# Mathematics Grade 12 Paper 2 KZN Prelim 2024 Memo

> Grade 12 Mathematics Paper 2 memorandum from the KZN Prelim Exam 2024, 14 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/kzn-gr-12-maths-sept-2024-memo-p2-final/b652502c-ee92-4539-aba2-e09338d41a69
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2024
- Exam period: KZN Prelim Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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### Page 1

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                                         FINAL

                                MATHEMATICS P2

                        PREPARATORY EXAMINATION

                                 SEPTEMBER 2024

                            MARKING GUIDELINES

                                  NATIONAL
                             SENIOR CERTIFICATE

                                       GRADE 12

 MARKS:       150

                        These marking guidelines consist of 14 pages.

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Mathematics P2                                            2       KZN September 2024 Preparatory Examinations
                                                    GRADE 12
                                                 Marking Guidelines

       •     If a candidate answered a QUESTION TWICE, mark only the FIRST attempt.
       •     If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.
       •     Consistent accuracy applies to ALL aspects of the marking guidelines.
       •     Assuming values/answer in order to solve a problem is unacceptable.

                                                       GEOMETRY
                        A mark for a correct statement
               S        (A statement mark is independent of a reason.)

               R        A mark for a correct reason
                        (A reason mark may only be awarded if the statement is correct.)

             S/R        Award a mark if the statement AND reason are both correct.

QUESTION 1
Penalise only once for incorrect rounding in Question 1.
                165500
 1.1.1 Mean =                                                                 ✓A 165 500 in numerator
                  12         Also accept: 13,79       Answer only:
              = R13792       thousand rand            Full marks              ✓CA answer

             If answer is given as 13,79 instead of R13 792, penalise in
             1.1.1, but don’t penalise again for this mistake in 1.1.2, 1.2
             and 1.5.
                                                                                                          (2)
 1.1.2 Standard deviation = R4 404                                            ✓A answer
                                                                                                          (1)
 1.2    R13 792 +R4 404 = R18 196                                             ✓CA R18 196
        2 employees earn a salary more than one standard deviation            ✓CA 2 employees
        above the mean.
                                                       Answer only:
                                                       Full marks                                         (2)
 1.3    a = 8, 45                                                             ✓A correct a value
        b = 0, 45                                                             ✓A correct b value
                                                              Answer only:
        yˆ = 0, 45 x + 8, 45                                  Full marks      ✓ CA answer
                                                                                                          (3)
 1.4       r = 0,94                                                           ✓A answer
                                                                                                          (1)
 1.5       yˆ = 0, 45 ( 30 ) + 8, 45                                          ✓CA substitution
        yˆ = 21,95
         R21 950                                                             ✓CA answer                  (2)
        OR                                                                    OR
        R21 804 (calculator)                                                  ✓✓ CA CA                    (2)

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Mathematics P2                                        3       KZN September 2024 Preparatory Examinations
                                                GRADE 12
                                             Marking Guidelines

 1.6    Yes.                                                              ✓CA answer
        r = 0,94 implies a strong correlation between employee            ✓CA justification
        experience and monthly salary and therefore a prediction would
        be reliable.                                                                                   (2)

        OR                                                                OR

        Yes.                                                              ✓CA answer
        r = 0,94 , which is close to 1, and therefore implies a strong    ✓CA justification
        correlation between employee experience and monthly salary
        and therefore a prediction would be reliable.                                                  (2)
                                                                                                      [13]

QUESTION 2

 2.1   5500                                                                 ✓A answer
                                                                                                       (1)
 2.2    Q1 = 29      (accept 28 – 29)                                       ✓A value of Q1
        Q3 = 39      (accept 38 – 39)                                       ✓A value of Q3
        IQR = 10      (accept 9 – 11)                                       ✓CA answer
                                                                                                       (3)
 2.3       650
                     (accept 620 – 700)                                     ✓A numerator in range 620
          5500                                                              to 700
        = 11,82% (accept 11,27% – 12,73%)                                   ✓CA answer
                                                                                                       (2)
                                                                                                       [6]

QUESTION 3

               y2 − y1
 3.1    mAB =
               x2 − x1
               −2 − 4
             =                                                            ✓A substitution
               −6 − 0
             =1                                                           ✓CA answer
                                                                                                       (2)
 3.2     mCD = mAB = 1                                                    ✓CA mCD = 1
            y = mx + c
        Substitute (10 ; −1) and mCD = 1 :
                                                                          ✓CA substitution of gradient
           −1 = 1(10 ) + c
                                                                          and point
             c = −11
              y = 1x − 11                                                 ✓CA answer
                                                                                                       (3)

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Mathematics P2                                   4        KZN September 2024 Preparatory Examinations
                                            GRADE 12
                                         Marking Guidelines

 3.3   Midpoint of AC is the same as the midpoint of BD               ✓A midpoint of BD
       [diagonals of parm. bisect each other]
        Midpoint of AC
             −6 + 10 −2 − 1 
       = M            ;      
             2           2 
                 −3                               Answer only:      ✓CA x coordinate
       = M2 ;       
                  2                               Full marks        ✓CA y-coordinate
                                                                                                  (3)
       OR                                                             OR
                                                                      ✓A coordinates of C
        C (4 ; − 7)
        Midpoint of AC
          0+4 4−7
       = M       ;     
           2         2 
               −3                                                   ✓CA x coordinate
        = M2 ;                                                      ✓CA y-coordinate
               2 
                                                                                                  (3)
 3.4    C (4 ; − 7)                                                   ✓CA x coordinate
                                                                      ✓CA y-coordinate
                                                                                                  (2)
 3.5             mAB = 1
               ˆ =1
          tan AFG                                                     ✓CA tan ˆ
                                                                             AFG = 1
               ˆ = 45
              AFG                                                           ˆ = 45
                                                                      ✓CA AFG

                      −1 − 4
                 mAD =
                      10 − 0
                         1                                                         1
                     =−                                               ✓A mAD = −
                         2                                                         2
                ˆ =−    1
           tan AHJ
                        2
                 ˆ
               AHJ = 153, 43                                              ˆ = 153, 43
                                                                      ✓CA AHJ
               BADˆ = 153, 43 − 45 [exterior  of HAF ]
                      = 108, 43                                      ✓CA ˆ
                                                                         BAD = 108, 43
               BCDˆ = 108, 43      [opp  s of a parm.]                  ˆ = 108, 43
                                                                      ✓CA BCD
                                                                                                  (6)

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Mathematics P2                                                                    5        KZN September 2024 Preparatory Examinations
                                                                             GRADE 12
                                                                          Marking Guidelines

         OR                                                                                            OR

                     (10 − 4 ) + ( −1 + 7 ) = 6 2
                                  2                       2
         CD =                                                                                          ✓CA length of CD

                  ( −6 − 4 ) + ( −2 + 7 ) = 5 5
                                  2                       2
         BC =                                                                                          ✓CA length of BC

                     ( −6 − 10 ) + ( −2 + 1) =
                                      2                       2
         BD =                                                     257                                  ✓A length of BD
                                        ˆ
         BD 2 = BC2 + CD2 − 2.BC.CD.cosBCD                                                             ✓A use of cosine rule
         ( 257 ) = (5 5 ) + ( 6 2 ) − 2.(5 5 ). (6 2 ).cosBCD
                  2                   2                   2
                                                           ˆ
                                                                                                       ✓CA substitution into cosine

                    (5 5 ) + ( 6 2 ) − ( 257 )
                                          2                   2             2                          rule
                ˆ
          cosBCD=
                          2. ( 5 5 ) . ( 6 2 )
               ˆ = 108, 43
              BCD                                                                                      ✓CA answer
                                                                                                                                       (6)
                                                                                                                                      [16]

QUESTION 4

          r 2 = OJ 2 = 22 + ( −1)
                                              2
 4.1.1                                                                                                  ✓A substitution
          r=     5                                                                                      ✓A length of OJ
                                                                                                                                       (2)
 4.1.2   OK = OJ + JK = 5 + 2 5 = 3 5                                                                   ✓A length of OK

                     (3 5 ) = ( a − 0 ) + ( −3 − 0 )
                              2                       2               2
                                                                                                        ✓A substitution
                          45 = a 2 + 9
                          a 2 = 36                                                                      ✓A a 2 subject of formula
                           a = −6 or a = 6
                            N/A                                                                                                        (3)
         OR                                                                                             OR
                       OJ = 5
                       JK = 2 5                                                                        ✓A length of JK

                 ( 2 5 ) = ( a − 2) + ( −3 + 1)
                          2                       2               2
                                                                                                        ✓A substitution
                       20 = a 2 − 4a + 4 + 4
            a 2 − 4a − 12 = 0                                                                           ✓A standard form
         ( a − 6 )( a + 2 ) = 0
                          a = 6 or a = −2
                                     N/A                                                                                               (3)
         ( x − 6 ) + ( y + 3) = 20                                                                      ✓A ( x − 6 ) + ( y + 3)
                 2                    2                                                                             2             2
 4.1.3
                                                                                                        ✓CA = 20
                                                                                                                                       (2)

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Mathematics P2                                                    6        KZN September 2024 Preparatory Examinations
                                                             GRADE 12
                                                          Marking Guidelines

 4.1.4 Substitute (10; − 4 ) :
           (10 − 6 ) + ( −4 + 3)                                                        ✓CA substitution
                     2             2

        = 17
       17  20 ,                                                                        ✓CA 17  20
        the point lies inside the circle                                               ✓CA conclusion               (3)
 4.1.5  KO = 5 + 2 5 = 3 5                                                              ✓ CA length of KO
         In POR and PKS :
           1.      Pˆ = Pˆ            [common]
                 ˆ
           2. PRO = PSK    ˆ          [= 90 ; tangent ⊥ radius]
           3. PORˆ = PKS   ˆ          [remaining s ]
                 POR ||| PKS              [  ]
                 PO OR
                    =                        [ ||| s ]
                 PK KS
                       OR 1
                     =    =
                       2OR 2
                        1
                  PO = PK                                                              ✓A PO =
                                                                                                  1
                                                                                                    PK
                        2                                                                         2
                     PO = OK = 3 5
                              ( )
                     PK = 2 3 5 = 6 5                                                   ✓ CA length of PK
                    ˆ
                  PSK = 90                                [radius ⊥ tangent]
                   PS2 = PK 2 − KS2                        [Theorem of Pythagoras]      ✓ CA substitution in Theorem
                          (      ) − (2 5 )
                                   2            2
                         = 6 5                                                          of Pythagoras

                         = 160
                    PS = 160 = 4 10                                                    ✓CA answer                   (5)
                                   25            25
 4.2.1 x 2 − 4 x + 4 + y 2 + 5 y +    = −d + 4 +                                        ✓A completing the square
                                   4             4
                             2

       ( x − 2 ) +  y +  = −d +
                                                                                                                 2
                          5            41                                                                  5
                                                                                        ✓A ( x − 2 ) +  y + 
                2                                                                                  2

                         2           4                                                                   2
                       5                                            Answer only:
       Centre  2 ; −                                                Full marks        ✓CA x coordinate
                       2                                                              ✓CA y coordinate
                                                                                                                      (4)

 4.2.2   diameter = 24 units,           radius =12 units                               ✓A radius =12 units
                                            41
                                       −d + = 144                                       ✓CA equating
                                             4
                                                   535
                                             d =−                                       ✓CA answer
                                                    4
                                                                                                                      (3)
                                                                                                                     [22]

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Mathematics P2                                               7        KZN September 2024 Preparatory Examinations
                                                        GRADE 12
                                                     Marking Guidelines

QUESTION 5

 5.1.1
         tan 58 = n

                r 2 = x2 + y 2           [Theorem of Pythagoras]             32°       ✓A subst. in Theorem of
                r =1 + n
                    2       2        2                                                 Pythagoras
                                                                                   n
                    r = n2 + 1                                                         ✓A r = n 2 + 1

                                 n
          sin 58 =
                                1 + n2                             58°                 ✓CA answer
                                                                         1
                                                                                                               (3)
 5.1.2     sin 296
         = − sin 64                                                                   ✓A − sin 64
         = − sin 2 ( 32 )
         = −2sin 32 cos 32                                                           ✓CA expansion
         = −2 cos 58 sin 58                                                          ✓CA co-functions
               1  n 
         = −2                
               1 + n  1 + n 
                     2        2

            −2n                                                                        ✓CA answer
         =
           1 + n2                                                                                              (4)
 5.1.3     cos 2
         = cos ( 60 − 58 )                                                           ✓A cos ( 60 − 58 )
         = cos 60 cos 58 + sin 60 sin 58                                           ✓A expansion
          1   1        3     n
         =         +    
          2  1+ n 2   2    1 + n2
                                                                                       ✓CA answer
             1 + 3n
         =
             2 1 + n2                                                                                          (3)

         OR                                                                            OR

             cos 2
         = cos ( 32 − 30 )                                                           ✓A cos ( 32 − 30 )
         = cos 32 cos 30 + sin 32 sin 30                                           ✓A expansion
                n                3    1       1
         =                        +        
              1+ n      2       2    1+ n 2   2
                                                                                       ✓CA answer
               3n + 1
         =
             2 1 + n2                                                                                          (3)

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Mathematics P2                                              8        KZN September 2024 Preparatory Examinations
                                                       GRADE 12
                                                    Marking Guidelines

 5.2.1       LHS
             1 − (1 − 2sin 2 x )                                                     ✓A 1 − 2sin 2 x
         =
             2sin x cos x                                                            ✓A 2sin x cos x
             2sin 2 x
         =
           2sin x cos x
           sin x
         =
           cos x                                                                     ✓A simplification
         = tan x
         = RHS                                                                                               (3)

         OR                                                                          OR

           LHS

          =
                                   (
            sin 2 x + cos 2 x − 2 cos 2 x − 1 )
                                                                                     ✓A 2cos 2 x − 1
                       2sin x cos x
                                                                                     ✓A 2sin x cos x
            sin x − cos 2 x + 1
                2
          =
                2sin x cos x
            sin 2 x − cos 2 x + sin 2 x + cos 2 x
          =
                        2sin x cos x
                    2
              2sin x
          =
            2sin x cos x
            sin x
          =
            cos x
                                                                                     ✓A simplification
          = tan x
          = RHS
                                                                                                             (3)
         OR
           LHS                                                                       OR

          =
                                   (
            sin 2 x + cos 2 x − cos 2 x − sin 2 x   )
                        2sin x cos x                                                 ✓A cos 2 x − sin 2 x
              2sin x2                                                                ✓A 2sin x cos x
          =
            2sin x cos x
            sin x
          =
            cos x                                                                    ✓A simplification
          = tan x
          = RHS
                                                                                                             (3)

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Mathematics P2                                                     9      KZN September 2024 Preparatory Examinations
                                                            GRADE 12
                                                         Marking Guidelines
 5.2.2    tan15
             1 − cos 2 (15 )                                                                  1 − cos 2 (15 )
         =                                                                               ✓A
               sin 2 (15 )                                                                      sin 2 (15 )
            1 − cos 30
         =
              sin 30
                  3
            1−
                 2                                                                       ✓A substitution of special
         =
               1                                                                         angle values
               2
                  3 2
         = 1 −     
                 2  1
         = 2− 3                                                                          ✓CA answer                     (3)
                                                                sin x
 5.3         sin ( 360 + x ) .cos ( 90 + x ) −
                                                   cos ( − x ) .tan ( 360 − x )

         = sin x. ( − sin x ) −
                                       sin x                                             ✓A sin x        ✓A − sin x
                                  cos x. ( − tan x )                                     ✓A cos x        ✓A − tan x

         = − sin 2 x + 1                                                                 ✓CA 1
         = cos x 2
                                                                                         ✓CA answer                     (6)
                                 1 1
 5.4                   cos 2 x − = sin x
                                 3 3
                                 1 1
                  1 − 2sin 2 x − = sin x                                                 ✓A 1 − 2sin 2 x
                                 3 3
                   3 − 6sin x − 1 = sin x
                            2

               6sin 2 x + sin x − 2 = 0                                                  ✓A standard form
          (3sin x + 2)(2sin x − 1) = 0                                                   ✓CA factors
                   2
         sin x = −
                   3
                                                                                         ✓CA x = 221,81 + k.360
          x = 221,81 + k.360 or x = 318,19 + k.360 , k  Z
                                                                                         or x = 318,19 + k.360
                      1
         or sin x =
                      2
                                                                                         ✓CA x = 30 + k.360 or
          x = 30 + k.360 or x = 150 + k.360 , k  Z
                                                                                         x = 150 + k.360
                                                                                         ✓A k  Z                (6)
 5.5     sin ( 2 x + 30 ) + k = 3
         sin ( 2 x + 30 ) = 3 − k                                                       ✓A sin ( 2 x + 30 ) = 3 − k

         sin ( 2 x + 30 )  −1 or sin ( 2 x + 30 )  1                                 ✓A sin ( 2 x + 30 )  −1 or
                                                                                          sin ( 2 x + 30 )  1
                     3 − k  −1           or           3− k 1                           ✓CA 3 − k  −1 or 3 − k  1
                           k 4           or             k2                             ✓CA k  4
                                                                                         ✓CA k  2                   (5)
                                                                                                                    [33]

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Mathematics P2                                                    10    KZN September 2024 Preparatory Examinations
                                                          GRADE 12
                                                       Marking Guidelines

QUESTION 6

             1
 6.1    b=                                                                             ✓A answer
             2
                                                                                                                  (1)
 6.2    period = 360                                                                  ✓A answer
                                                                                                                  (1)
 6.3    A ( 30;1)                                                                     ✓A 30 ✓A 1
                                                                                                                  (2)
 6.4    x = 160                                                                       ✓A answer
                                                                                                                  (1)
 6.5    −3  y  1                                                                     ✓✓ AA
                                                                                                                  (2)
        OR
                                                                                       OR
        y   −3 ; 1                                                                  ✓✓ AA
                                                                                                                  (2)
                                                                                                                  [7]

QUESTION 7

        sin y sin x
 7.1          =                                                                      ✓A substitution in sine rule
          2b      b
                2b sin x                                                                           2b sin x
        sin y =                     OR       b sin y = 2b sin x                      ✓A sin y =             OR
                   b                                                                                   b
        sin y = 2sin x                                                                b sin y = 2b sin x
                                                                                                                  (2)
                 AB                                                                     AB
 7.2                = tan                                                           ✓A    = tan 
                 BC                                                                     BC
              AB = BC.tan                                                          ✓A AB = BC.tan
              D̂ = 180 − ( x + y )                                                  ✓A D̂ = 180 − ( x + y )
                                           ˆ
                 BC = BD + CD − 2BD.CD cos D
                     2          2        2

                 BC2 = (2b) 2 + b 2 + 2(2b)(b) cos 180 − ( x + y )              ✓A substitution in cosine rule
                 BC2 = (2b)2 + b2 + 2(2b)(b) cos ( x + y )                           ✓A + cos ( x + y )
                 BC = 5b + 4b cos ( x + y )
                     2      2        2

                 BC2 = b 2 ( 5 + 4 cos ( x + y ) )                                   ✓A simplification
                                                                                     ✓A taking square root on
                 BC = b ( 5 + 4cos ( x + y ) )
                                                                                     LHS and RHS
              AB = b tan          (5 + 4cos ( x + y ))                                                          (7)
 7.3    AB = 54,8 tan 42, 6 5 + 4 cos ( 31 + 75,84 )                              ✓A substitution
        AB = 98, 76 metres                                                           ✓A answer
                                                                                                                  (2)
                                                                                                                 [11]

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Mathematics P2                                         11      KZN September 2024 Preparatory Examinations
                                                 GRADE 12
                                              Marking Guidelines

QUESTION 8

 8.1.1           (
         ˆ = 1 COE
         A1
             2
                ˆ       )         [  at centre = 2  at circumference]    ✓R
            = 68                                                           ✓A answer
                                                                                                        (2)
 8.1.2         ˆ
         Ê1 = A                  [tan chord theorem]                       ✓R
                 1

             = 68                                                          ✓CA answer
                                                                                                        (2)
 8.1.3    ˆ = Eˆ
         BCE                        [ alt  s; DF || CA]                    ✓R
                1

             = 68                                                          ✓CA answer
                                                                                                        (2)
 8.1.4   ˆ = 180 − BCE
         G           ˆ              [opp.  s of cyclic quad]               ✓R
           = 112                                                           ✓CA answer
                                                                                                        (2)
 8.2       ˆ = 90
         BED                         [radius ⊥ tangent]                     ✓S✓R
              = B̂1                 [co-interior  s ; DF || CA]            ✓ S/R
                1
          AB = AC                  [line from centre ⊥ to chord]           ✓R
                2
              = 7 units
                                                                            ✓A answer                   (5)
         OR                                                                 OR
           ˆ = 90
         BED                               [radius ⊥ tangent]               ✓S✓R
          Ê 2 = 22
                            (
                         ˆ + Eˆ
          ˆB1 = 180 − BCE     2   )     [sum of  s of  BCE]
             = 180 − ( 68 + 22 )
             = 90                                                          ✓ S/R
                1
          AB = AC                  [line from centre ⊥ to chord]           ✓R
                2
              = 7 units                                                     ✓A answer                   (5)
                                                                                                       [13]

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Mathematics P2                                      12      KZN September 2024 Preparatory Examinations
                                              GRADE 12
                                           Marking Guidelines

QUESTION 9

 9.    In QVS:
        QU QT
           =         [prop. theorem; UT||VS] OR [line || one side of  ]   ✓S ✓R
        UV TS
             5
           =
             2
             5k
           =
             2k
                            5                15
        5k = 2 x ; or: x = k . And: 3x =       k                          ✓ x i.t.o. k
                            2                 2
       In UPR:
        PS UV
           =       [prop. theorem; UP||VS] OR [line || one side of  ]     ✓S
        PR UR
               2k
           =
              15
                 k
               2
               4
            =                                                              ✓S
              15
         PS 4
           =
         SR 11                                                             ✓answer
                                                                                                       (6)
                                                                                                       [6]

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### Page 13

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Mathematics P2                                  13       KZN September 2024 Preparatory Examinations
                                           GRADE 12
                                        Marking Guidelines

QUESTION 10

 10.1
                          A
                                                                        D

                     1
                 P                          Q

                                                           E                               F
          B                                          C

          Construct AP = DE and AQ = DF                                     ✓construction
          In APQ and DEF:
             1. AP = DE                 [from construction]
             2. AQ = DF                 [from construction]
             3.    ˆ
                  A= ˆ D                [given]                             ✓ APQ  DEF
            APQ  DEF                [SAS]
                 Pˆ = Eˆ
                     1                   [from   s ]                      ✓ Pˆ1 = Eˆ
             But: B ˆ = Eˆ                [given]
                 Pˆ1 = Bˆ
                 PQ || BC                [corresponding  s are =]         ✓S✓R
                   AP AQ
                       =                 [prop. theorem; PQ || BC]         ✓S/R
                   AB AC
                   DE DF
                       =                 [DE = AP; DF = AQ]
                   AB AC                                                                          (6)
 10.2.1   In MKL and MNP:
                 MK 40
             1.        =   = 2,5                                            ✓S
                 MN 16
                 KL 25
             2.       =    = 2,5                                            ✓S
                 NP 10
                 ML 30
             3.        =   = 2,5                                            ✓S
                 MP 12
                MKL||| MNP            [sides of  s in proportion]       ✓R
                                                                                                  (4)

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Mathematics P2                                       14       KZN September 2024 Preparatory Examinations
                                                GRADE 12
                                             Marking Guidelines

 10.2.2      ˆ = Lˆ
            NPM                                [from |||  s ]                   ✓S ✓R
           KLNP is a cyclic quadrilateral     [converse: ext.  of cyclic
                                                                                 ✓R
                                                 quadrilateral] OR
                                               [ext.  of quad = int. opp.  ]                            (3)
          OR                                                                     OR
             ˆ =K
            PNM ˆ                              [from |||  s ]                   ✓S ✓R
           KLNP is a cyclic quadrilateral     [converse: ext.  of cyclic
                                                quadrilateral] OR                ✓R
                                               [ext.  of quad = int. opp.  ]                            (3)
 10.3.1   In BCE and ADE:                                                      ✓ selecting triangles
             1. Eˆ 1 = Eˆ 3                     [vertically opp.  s]            ✓S
             2. Cˆ =D   ˆ                       [  s in the same segment]       ✓S/R
                     1     2
                     ˆ
             3. B̂ = A                          [sum of  s of  s ]                    ˆ
                                                                                 ✓ B̂ = A
            BCE||| ADE                       [   ]                         OR
                 BC AD                                                           [   ]
                   =                           [from |||  s ]
                 CE DE                                                           ✓R
                       AD.CE
                BC =                                                            ✓S/R
                        DE
                                                                                                          (5)
 10.3.2   In ADE and BDC:                                                      ✓ selecting triangles
                ˆ =D
             1. D   ˆ                            [given]
                  2   1

             2. Aˆ =B ˆ                         [  s in the same segment]       ✓S/R
                        ˆ
             3. Ê3 = BCD                       [sum of  s of  s ]
            ADE||| BDC                       [   ]                         ✓R
                AD DE                                                            ✓S
                   =                            [from |||  s ]
                BD CD
            AD.CD = DE.BD
                    = DE. ( DE + BE )                                            ✓ substitute DE + BE
                         = DE 2 + DE.BE
                                                                                                          (5)
                                                                                                         [23]

                                                                                      TOTAL:             150

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