WISKUNDE
GRAAD 10
NOG OEFENINGE
  
Pythagoras se stelling : antwoorde.
  
MATHEMATICS
GRADE 10
MORE EXERCISES
  
Theorem of Pythagoras : answers.
  

Antwoorde / Answers 1.1 - 1.3

    1.1  a2 = 32 + 42                                                         1.2  b2 = 122 + 52                                                        1.3  102 = c2 + 82
                = 9 + 16                                                                     = 144 + 25                                                                  c2 = 102 − 82
                = 25                                                                           = 169                                                                               = 100 − 64  = 36
             a = √25                                                                      b = √169                                                                          c =  √36
             a = 5                                                                          b = 13                                                                              c =  6
  
  
Vraag / Question 1.

  

Antwoorde / Answers 1.4 - 1.6

    1.4  262 = d2 + 242                                                         1.5  202 = e2 + 162                                                        1.6  f2 = 0,52 + 1,22
             d2 = 262 − 242                                                                e2 = 202 − 162                                                                 = 0,25 + 1,44
                  = 100                                                                              = 144                                                                           = 1,69
               d = √100                                                                         e = √144                                                                      f =  √1,69
               d = 5                                                                               e = 12                                                                          f =  1,3
  
  
Vraag / Question 1.

  

Antwoorde / Answers 1.7 - 1.9

    1.7  g2 = 0,82 + 0,62                                                         1.8  1,32 = h2 + 1,22                                                        1.9  22 = 12 + k2
                = 0,64 + 0,36                                                                  h2 = 1,32 − 1,22                                                             k2 = 22 − 12
                = 1,00                                                                                  = 1,69 − 1,44                                                                 = 3
             d = √1,00                                                                             h = √0,25                                                                       k =  √3
             d = 1                                                                                    h = 0,5
  
  
Vraag / Question 1.

  
    In 'n reghoekige driehoek is die skuinssy die langste sy. Kry nou
    die som van die vierkante op die twee korter sye en bepaal of dit
    gelyk is aan die vierkant van die langste sy. As dit waar is,
    is die driehoek 'n reghoekige driehoek.
    In a right angled triangle the hypotenuse is the longest side. Find the
    sum of the squares of the two shorter sides and determine whether it
    is eual to the square of the longest side. If this is true, then the triangle
    is a right-angled triangle.

Antwoorde / Answers 2.1 - 2.3

    2.1   92 + 122  = 81 + 144                                              2.2   152 + 362  = 225 + 1 296                                           2.3   102 + (√69)2  = 100 + 69
                            = 225                                                                                = 1 521                                                                                   = 169
                     152 = 225                                                                         392 = 1 521                                                                             132 = 169
                  152 = 92 + 122                                                               392 = 152 + 362                                                                    392 = 102 + (√69)2
                     a = 90°     . . .  Pythagoras                                              b = 90°     . . .  Pythagoras                                                    c = 90°     . . .  Pythagoras
  
  
Vraag / Question 2.

  

Antwoorde / Answers 2.4 - 2.6

    2.4   142 + (√93)2  = 196 + 93                                              2.5   82 + 112  = 64 + 121                                           2.6   102 + 132  = 100 + 169
                                  = 289                                                                              = 185                                                                              = 269
                           172 = 289                                                                 (√185)2 = 185                                                                       172 = 289
                        172 = 142 + (√93)2                                                 (√185)2 = 82 + 112                                                            172 ≠ 102 + 132
                            d = 90°     . . .  Pythagoras                                            e = 90°     . . .  Pythagoras                                          f ≠ 90°     . . .  Pythagoras
  
  
Vraag / Question 2.

  
  
Antwoord / Answer 3  

          AC2 = AB2 + BC2
          AB2 = AC2 − BC2
                 = 42 − 0,52
                 = 16 − 0,25    = 15,75
          AB = √15,75      = 3,969
          Die leer reik 3,969 m tee die muur.    /
          The ladder reaches a height of 3,969 m.
  
  
Vraag / Question 3.

  
Antwoord / Answer 4  

          PR2 = PQ2 + QR2
                 = 52 + 42
                 = 25 + 16    = 41
          AB = √41      = 6,40
  
          Die ankertou is 6,40 m lank.    /
          The length of the rope is 6,40 m.
  
  
  
Vraag / Question 4.

  
Antwoord / Answer 5  

          AC2 = AB2 + BC2
                 = 5002 + 5002
                 = 250 000 + 250 000    = 500 000
          AB = √500 000      = 707,107
          Die vliegtuig is 707,107 km vanaf die beginpunt.    /
          The aircraft is 707,107 km from the starting point.
  
  
  
Vraag / Question 5.

  
Antwoord / Answer 6  

          KM2 = KL2 + ML2
          KL2 = KM2 − ML2
                 = 5002 − 3002
                 = 250 000 − 90 000    = 160 000
          AB = √160 000      = 400
  
          Die krans is 400 m hoog.    /
          The cliff is 400 m high.
  
Vraag / Question 6.

  
Antwoord / Answer 7  

          AE = ED  =  ½BD  = 3 
          EC = AC − AE  = 12 − 4   =  8
          In ΔABE :  AB2 = AE2 + BE2
                                   = 42 + 32   = 25
                             AB = 5
          In ΔBEC :  BC2 = BE2 + EC2
                                   = 32 + 82   = 73
                             BC = √73
                             AB = AD = 5
                             BC = DC = √73  = 8,544
  
  
Vraag / Question 7.

  
Antwoord / Answer 8  

          Gestel / Suppose  BD = 18 cm.
          BE = ½BD = 9 cm
          In ΔABE :  AB2 = AE2 + BE2
                            AE2 = AB2 − BE2
                                    = 112 − 92    = 160 000
                                    = 121 − 81    = 40
                              AE = √40   = 6,3243....
                              AC = 2AE    =  2√40   = 12,6
  
  
  
  
  
Vraag / Question 8.

  
Antwoord / Answer 9  

          In ΔABD :  AB2 = AD2 + BD2
                            AD2 = AB2 − BD2   = 25
                                   = 52 − 32   = 16
                             AD = √16   = 4
  
  
  
  
  
  
  
  
  
  
Vraag / Question 9.

  
Antwoord / Answer 10  

          In ΔPQT :  PT2 = PQ2 + QT2
                             Maar PQ is onbekend en moet dus eers bereken word.  /
                             But PQ is unknown and we must therefore calculate it.
          In ΔPQR :  PR2 = PQ2 + QR2
                             PQ2 = PR2 − QR2
                                      = 172 − 82
                                      = 289 − 64    = 225
                             PQ = √225  = 15
                             QT = QR + RT   = 8 + 4   = 12
          In ΔPQT :  PT2 = PQ2 + QT2
                                    = 152 + 122    = 225 − 144    = 369
                             PT = √369  = 19,21
  
  
Vraag / Question 10.

  
Antwoord / Answer 11.1  

          Die langste sy is m2 + n2  /   The longest side is m2 + n2
          (m2 + n2)2  =  m4  +  2m2n2  +  n4
          (m2 − n2)2  +  (2mn)2  =  m4  −  2m2n2  +  n4  +  4m2n2
                                               =  m4  +  2m2n2  +  n4
          (m2 + n2)2  =  (m2 − n2)2  +  (2mn)2
          Driehoek is 'n reghoekige driehoek.  /
          The triangle is a right-angled triangle.
  
  
Vraag / Question 11.1

  
Antwoord / Answer 11.2  

          Die langste sy is 4n2 + 1  /   The longest side is 4n2 + 1
          (4n2 + 1)2  =  16n4  +  8n2  +  1
          (4n2 − 1)2  +  (4n)2  =  16n4  −  8n2  +  1  +  16n2
                                           =  16n4  +  8n2  +  1
          (4n2 + 1)2  =  (4n2 − 1)2  +  (4n)2
          Driehoek is 'n reghoekige driehoek.  /
          The triangle is a right-angled triangle.
  
  
Vraag / Question 11.2

  
Antwoord / Answer 12.  

          Trek PC || AB  /  Draw PC || AB
            PC = AB = 240 m   en / and  BC = PA = 60 m
              TC = TB − CB   =  120 − 60  =  60
          In ΔPCT :  PT2 = PC2 + CT2
                                   = 2402 + 602
                                   = 57 600 − 3 600    = 61 200
                             PT = √61 200  = 247,39
  
Vraag / Question 12

            
  
  
Antwoord / Answer 13.  

                                             LN = KP = 5 m       . . .  lengte van leer  /  length of ladder
             KL = 800 mm = 0,8 m;   MN = 3 m       . . .  gegee  /  given
          In ΔLMN :  LN2 = LM2 + MN2         . . .  Pythagoras
                            LM2 = LN2 − MN2         . . .  hoogte teen muur  /  height above floor
                                    = 52 − 32    =  16
                              LM = 4 m
                              KM = LM + KL   = 4 + 0,8 m   = 4,8 m
          In ΔKMP :  KP2 = KM2 + MP2         . . .  Pythagoras
                            MP2 = KP2 − KM2         . . .  afstand vanaf muur  /  distance from wall
                                    = 52 − 4,82    =  1,96
                             MP = √1,96   = 1,4  m
                          NP = 3 − 1,4 = 1,6  m
                          Leer is 1,6 m nader aan die muur geskuif.  /  Ladder is moved 1,6 m closer to the wall.
  
Vraag / Question 13

Antwoord / Answer 14.  

          In ΔPST :  ∠T + ∠S + ∠SPT = 180°       . . .  binneoeke van Δ  /  interior angles of Δ  
                                 ∠T + 40° + 90° = 180°       . . .  gegee  /  given
                                                   ∠T  =  50°
                                           ∠TPM  =  50°       . . .  PM = MT  (gegee  /  given)
                                           ∠SPM  =  40°       . . .  ∠P = 90°  (gegee  /  given)
                                           ∠SPM  =  ∠S       . . .  ∠S = 40°  (gegee  /  given)
                                                 SM  =  PM       . . .  basis ∠'e gelyk  /  base ∠'s equal
                                                 SM  =  MT
  
          In ΔPST :  ST2 = PS2 + PT2       . . .  Pythagoras
                                   = 482 + 642    =  2 304 + 4 096 
                                   = 6 400
                             ST = √6 400    =  80
                             MT = ½(80)    =  40      . . .  SM = MT  = ½ST
                             PM = 40                        . . .  PM = MT
  
Vraag / Question 14

Antwoord / Answer 15.  

                                 CE2  =  AE2 + AC2       . . .  Pythagoras
                                 BD2  =  AB2 + AD2       . . .  Pythagoras
                                 ED2  =  AE2 + AD2       . . .  Pythagoras
                                 BC2  =  AB2 + AC2       . . .  Pythagoras
                                 CE2  +  BD2  =  AE2 + AC2  +  AB2 + AD2
                                                       =  AE2 + AD2  +  AB2 + AC2
                                                       =  ED2 + BC2
  
Vraag / Question 15

Antwoord / Answer 16.  

          In ΔDEP :    DE2  =  EP2  +  DP2         . . .  Pythagoras
          In ΔDFP :    DF2  =  DP2  +  PF2         . . .  Pythagoras
                              PF2  =  DF2  −  DP2
                              DE2  +  PF2  =  EP2  +  DP2  +  DF2  −  DP2
                                                   =  EP2  +  DF2   
Vraag / Question 16

Antwoord / Answer 17.  

      Konstruksie / Construction  :  Trek / Draw FEG || AD
        AD = FEG = BC  en / and FEG ⊥ AB;  FEG ⊥ CD
        AF = DG;  BF = CG
      
      In ΔAFE :    AE2  =  AF2  +  FE2  . . .  Pythagoras
        In ΔCGE :    EC2  =  CG2  +  EG2  . . .  Pythagoras
      In ΔBFE :    BE2  =  BF2  +  FE2  . . .  Pythagoras
        In ΔDGE :    ED2  =  EG2  +  GD2  . . .  Pythagoras
  
             EA2  +  EC2  =  AF2  +  FE2  +  CG2  +  EG2
                                  =  DG2  +  FE2  +  BF2  +  EG2
                                  =  BF2  +  FE2  +  EG2  +  GD2
                                  =  EB2  +  ED2
  
  
Vraag / Question 17