WISKUNDE
GRAAD 10
NOG OEFENINGE
Vierhoeke - parallelogramme, ens. : antwoorde
MATHEMATICS
GRADE 10
MORE EXERCISES
Quadrilaterals, parallelograms, etc. : answers.
Antwoord / Answer 1.1
∠P + ∠S
1 + 2 = 180°
. . . ko-binnehoeke, PQ || RS / co-int. angles; PQ || RS
x = 180° − (34° + 46°)
. . . gegee / given
= 100°
∠R = ∠P
. . . regoorst. ∠'e, ||
mPQRS / vert. opp. ∠'s, ||
mPQRS
y = 100°
Antwoord / Answer 1.2
QR = PS
. . . oorstaande sye, ||
m PQRS / opp. sides, ||
m PQRS
= QT
. . . gegee / given
∠T
1 + T
2 = 180°
. . . STR 'n reguit lyn / STR a line
∠T
1 = 180° − 130°
. . . gegee / given
= 50°
x = ∠P = ∠R
. . . oorst. ∠'e, ||
m PQRS / opp. ∠'s, ||
m PQRS
= ∠T
1 . . . QT = QR / QT = QR
= ∠50°
∠S + ∠R = 180°
. . . ko-binne ∠'e, PS || QR / co-int. ∠'s, PS || QR
y = ∠S = 180° − 50°
= 130°
Antwoord / Answer 1.3
SR = PQ
. . . oorstaande sye, ||
m PQRS / opp. sides, ||
m PQRS
= ST
. . . gegee / given
∠R
1 = ∠P = 115°
. . . oorstaande ∠'e, ||
m PQRS; / opp. ∠'s, ||
m PQRS;
∠R
1 + ∠R
2 = 180°
. . . QRS 'n rt. lyn / QRS a str. line
∠R
2 = 180° − 115°
= 65°
x = ∠T = ∠R
2 . . . SR = ST
= 65°
∠R
1 = ∠S
3 + ∠T
. . . buitehoek van ΔSRT / ext. ∠ of ΔSRT
y = ∠S
3 = R
1 − ∠T
= 115° − 65°
= 50°
Antwoord / Answer 1.4
∠T + ∠P + ∠M = 180°
. . . binnehoeke van ΔTPM / int. ∠'s of ΔTPM
∠P = 180° − (∠T + ∠M)
x = 180° − (32° + 22°)
. . . gegee / given
= 126°
∠S
1 + ∠P = 180°
. . . binnehoeke van ||
m PQRS / int. ∠'s of ||
m PQRS
∠S
1 = 180° − ∠P
= 180° − 126°
= 54°
∠S
1 = R
1 + ∠M
. . . buitehoek van ΔRSM / ext. ∠ of ΔRSM
∠R
1 = S
1 − ∠M
y = 54° − 22°
= 32°
OF / OR
PQ || RS
. . . teenoorst. sye van ||
m PQRS / opp. sides of ||
m PQRS
∴ TQP || RS
. . . TQP 'n rt. lyn / TQP a str. line
∠R
1 = ∠T
. . . ooreenkomstige ∠'e;; TQP || RS / corresponding ∠'s; TQP || RS
y = 32°
Antwoord / Answer 2.
2.1
AD || BC
. . . teenoorst. sye van ||
m ABCD / opp. sides of ||
m ABCD
∴ AD || CE
. . . BCE 'n rt. lyn / BCE a str. line
AC || DE
. . . gegee / given
∴ ACED is 'n parallelogram.
. . . oorst. pare sye ewewydig.
∴ ACED is a parallelogram.
. . . opp. pairs of sides parallel.
2.2
BC = AD
. . . teenoorst. sye van ||
m ABCD / opp. sides of ||
m ABCD
CE = AD
. . . teenoorst. sye van ||
m ACED / opp. sides of ||
m ACED
∴ BC = CE
Antwoord / Answer 3.
3.1
PT = TR en / and ST = TQ
. . . hoeklyne van ||
m PQRS halveer mekaar /
. . . diagonals of ||
m PQRS bisect one another.
A is die middelpunt van PT en B van TR
. . . gegee
A is the midpoint of PT and B of TR
. . . given
∴ AT = TB
AT = TB en / and ST = TQ
∴ AQBS is 'n ||
m / a ||
m . . . hoeklyne halveer mekaar / diagonals bisect each other.
3.2
AS || QB
. . . teenoorst. sye van ||
m AQBS / opp. sides of ||
m AQBS
∠A
1 = ∠B
2 . . . verwis. ∠'e; AS || QB / alt. ∠'s; AS || QB
Antwoord / Answer 4.
{Om te bewys dat KPMQ 'n parallelogram is, kan ons bewys dat een paar oorstaande sye
gelyk en ewewydig is.
To prove that KPMQ is a parallelogram, we can prove that one pair of opposite sides
are equal and parallel. }
KL = NM en / and KL || NM
. . . teenoorstaande sye van ||
m KLMN /
. . . opp. sides of ||
m KLMN.
PL = NQ
. . . gegee / given.
∴ KL − PL = NM − NQ
∴ KP = QM en / and KP || MQ
∴ KPMQ is 'n parallelogram/ a parallelogram
. . . teenoorstaande sye = en ||
. . . opp. sides = and ||
Antwoord / Answer 5.
{ ∠E = ∠DCE as DC = DE . . . basishoeke van gelykbenige driehoek is gelyk.
∠E = ∠DCE if DC = DE . . . base angles of an isosceles triangle are equal. }
DE = AB
. . . gegee / given.
CD = AB
. . . teenoorstaande sye van ||
m ABCD / opp. sides of ||
m ABCD.
∴ DC = DE
∴ ∠E = ∠DCE
. . . basis∠'e van gelykbenige Δ DCE / base ∠'s of isosceles Δ DCE
Antwoord / Answer 6.
{ EM = MF as / if Δ AEM ≡ ΔCFM.
AD || BC }
In Δ AEM en / and ΔCFM
1. ∠ A
1 = ∠C
1 . . . verwis. ∠'e, AD || BC / alt. ∠'s, AD || BC.
2. ∠ E
1 = ∠F
1 . . . verwis. ∠'e, AD || BC / alt. ∠'s, AD || BC.
3. AM = MC
. . . hoeklyne van ||
m ABCD halveer mekaar. /
diagonals of ||
m ABCD bisect each other.
∴ ΔAEM ≡ ΔCFM
. . . HHS / AAS.
EM = MF
6.2
AM = MC
. . . hoeklyne van ||
m ABCD halveer mekaar. /
diagonals of ||
m ABCD bisect each other.
EM = MF
. . . bewys in 6.1 / proved in 6.1
∴ AFCE is 'n parallelogram / a parallelogram
. . . hoeklyne halveer mekaar. /
diagonals bisect each other.
Antwoord / Answer 7.
{ KP = QM as / if Δ KLP ≡ ΔMNQ. }
7.1 In Δ KLP en / and ΔMNQ
1. ∠ K = ∠M
. . . oorst. ∠'e, parm KLMN || BC / opp. ∠'s,&nbs;parm. KLMN.
2. ∠ L
1 = ∠N
1 . . . gegee / given
3. KL = MN
. . . oorstaande sye van ||
m KLMN. / opp. sides of ||
m KLMN.
∴ ΔKLP ≡ ΔMNQ
. . . HHS / AAS.
KP = QM
7.2
KN = LM en / and KN || LM
. . . oorst. sye van ||
m KLMN. / opp. sides of ||
m KLMN.
KP = QM
. . . bewys in 7.1 / proved in 7.1
∴ KN − KP = LM − QM
∴ PN = LQ en / and PN || LQ
∴ PLQN is 'n parm. / a parm.
. . . paar oost. sye = en || / pair of opp. sides = and ||