WISKUNDE
NOG OEFENINGE

Pythagoras se stelling : antwoorde.

MATHEMATICS
MORE EXERCISES

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.

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.

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.

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.

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.

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
BC = DC = √73  = 8,544

Vraag / Question 7.

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.

In ΔABD :  AB2 = AD2 + BD2
AD2 = AB2 − BD2   = 25
= 52 − 32   = 16

Vraag / Question 9.

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.

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

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

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

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

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

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

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

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