WISKUNDE
GRAAD 10
NOG OEFENINGE
Hoeke en driehoeke : antwoorde.
MATHEMATICS
GRADE 10
MORE EXERCISES
Angles and triangles : answers.
Antwoord / Answer 2.1
∠A = ∠C
. . . AB = BC
∠DBC = ∠A + ∠C
. . . buitehoek ΔABC / ewt. ∠ of ΔABC
= 2x
2x = 114°
. . . ∠DBC = 114°
x = 57°
∠C = 57°
Antwoord / Answer 2.2
In ΔDBC
: ∠BCD = ∠D
. . . BD = BC
∠ABC = ∠D + ∠BCD
. . . Buitehoek ΔDBC / Ext. ∠ ΔDBC
= 2x
In ΔABC
:
∠ABC + ∠A + ∠ ACB = 180°
. . . Binnehoeke ΔABC / Int. ∠'s ΔABC
2x + 2x − 4° + x + 4° = 180°
. . . gegee / given
5x = 180°
x = 36°
∠A = 2(36°) − 4° = 68°
∠ABC = 2(36°) = 72°
∠ACB = 36° + 4° = 40°
∠D = 36°
Antwoord / Answer 2.3
∠ABE + ∠EBC = 180°
. . . ABC rt. lyn / ABC a str. line
∠EBC = 180° − (x + 60°)
= 120° − x
∠ECD = ∠E + ∠EBC
. . . Buitehoek ΔEBC / Ext. ∠ ΔEBC
2x + 40° = 70° + 120° − x
3x = 150°
x = 50°
∠ABE = 50° + 60° = 110°
∠ECD = 2(50°)+ 40° = 140°
Antwoord / Answer 2.4
∠PQB = ∠C
. . . ooreenk. ∠'e, PQ || AB / correspon. ∠'s, PQ || AC
∠APQ = ∠B + ∠PQB
. . . buitehoek ΔPBQ / ext. ∠ ΔPBQ
125° − x = 62° + 2x + 30°
125° − 92° = 2x + x
x = 11°
Antwoord / Answer 2.5
In ΔTQR
: ∠QTR + ∠TQR +∠TRQ = 180°
. . . binne ∠'e, ΔTQR / int. ∠'s, ΔTQR
∠TQR + ∠TRQ = 180° − 110°
x + y = 70°
In ΔPQR
: ∠QPR + ∠PQR +∠PRQ = 180°
. . . binne ∠'e, ΔPQR / int. ∠'s, ΔPQR
x + 2x + 2y = 180°
x + 2(x + y) = 180°
x + 2 × 70° = 180°
x = 40°
y = 70° − 40° = 30°
∠P = ∠PQT = ∠TQR = 40°
∠PRT = ∠TRQ = 30°
Antwoord / Answer 2.6
∠ACD = ∠BAC + ∠B
. . . buite ∠ ΔABC / ext. ∠ ΔABC
130° = x + 50°
x = 80°
∠A
1 = 80°
∠ADE = 2(80°) − 5° = 155°
Antwoord / Answer 3
∠ADB = ∠A + ∠ACB
. . . buite ∠ ΔABC / ext. ∠ ΔABC
p = r + t
∠ACE = ∠A + ∠ABC
. . . buite ∠ ΔABC / ext. ∠ ΔABC
q = r + s
p + q = r + t + r + s
= r + t + s + r
= 180° + r
Antwoord / Answer 4
4.1 In ΔABD en / and ΔCDB
(i) AB = CD
. . . gegee / given
(ii) AD = BC
. . . gegee / given
(iii) BD = BD
ΔABD ≡ ΔCDB
. . . SSS / SSS
4.2 ∠B
1 = ∠D
2 . . . ΔABD ≡ ΔCDB
4.3 ∠D
1 = ∠B
2 . . . ΔABD ≡ ΔCDB
Hulle is verwis. ∠'e / They are alt. 's
 
∴ AD || BC
Antwoord / Answer 5
5.1 ∠B
1 = ∠B
2 as / if ΔABD ≡ ΔCBD
Bewys dus dat ΔABD ≡ ΔCBD / Therefore prove that ΔABD ≡ ΔCBD
In ΔABD en / and ΔCDB
(i) AB = BC
. . . gegee / given
(ii) AD = CD
. . . gegee / given
(iii) BD = BD
ΔABD ≡ ΔCDB
. . . SSS
∴ ∠B
1 = ∠B
2
5.2 AE = EC as / if ΔABD ≡ ΔCDB
Bewys dus dat / Therefore prove that ΔABD ≡ ΔCDB
In ΔABE en / and ΔCBE
(i) AB = BC
. . . gegee / given
(ii) ∠B
1 = ∠B
2 . . . bewys in 5.1 / proved in 5.1
(iii) BE = BE
ΔABE ≡ ΔCBE
. . . S, ∠, S
∴ AE = EC
5.3 ∠E
1 = ∠E
2 . . . ΔABE ≡ ΔCBE
∠E
1 + ∠E
2 = 180°
. . . AEC 'n rt. lyn / AEC a straight line.
∴ ∠E
1 = ∠E
2 = 90° en / and BD ⊥ AC
Antwoord / Answer 6
6.1 PQ = RS as / if ΔPOQ ≡ ΔSOR
Bewys dus dat ΔPOQ ≡ ΔSOR / Therefore prove that ΔPOQ ≡ ΔSOR
In ΔPOQ en / and ΔSOR
(i) PO = OR
. . . radii
(ii) ∠O
1 = ∠O
3 . . . gegee / given
(iii) OQ = OS
. . . radii
ΔPOQ ≡ ΔSOR
. . . S, ∠, S
∴ PQ = RS
6.2 ∠P = ∠S
. . . ΔPOQ ≡ ΔROS
Antwoord / Answer 7
7.1 LM = MN as / if ΔKML ≡ ΔKMN
Bewys dus dat ΔKML ≡ ΔKMN / Therefore prove that ΔKML ≡ ΔKMN
In ΔKML en / and ΔKMN
(i) KL = KN
. . . gegee / given
(ii) ∠K
1 = ∠K
2 . . . KM halveer ∠LKN / KM bisects ∠LKN
(iii) KM = KM
ΔKML ≡ ΔKMN
. . . S, ∠, S
∴ LM = MN
7.2 ∠M
1 = ∠M
2 . . . ΔKLM ≡ ΔKNM
∠M
1 + ∠M
2 = 180°
. . . KMN 'n rt. lyn / KMN a straight line.
∴ ∠M
1 = ∠M
2 = 90° en / and KM ⊥ LN
Antwoord / Answer 8
8.1
In ΔKLM en / and ΔKNM
(i) ∠K
1 = K
2 . . . KM halveer ∠LKN / KM bisects ∠LKN
(ii) ∠L = ∠N
. . . gegee / given
(iii) KM = KM
ΔKLM ≡ ΔKNM
. . . ∠, ∠, S
8.2 LM = MN
Antwoord / Answer 9
9.1
In ΔPQR en / and ΔPQS
(i) ∠P
1 = ∠Q
1 . . . gegee / given
(ii) ∠R = ∠S
. . . gegee / given
(iii) PQ = PQ
ΔPQR ≡ ΔPQS
. . . ∠, ∠, S
9.2 PR = QS
. . . ΔPQR ≡ ΔPQS
PT + TR = QT + TS
In ΔPQT
:
PT = QT
. . . ∠P
1 = ∠Q
1
∴ TR = TS
OF / OR
In ΔQRT en / and ΔPST
(i) ∠T
1 = ∠T
3 . . . regoorstaande ∠'e / vert. opp. ∠'s
(ii) ∠R = ∠S
. . . gegee / given
(iii) QR = PS
. . . ΔPQR ≡ ΔPQS
ΔQRT ≡ ΔPST
. . . ∠, ∠, S
∴ TR = TS