Antwoorde / Answers 1.1 - 1.3
1.1 a
2 = 3
2 + 4
2
1.2 b
2 = 12
2 + 5
2
1.3 10
2 = c
2 + 8
2
= 9 + 16
= 144 + 25
c
2 = 10
2 − 8
2
= 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 26
2 = d
2 + 24
2
1.5 20
2 = e
2 + 16
2
1.6 f
2 = 0,5
2 + 1,2
2
d
2 = 26
2 − 24
2
e
2 = 20
2 − 16
2
= 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 g
2 = 0,8
2 + 0,6
2
1.8 1,3
2 = h
2 + 1,2
2
1.9 2
2 = 1
2 + k
2
= 0,64 + 0,36
h
2 = 1,3
2 − 1,2
2
k
2 = 2
2 − 1
2
= 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 9
2 + 12
2 = 81 + 144
2.2 15
2 + 36
2 = 225 + 1 296
2.3 10
2 + (√69)
2 = 100 + 69
= 225
= 1 521
= 169
15
2 = 225
39
2 = 1 521
13
2 = 169
∴ 15
2 = 9
2 + 12
2
∴ 39
2 = 15
2 + 36
2
∴ 39
2 = 10
2 + (√69)
2
∴ a = 90° . . . Pythagoras
∴ b = 90° . . . Pythagoras
∴ c = 90° . . . Pythagoras
Vraag / Question 2.
Antwoorde / Answers 2.4 - 2.6
2.4 14
2 + (√93)
2 = 196 + 93
2.5 8
2 + 11
2 = 64 + 121
2.6 10
2 + 13
2 = 100 + 169
= 289
= 185
= 269
17
2 = 289
(√185)
2 = 185
17
2 = 289
∴ 17
2 = 14
2 + (√93)
2
∴ (√185)
2 = 8
2 + 11
2
∴ 17
2 ≠ 10
2 + 13
2
∴ d = 90° . . . Pythagoras
∴ e = 90° . . . Pythagoras
∴ f ≠ 90° . . . Pythagoras
Vraag / Question 2.
Antwoord / Answer 3
AC
2 = AB
2 + BC
2
AB
2 = AC
2 − BC
2
= 4
2 − 0,5
2
= 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 5
AC
2 = AB
2 + BC
2
= 500
2 + 500
2
= 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 7
AE = ED = ½BD = 3
EC = AC − AE = 12 − 4 = 8
In ΔABE
: AB
2 = AE
2 + BE
2
= 4
2 + 3
2 = 25
AB = 5
In ΔBEC
: BC
2 = BE
2 + EC
2
= 3
2 + 8
2 = 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
: AB
2 = AE
2 + BE
2
AE
2 = AB
2 − BE
2
= 11
2 − 9
2 = 160 000
= 121 − 81 = 40
AE = √40 = 6,3243....
AC = 2AE = 2√40 = 12,6
Vraag / Question 8.
Antwoord / Answer 10
In ΔPQT
: PT
2 = PQ
2 + QT
2
Maar PQ is onbekend en moet dus eers bereken word. /
But PQ is unknown and we must therefore calculate it.
In ΔPQR
: PR
2 = PQ
2 + QR
2
PQ
2 = PR
2 − QR
2
= 17
2 − 8
2
= 289 − 64 = 225
PQ = √225 = 15
QT = QR + RT = 8 + 4 = 12
In ΔPQT
: PT
2 = PQ
2 + QT
2
= 15
2 + 12
2 = 225 − 144 = 369
PT = √369 = 19,21
Vraag / Question 10.
Antwoord / Answer 11.1
Die langste sy is m
2 + n
2 /
The longest side is m
2 + n
2
(m
2 + n
2)
2 = m
4 + 2m
2n
2 + n
4
(m
2 − n
2)
2 + (2mn)
2 = m
4 − 2m
2n
2 + n
4 + 4m
2n
2
= m
4 + 2m
2n
2 + n
4
∴ (m
2 + n
2)
2 = (m
2 − n
2)
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 4n
2 + 1 /
The longest side is 4n
2 + 1
(4n
2 + 1)
2 = 16n
4 + 8n
2 + 1
(4n
2 − 1)
2 + (4n)
2 = 16n
4 − 8n
2 + 1 + 16n
2
= 16n
4 + 8n
2 + 1
∴ (4n
2 + 1)
2 = (4n
2 − 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
: PT
2 = PC
2 + CT
2
= 240
2 + 60
2
= 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
: LN
2 = LM
2 + MN
2
. . . Pythagoras
LM
2 = LN
2 − MN
2
. . . hoogte teen muur / height above floor
= 5
2 − 3
2 = 16
LM = 4 m
KM = LM + KL = 4 + 0,8 m = 4,8 m
In ΔKMP
: KP
2 = KM
2 + MP
2
. . . Pythagoras
MP
2 = KP
2 − KM
2
. . . afstand vanaf muur / distance from wall
= 5
2 − 4,8
2 = 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
: ST
2 = PS
2 + PT
2
. . . Pythagoras
= 48
2 + 64
2 = 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.
CE
2 = AE
2 + AC
2
. . . Pythagoras
BD
2 = AB
2 + AD
2
. . . Pythagoras
ED
2 = AE
2 + AD
2
. . . Pythagoras
BC
2 = AB
2 + AC
2
. . . Pythagoras
CE
2 + BD
2 = AE
2 + AC
2 + AB
2 + AD
2
= AE
2 + AD
2 + AB
2 + AC
2
= ED
2 + BC
2
Vraag / Question 15
Antwoord / Answer 16.
In ΔDEP
:
DE
2 = EP
2 + DP
2
. . . Pythagoras
In ΔDFP
:
DF
2 = DP
2 + PF
2
. . . Pythagoras
PF
2 = DF
2 − DP
2
DE
2 + PF
2 = EP
2 + DP
2 + DF
2 − DP
2
= EP
2 + DF
2
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