Mathematics-GR-10-MEMO-Paper-2__hlayiso.com_.pdf

Downloaded from hlayiso.com GRADE 10 MATHEMATICS PAPER 2 MID – YEAR EXAMINATION 2019 MEMORANDUM MARKS: 75 This memorandum consists of 6 pages Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/P2 NSC – Grade 10 NW/June 2019 Question 1  formula 1.1 PQ  x 2  x 1 2  y 2  y1 2  7  62  4  62  substitution  5  answer (3) 1.2 6 x 6 y 7 7  ;  ;  6 x 7  2 2  2 2   6x 7 6 y 7 2 2   2 2 2 2 6 y 7 x 1 y 1   2 2 S(1; 1)  answer (3) 1.3 PR  x  x   y  y  P Q 2 P Q 2  7  02  4  32  substitution  50 5 2  answer  7.07 QS  x  x   y  y  S Q 2 S Q 2  1  62  1  62  substitution  50 5 2  7.07  answer  PR  QS  conclusion (5) 1.4 63 m QR   substitution 60 1  1 2  m QR  2 3 1 m RS  0 1  m RS  2  2 m RS  m QR 1 1  2    2 2 2  1 Demo NW/JUNE/MATH/ EMIS/6******* 2

Downloaded from hlayiso.com Mathematics/P2 NSC – Grade 10 NW/June 2019 m RS  m QR  1  QR  RS  m RS  m QR  1 (5) 1.5 Rectangle  Rectangle The diagonals are equal and one of the interior angles is equal  reason to 90◦ 1.6 ^ 5 ^ 5 cosR S Q   cosR S Q  5 2 5 2 ^ RSQ  71.57  answer (3) Question 2 2.1.1 1  substitution  sin112.4 2  0.46  answer (2) 2.1.2   cosec112,4  48,6     substitution 1   reciprocal  sin 112,4  48,6   3.07  answer (3) 2.1.3  112,4  48,6     2cos   substitution  2     0,33  answer (2) 2.1.4  112.4   tan    substitution  3     0,77  answer (2) 2.2.1 tanθ  2,736 θ  69,92  answer (1) 2.2.2 3sin(3θ  60 )  0,531 0  division by 3 sin(3θ  600 )  0,177 3θ  10,1950  600  simplification 3θ  70,1950 θ  23,400  answer (3) Demo NW/JUNE/MATH/ EMIS/6******* 3

Downloaded from hlayiso.com Mathematics/P2 NSC – Grade 10 NW/June 2019 Question 3 3.1.1 x 2  y 2  r 2 .......... pythagoras  substitution x 2  (5) 2  132  simplification x 2  169  25 x 2  144  answer x  12 (3) 3.1.2 y tan   x 5 tan   12  answer (1) 3.1.3 sin 2   cos 2  2 2  12   5   substitution =      13   13  =1  answer (2) 3.1.4 1 sec  cos r sec  r  sec  x x 13   answer 12 (2) 3.2 cos ec20 . sin 20  tan 45 . sec 60     cot 45. sin 90 1 1   . sin 20  1.2 sin 20 0  sin 20 1.1 1 1  1.2 = 2 1 1 3 1  answer (6) Demo NW/JUNE/MATH/ EMIS/6******* 4

Downloaded from hlayiso.com Mathematics/P2 NSC – Grade 10 NW/June 2019 Question 4 4.1.1  S O1  90 diagonals bisect at 90◦ R (2) 4.1.2 ^ ^ ^ L1  O1  LKM  180 sum of angles of ∆  S/R   ^ L1  180  34  90 ^ L1  560 answer (2) 4.1.3 ^ ^ L1  L 2  56 Diagonals bisect the angles  S/R ^ ^ ^ ^ L1  L 2  N1  N 2 opp s of a rhombus  S/R ^ ^ N1  N 2  56  56  Substitution ^  KNM  112  answer (4) 4.2.1 ^ ^  S/R ABF  BFE  2x alt s AB//BF ^ AFE  x  2x  S ^  AFE  3x  answer (3) 4.2.2 ^ S B FE  2 x proven above ^ ^ A F E  F E B  180 co  int s AF//FEB  S/R 3 x  7 x  1800 10 x  1800  simplification x  180  x – value ^ ^ SFA  SAF  3x s opp  sides  S/R  3(18 )  54    540  540  y  1800 sum of angles of ∆SAF  y  1800  1080  y – value  720 (6) 5.1 ^ ^  S/R A B E  F D C [alts; AB // DC ] BE = FD given  S/R AB = DC opp sides of parm S/R  conclusion Demo NW/JUNE/MATH/ EMIS/6******* 5

Downloaded from hlayiso.com Mathematics/P2 NSC – Grade 10 NW/June 2019 ∆𝐴𝐵𝐶 ≡ ∆𝐶𝐷𝐹 S ; ; S (4) 5.2 ^ ^ A E B  D FC proven above from congruency ^ ^ A E B  A E F  180 s on a str. line R ^ ^ C F D  C F E  180 s on a st. line R ^ ^ A E F  CFE AEB  DFC  AE//EF [alt s ]  conclusion (3) 5.3 AE//FC proven above  R AE = FC ΔABE  CDF  R  AECF is a parm [Pair of opp sides  and//]  conclusion (3) Demo NW/JUNE/MATH/ EMIS/6******* 6