WISKUNDE
GRAAD 11
NOG OEFENINGE
Loodlyne op koorde : antwoorde.
GRAAD 11
NOG OEFENINGE
Loodlyne op koorde : antwoorde.
MATHEMATICS
GRADE 11
MORE EXERCISES
Perpendicular lines to a chord : answers.
GRADE 11
MORE EXERCISES
Perpendicular lines to a chord : answers.
1.1 OQ ⊥ PR ⋯ gegee / given
∴ PQ = QR ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
PQ = QR = 4 cm ⋯ PR = 8 cm (gegee / given)
In ΔOPQ : OP2 = OQ2 + PQ2 ⋯ Pythagoras
52 = OQ2 + 42
OQ2 = 25 ─ 16
= 9 = 32
OQ = √9 = 3
OQ = 3 cm [ 1.1 ]
1.2 OQ ⊥ PR ⋯ gegee / given
∴ PQ = QR ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
OP = 13 cm ⋯ OQ = 5 cm (gegee / given)
In ΔOPQ : OP2 = OQ2 + PQ2 ⋯ Pythagoras
132 = 52 + PQ2
PQ2 = 169 ─ 25
= 144 = 122
PQ = 12
PR = 24 cm [ 1.2 ]
1.3 OQ ⊥ PR ⋯ gegee / given
∴ PQ = QR ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
PR = 15 cm; PQ = 7,5 cm (gegee / given)
In ΔOPQ : OP2 = OQ2 + PQ2 ⋯ Pythagoras
= 82 + 7,52
= 64 + 56,25
= 120,25
OP = √120,25 = 3
√481
OP = ───── cm = 10,966 cm
2 [ 1.3 ]
2.1 OM ⊥ AB ⋯ gegee / given
∴ AM = MB ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
AB=8 cm en/and OA=5 cm ⋯ gegee/given
∴ AM = 4 cm
x = 4 cm
In ΔOAM : OA2 = OM2 + AM2 ⋯ Pythagoras
52 = y2 + 42
y2 = 25 ─ 16
= 9 = 32
y = 3 cm [ 2.1 ]
2.2 ME ⊥ CD ⋯ gegee / given
∴ CE = ED ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
ME = 5 mm; MC = 13 mm; ⋯ gegee/given
In ΔMCE : MC2 = ME2 + CE2 ⋯ Pythagoras
132 = 52 + x2
x2 = 169 ─ 25
= 144 = 122
x = 12 mm
CD = 2 × CE
= 24
y = 24 mm [ 2.2 ]
2.3 MO ⊥ AB ⋯ gegee / given
∴ AM = MB ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
AB = 18 cm; OC = 15 cm; ⋯ gegee/given
OM = x; MON = y ⋯ gegee/given
AB = 18 cm ∴ AM = 9 cm
OA = OC = 15 cm
In ΔOAM : OA2 = OM2 + AM2 ⋯ Pythagoras
152 = x2 + 92
x2 = 225 ─ 81
= 144 = 122
x = 12 cm
ON ⊥ CD ⋯ gegee / given
∴ CN = ND ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
CD = 24 cm; CN = 12 cm
In ΔOCN : OC2 = ON2 + CN2 ⋯ Pythagoras
152 = ON2 + 122
ON2 = 225 ─ 144
= 81 = 92
ON = 9 cm
y = MON = MO + ON
= 12 + 9 cm
= 21 cm [ 2.3 ]
3.1 OM ⊥ AB ⋯ gegee / given
∴ AM = MB ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
AB=14 cm en/and OA=12 cm ⋯ gegee/given
∴ AM = 7 cm
In ΔOAM : OA2 = AM2 + OM2 ⋯ Pythagoras
122 = 72 + OM2
OM2 = 144 ─ 49
= 95
OM = √95 cm = 9,747 cm [ 3.1 ]
3.2 OB = AO = 12 cm⋯ radii en gegee /
radii and given
ON ⊥ BC ⋯ gegee / given
∴ BN = NC ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
In ΔONB : OB2 = BN2 + ON2 ⋯ Pythagoras
122 = BN2 + (√44)2
BN2 = 144 ─ 44
= 100 = 102
BN = 10 cm
BC = 2 × BN = 2 × 10
= 20 cm [ 3.2 ]
3.3 OP ⊥ AC as / if ∠APO = 90 °
In ΔOAP : OA2 = 122 = 144
AP2 = 42 = 16
OP2 = (√128)2 = 128
AP2 + OP2 = 16 + 128
= 144 = OA2
∴ ΔOAP is 'n reghoekige driehoek /
a right angled triangle
∠ OPA = 90 °
∴ OP ⊥ AP [ 3.3 ]
OS ⊥ MN ⋯ gegee / given
MS = SN en / and MN = 48 mm ⋯ gegee / given
MS = SN = 24 mm
In ΔOMS : OM2 = MS2 + OS2 ⋯ Pythagoras
= 242 + 72
= 625 = 252
OM = 25
In ΔOPR : OP2 = PR2 + OR2 ⋯ Pythagoras
252= PR22 + 52 ⋯ OP = OM
PR2 = 625 ─ 25 = 600
PR = 10√6
PQ = 2 × PR ⋯ OR ⊥ PQ
= 2 × 10√6 = 20√6
= 48,90 mm [ 4 ]
MS = SN en / and MN = 48 mm ⋯ gegee / given
MS = SN = 24 mm
In ΔOMS : OM2 = MS2 + OS2 ⋯ Pythagoras
= 242 + 72
= 625 = 252
OM = 25
In ΔOPR : OP2 = PR2 + OR2 ⋯ Pythagoras
252= PR22 + 52 ⋯ OP = OM
PR2 = 625 ─ 25 = 600
PR = 10√6
PQ = 2 × PR ⋯ OR ⊥ PQ
= 2 × 10√6 = 20√6
= 48,90 mm [ 4 ]
AB = AC = 15 mm ⋯ gegee / given
∴ ΔABC is 'n gelykbenige driehoek /
∴ ΔABC is an isosceles triangle.
AMD ⊥ BC ⋯ gegee / given
BD = DC = 9 ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
In ΔABD : AB2 = AD2 + BD2 ⋯ Pythagoras
152 = AD2 + 92
AD2 = 225 ─ 81 = 144 = 122
AD = 12
AM = AD ─ MD
Gestel / Let MD = x
AM = 12 ─ x
In ΔMBD : MB = AM ⋯ radii
MB2 = MD2 + BD2 ⋯ Pythagoras
(12 ─ x)2 = x2 + 92
144 ─ 24x + x2 = x2 + 81
24x = 144 ─ 81
21
x = ──
8
21 75
AM = 12 ─ ── = ── = 9,375 mm
8 8
75
= ── = 9,375 mm
8 [ 5 ]
∴ ΔABC is 'n gelykbenige driehoek /
∴ ΔABC is an isosceles triangle.
AMD ⊥ BC ⋯ gegee / given
BD = DC = 9 ⋯ lyn uit midpt ⊥ koord /
line from centre ⊥ chord
In ΔABD : AB2 = AD2 + BD2 ⋯ Pythagoras
152 = AD2 + 92
AD2 = 225 ─ 81 = 144 = 122
AD = 12
AM = AD ─ MD
Gestel / Let MD = x
AM = 12 ─ x
In ΔMBD : MB = AM ⋯ radii
MB2 = MD2 + BD2 ⋯ Pythagoras
(12 ─ x)2 = x2 + 92
144 ─ 24x + x2 = x2 + 81
24x = 144 ─ 81
21
x = ──
8
21 75
AM = 12 ─ ── = ── = 9,375 mm
8 8
75
= ── = 9,375 mm
8 [ 5 ]
Trek OE ⊥ ABCD / Draw OE ⊥ ABCD
OE ⊥ ABCD en dus / and thus
BE = EC; AE = ED
BE = EC = 3 ⋯ BC = 6
In ΔOBE : OB2 = OE2 + BE2 ⋯ Pythagoras
52 = OE2 + 32
OE2 = 25 ─ 9 = 16 = 42
OE = 4
In ΔOED : OD2 = ED2 + OE2 ⋯ Pythagoras
8,52 = ED2 + 42
ED2 = 72,25 ─ 16 = 56,25 = 7,52
ED = 7,5
AD = 2 × ED = 2 × 7,5
AD = 15 cm [ 6 ]
OE ⊥ ABCD en dus / and thus
BE = EC; AE = ED
BE = EC = 3 ⋯ BC = 6
In ΔOBE : OB2 = OE2 + BE2 ⋯ Pythagoras
52 = OE2 + 32
OE2 = 25 ─ 9 = 16 = 42
OE = 4
In ΔOED : OD2 = ED2 + OE2 ⋯ Pythagoras
8,52 = ED2 + 42
ED2 = 72,25 ─ 16 = 56,25 = 7,52
ED = 7,5
AD = 2 × ED = 2 × 7,5
AD = 15 cm [ 6 ]
