37        4                                                                       3 3
                                                                                             mradius      
        mtangent  
                     4
                          [radius  tangent/raaklyn  radius ]                                          4 4
                     3                                                                                     4
                                                                                             mtangent  
         2   7   c                          y   2   x  7 
                  4                                              4                                         3
                                         OR
                  3                                              3                           substitution of m and
            22                                          4     22                              N(7 ; –2)
        c                                       y  x
             3                                          3      3
               4      22
        y  x
               3       3                                                                     equation
                                                                                                                        (5)
4.3                                                                                          subst m and P
        8  
              4
                 1  c
              3
                28                                                                           value of c
         c
                3
          28       22                                                                        answer
            k
          3         3                                                                                                   (4)

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Mathematics P2/Wiskunde V2                           10                                     DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

4.4.1   AB2  AM2  MB2

        AB2   t  3   t  5   52
                       2          2
                                                                                        substitution into
               t  6t  9  t  10t  25  25
                 2            2                                                            Pythagoras
                                                                                          simplification (A)
        AB 2t 2  4t  9
                                                                                                                     (2)
4.4.2        4                                                                           substitution into
        t 
             22                                                                          correct formula
            1                                                                           t  1
        Minimum at t   1
        AB 2 1  4 1  9
                      2
                                                                                          substitution
        AB 7                                                                             answer
                                                                                                                     (4)
        OR/OF

        4t  4  0                                                                        derivative = 0
                                                                                          t  1
        t  1
        Minimum at t   1
        AB 2 1  4 1  9
                      2
                                                                                          substitution
        AB 7                                                                             answer
                                                                                                                     (4)
        OR/OF
        Length of AB  2t 2  4t  9
                                         9
                           2  t 2  2t  
                                         2
                                        7
                           2  t  1  
                                       2
                                                                                          completing of the
                                        2                                                square
                           2  t  1  7
                                     2

        Minimum at t  1
                                                                                          t  1
        AB 2 1  4 1  9
                      2

        AB 7                                                                            substitution
                                                                                         answer
                                                                                                                 (4)
                                                                                                                [20]

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Mathematics P2/Wiskunde V2                              11                                     DBE/November 2022
                                     NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5

5.1.1      sin 360  x 
             sin x                                                                        +  sin x
                                                                                                                        (2)

                                     13    3
5.1.2                                     2                                                substitution
            x  coordinate 
                                                       2

                               2

                      3                                                                   method
            tan x 
                      2
                      3
                    
                      2                                                                                                 (3)
           OR/OF

                                     13   3
                                          2
            x  coordinate 
                                                   2
                                                                                           substitution
                             2
                    3
            tan x 
                    2                                                                      method
                                                                                                                        (3)
5.1.3       cos(180  x)

              cos x                                                                      –  cos x
                                                                                                                        (2)
5.2               cos(90   )
            sin(  180)  3 sin( )

                           sin                                                           – sin
            
                sin  180     3 sin                                               – 3sin
                      sin 
                                                                                          – sin
                 sin   3 sin 

                  sin 
            
                 4 sin                                                                   simplification
                1
                                                                                          answer
                4
                                                                                                                        (5)

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Mathematics P2/Wiskunde V2                                       12                                     DBE/November 2022
                                              NSC/NSS – Marking Guidelines/Nasienriglyne

5.3         (cos x  2 sin x)(3 sin 2 x  1)  0

            cos x  2sin x  0                             or    3sin 2 x 1  0                    both equations
                          1                                                  1                                  1
            tan x                                              sin 2 x                           tan x  
                          2                                                  3                                  2
            ref   26,565...                                   ref  19,471...                  sin 2 x 
                                                                                                               1
                                                                                                               3
           x = 153,43° + k.180° ; k  Z                          x = 9,74° + k.180°; k  Z
                                                                                                    x = 153,43° OR
                                OR/OF                                        or                      x = 153,43° &333,43°
           x = 153,43° + k.360° ; k  Z                         x = 80,26° + k.180°;                x = 9,74° & 80,26°
                                                                                                    + k.180° ;  kZ
                                                                k Z
                                or
           x = 333,43° + k.360° ; k  Z                                                                                          (6)
5.4.1       LHS  cos( x  y). cos( x  y)

                      cos x. cos y  sin x. sin ycos x. cos y  sin x. sin y
                                                                                                    expansion
                      cos x. cos y  sin x. sin y
                            2             2          2      2
                                                                                                    simplification
                                                  
                  1  sin 2 x 1  sin 2 y  sin 2 x.sin 2 y                                        square identity
                      1  sin 2 x.sin 2 y  sin 2 x  sin 2 y  sin 2 x.sin 2 y                    product
                      1  sin 2 x  sin 2 y = RHS
                                                                                                                                 (4)
5.4.2      1  sin 45  sin 15
                      2               2

            cos(45  15). cos(45  15)                                                         identifying x and y
             cos 60. cos 30
               1  3 
                                                                                               substitution
               2  2 
                 3
            
                4                                                                                   answer
                                                                                                                                 (3)
           OR/OF

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Mathematics P2/Wiskunde V2                                13                                     DBE/November 2022
                                       NSC/NSS – Marking Guidelines/Nasienriglyne

           1  sin 2 45  sin 2 15                                                         identity
             sin 2 15  cos2 15  sin 2 45  sin 2 15
                                   2
                           2
             cos 15  
                     2
                              
                           2 
                          1
             cos 2 15 
                          2

                2 cos 2 15  1
            
                       2

                cos 30
            
                   2

                  3 1                                                                        substitution
                  
                 2 2

                 3                                                                           answer
                                                                                                                         (3)
                4

           OR

           1  sin 2 45  sin 2 15
             cos 2 45  sin 2  45  30 
                         2
                                                                                             expansion
              1 
                  sin 45 cos 30  cos 45 sin 30 
                                                           2

              2                                                                            substitution
                                              2
                1  1   3 1 1
                       
                2  2 2    2 2 
                                        2
             1  3    1 
                     
             2  2 2 2 2 
                                        2
             1  3    1 
                     
             2  2 2 2 2 
                1 3    3 1
                     
                2  8 4 8 
                                                                                             answer
               3
                                                                                                                         (3)
              4

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Mathematics P2/Wiskunde V2                             14                                     DBE/November 2022
                                    NSC/NSS – Marking Guidelines/Nasienriglyne

5.5.1      16 sin x. cos 3 x  8 sin x. cos x
            8 sin x. cos x2 cos 2 x  1                                                factorisation
             4 sin 2 xcos 2 x 
                                                                                          4sin2x  cos2x
             2 sin 4 x
                                                                                          double angle
           OR/OF
                                                                                                                       (4)

           16 sin x. cos 3 x  8 sin x. cos x
                         1         1           
             16cos 2 x  sin 2 x   8  sin 2 x 
                         2         2                                                  factorisation
                                1        
             8  2cos 2 x  1  sin 2 x 
                                2        
             4sin 2 x.cos 2 x
                                                                                          4sin 2x  cos2x
             2sin 4x
                                                                                          double angle
                                                                                                                       (4)
5.5.2      16 sin x. cos x  8 sin x. cos x = 2sin4x
                          3

           Minimum at x = 67,5°                                                           answer
                                                                                                                    (1)
                                                                                                                   [30]

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Mathematics P2/Wiskunde V2                           15                                     DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6

                                                       y
                           f                                            f

                        B                                   g          A(60° ; k)
                                                            x
                                                                                                                 x
 –180°                     –90°                        0°                   90°                           180°

                                                                                      f

6.1        180                                                                     answer
                                                                                                                     (1)
6.2.1      k  3  1,73                                                             answer
                                                                                                                     (1)
6.2.2        
           B  120; 3                                                             x = –120°
                                                                                                                     (1)
6.3       Range of g :               y   2 ; 2                                  y   2 ; 2
          Range of 2 g ( x) :        y   4 ; 4                                  answer
                                                                                                                     (2)
          OR/OF                            ANSWER ONLY: Full marks

          Range of g : 2  y  2                                                   2  y  2
          Range of 2 g ( x) : 4  y  4                                            answer
                                                                                                                     (2)
6.4        x   65 ;  5                                                       x   65 ;  5
                                                                                                                     (2)
          OR/OF

            65  x  5                                                           65  x  5
                                                                                                                     (2)
6.5        sin x.cos x  p
           4 sin x.cos x  4 p
                                                                                    2 sin 2 x  4 p
           2 sin 2 x  4 p
                                                                                    4p  2
                 4p  2
                      1    1                                                        answers
           p =        or                 ANSWER ONLY: Full marks
                      2    2                                                                                      (3)
                                                                                                                 [10]
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Mathematics P2/Wiskunde V2                             16                                     DBE/November 2022
                                    NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 7
                                A

                                 B                            2p
                                                                                                45°
            x
                                                                                                                   D
      C
7.1        AD2  AB2  BD2

                    
           AD2  5 p  2 p 
                          2          2                                                substitution in Pythagoras

           AD2  9 p 2

           AD  3 p                                                                   answer
                                                                                                                       (2)
7.2             CD          3p                                                        correct use of sine rule
                         
           sin(135  x)   sin x

                  3 p sin 135  x                                                  135° – x
           CD 
                         sin x

                3 p sin 135 cos x  cos 135 sin x                                 compound angle
           CD 
                                sin x
                3 p sin 45 cos x  cos 45 sin x 
           CD 
                               sin x
                      2              2        
                3 p     cos x        sin x                                       special values
           CD        2             2         
                              sin x
                    2
                3p         cos x  sin x 
                      2                                                             factorisation
           CD 
                           sin x
                3 p  sin x  cos x 
           CD 
                        2 sin x
                                                                                                                       (5)

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Mathematics P2/Wiskunde V2                           17                                     DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

7.3                          1                                                      correct use of area rule
           Area ADC           AD  CD  sinADC
                                                ˆ
                             2

                            1       3 p sinx  cosx  
                            (3 p)                    (sin45)
                            2              2 sin x      

           
               1
                 30 30sin 110  cos110  sin 45
               2               2 sin 110                                         substitution in area rule

            143,11m2                                                               answer
                                                                                                                 (3)
                                                                                                                 [10]

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Mathematics P2/Wiskunde V2                           18                                     DBE/November 2022
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 8

8.1

                                                M

                                                        2
                                                1
                                                        64°

                                                        O           1   N
                                                                    2
                              S                     1       2

                                            1
                                            2
                                        R
                                                                P

8.1.1      P̂  116              [opp s of cyclic quad/teenoorst. e van kvh]               SR
                                                                                                                     (2)
8.1.2      M̂1  64  90        [ in semi-circle/ in halwe sirkel]                        R
           M̂1  26                                                                          S
                                                                                                                     (2)
8.1.3      Ô1  52              [ at centre = 2 x  at circumference/midpts.               S R
                                   = 2 x omtreks. ]
                                                                                                                     (2)

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Mathematics P2/Wiskunde V2                          19                                     DBE/November 2022
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

8.2
                                                    A

                                                               E

                                 D
                                                                                           H

                                                                       G

                B

                                                                   F
                                                                                      C

8.2.1      Midpt theorem/Midpt. Stelling                                                       R
                                                                                                                    (1)
           OR/OF

           Converse prop intercept theorem                                                     R
                                                                                                                    (1)
8.2.2      BG = 2DE or 6x – 2                  [Midpt theorem/Midpt. stelling]                 S R

           BG = 6x – 2
                                                                                                S R
           GH FC
                                             [line || one side of  OR
           BG BF
                                               prop theorem; FG || CH /
                                               lyn || een sy v.  ]
            x 1   1
                                                                                               equation into x
           6x  2 4
           4x + 4 = 6x – 2
           2x = 6
            x=3                                                                                 answer
                                                                                                                    (6)

           OR/OF

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Mathematics P2/Wiskunde V2                          20                                     DBE/November 2022
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

           BF   BG
              =                 [line || one side of  OR prop theorem; FG || CH /
           FC   GH                                                                           S R

                                 lyn || een sy v. ]
           AE   DE
              =                  [ADE ||| ABG]
           AG   BG
                                                                                              S R
           BG = 4x + 4
           1   3x  1
             =
           2   4x  4                                                                         equation into x
           4x + 4 = 6x – 2
           x = 3
                                                                                              answer
                                                                                                                     (6)

                                                                                                                    [13]

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Mathematics P2/Wiskunde V2                          21                                     DBE/November 2022
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 9

9.1

                                                     O

                                    A             1 2
                                                   D                B

9.1.1      Construction:
           Draw OA and OB                                                                     construction
           In ADO and BDO
           OA = OB                            [radii/radiusse]
           OD = OD                            [common side/gemeenskaplike sy]                 first pair of sides
           AD = DB                            [given/gegee]                                   other 2 pairs
            ΔADO  ΔBDO                      [S;S;S]                                         R
           ADB is a straight line
            ˆ D
               ˆ                                                                              R
           D1   2                             ΔADO  ΔBDO
            OD  AB                           [s on a str line/e op 'n reguitlyn]
                                                                                                                    (5)

           OR/OF
           Construction:
           Draw OA and OB                                                                     construction
           In ADO and BDO
           AD = DB                            [given/gegee]                                   first pair of sides
           Aˆ Bˆ                             [s opp; s sides /e teenoor
                                               gelyke sye]
                                                                                              other 2 pairs
           OA = OB                            [radii/radiusse]
            ΔADO  ΔBDO                      [S;;S]                                         R
           ADB is a straight line
            ˆ D
               ˆ                                                                              R
           D1   2                             ΔADO  ΔBDO
            OD  AB                           [s on a str line/e op 'n reguitlyn]                                (5)

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Mathematics P2/Wiskunde V2                          22                                     DBE/November 2022
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

9.2
                                                                  B

        G
                                              E
                                                                          T
                                                                                                F

                                                                      O

                                                  P
                                                                                   S

9.2.1       OT̂G  90                         [line from centre to midpt of chord/ S  R
                                                midpt. sirkel; midpt. koord]
            OB̂G  90                         [tan  radius/raaklyn  radius]               S R
            OT̂G  OB̂G  90
            OTBG is a cyclic quadrilateral [line subtends equal s OR                       R
                                             converse s in the same segment/
                                             lyn onderspan gelyke e]
                                                                                                                     (5)
9.2.2       Ŝ  BT̂G                        [corresp s; GF || PS /                         S R
                                              ooreenk. s; GF || PS]
            But BT̂G  GÔB                  [s in the same segment/e in dies.             S R
                                               sirkelsegment ]
            GÔB  Ŝ                                                                                                (4)
                                                                                                                    [14]

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