WISKUNDE
GRAAD 11
NOG OEFENINGE
Koordevierhoeke : antwoorde.
MATHEMATICS
GRADE 11
MORE EXERCISES
Cyclic quadrilaterals : answers.
Antwoord/Answer 1.
∠A = 180° − (∠B + ∠C) (binnehoeke van / int. angles of ΔABC.)
= 180° − (45° + 30°)
= 105°
∠D1 = ∠E2 + ∠C (buite ∠ / ext. ∠ of ΔEDC.)
= 45° + 30°
= 75°
∠A + ∠D1 = 105° + 75°
= 180°
∴ ABDE is 'n koordevierhoek / ABDE is a cyclic quadrilateral . . .
som teenoorst. ∠'e = 180° / sum int. opp. ∠ = 180°
Antwoord/Answer 2.
Punte P, S, T en R is konsiklies as hulle op
Points P, S, T and R are concyclic if they lie on
die omtrek van 'n sirkel lê, d.w.s. as PSTR
the circumference of a circle, i.e. if PSTR is
'n koordevierhoek is.
a cyclic quadrilateral.
Dus, bewys PSTR is 'n koordevierhoek.
Therefore, prove PSTR is a cyclic quadrilateral.
∠P = 90° (∠ in halwe sirkel / ∠ in semi-circle)
∠STR = 90° (ST ⊥ QR)
∠P + ∠STR = 180°
∴ PSTR is 'n koordevierhoek / PSTR is a cyclic quadrilateral& . . .
(oorstaande ∠'e suppl. / opp. int. ∠'s suppl.)
Antwoord/Answer 3.
∠A1 = B1 (AE = BE)
∠A1 = ∠C2
OF / OR
∠B1 = ∠D2
(verwis. ∠'e / alt. ∠'s, AB || CD)
(verwis. ∠'e / alt. ∠'s, AB || CD)
∠A1 = ∠D2 (albei / both = ∠B1)
∠B1 = ∠C2 (albei / both = ∠A1)
Albei onderspan deur BC /
Albei onderspan deur AD /
Both subtended by∠BC
Both subtended by∠AD
∴ ABCD 'n koordevierhoek / a cyclic quadrilateral
Antwoord/Answer 4.
∠D + ∠CED + ∠C = 180° (Hoeke van / Angles of ΔCED = 180°)
∠D =180° − (∠CED + ∠C) = 180° − (100° + 30°)
= 50°
= ∠A
BC onderspan gelyke hoeke / BC subtends equal angles
∴ ABCD is 'n koordevierhoek / a cyclic quadrilateral
Antwoord/Answer 5.
∠P + ∠Q + ∠R + ∠S = 360° (Hoeke van / Angles of vierhoek / quadrilateral PQRS = 360°)
3x° + x° + 2x° + 4x° = 360°
10x° = 360°
x = 36°
∠P = 108°, ∠Q = 36°, ∠R = 72°, ∠S = 144°
∠P + ∠R = 108° + 72° OF / OR
∠Q + ∠S = 36° + 144°
= 180°
= 180°
∴ Teenoorstaande binnehoeke = 180° /
Opposite interior angles = 180°
∴ PQRS is 'n koorde vierhoek. /
PQRS is a cyclic quadrilateral.
Antwoord/Answer 6.
∠DAB + ∠DAE = 180°
OF / OR
∠DCB + ∠DCF = 180°
(BAE 'n rt. lyn / BAE a str. line)
(BCF 'n rt. lyn / BCF a str. lin)
∠DAB + 34° = 180°
∠DCB + 146° = 180°
∠DAB = 146°
∠DCB = 34°
∠DAB = 146° = ∠DCF
∠DCB = 34°° = ∠DAE
∴ Bnnehoek = buitehoek / exterior angle = interior angle
∴ ABCD is 'n koordevierhoek / ABCD is a cyclic quadrilateral
Antwoord/Answer 7.
∠MKN + ∠KMN + ∠KNM = 180°
(Binnehoeke van ΔKNM = 180° / Int. ∠'s of ΔKNM = 180°)
∠MKN + 35° + 70° = 180°
∠MKN = 75°
∠MKN = 75° = ∠MLN
NM onderspan gelyke hoeke / NM subtends equal angles.
KLMN is 'n koordevierhoek / KLMN is a cyclic quadrilateral.
Antwoord/Answer 8.
∠ABD = ∠ACD = x°
(gegee / given)
AD onderspan gelyke hoeke / AD subtends equal angles
∴ ABCD is 'n koordevierhoek. / ABCD is a cyclic quadrilateral.
∠ACB = ∠ABC (AB = AC)
∠ACB = ½ (180° − ∠BAC)
= 90° − ½ ∠BAC
∠ADB = ∠ACB (AB onderspan gelyke ∠'e / subtands equal ∠'s)
∠BDC = ∠BAC (BC onderspan gelyke ∠'e / subtands equal ∠'s)
∠ADC = ∠ADB + ∠BDC
= (90° − ½ ∠BAC) + ∠BAC
= 90° + ½ ∠BAC
Antwoord/Answer 9.
∠KLN = ∠NOM (buitehoek van kv. LNOM / ext. ∠ of Cyc. quad. LNOM)
∠NOM = ∠PNO + ∠P (buitehoek van ΔPNO / ext. ∠ of ΔNPO)
∠K + ∠KLN + ∠KNL = 180° (binnehoeke van ΔKLN / int. ∠ of ΔKLN)
42° + ∠P + ∠PNO + ∠KNL = 180°
42° + 42° + 2 ∠KNL = 180° (∠PNO = ∠KNL, regoorstaande hoeke / vert. opp. ∠'s)
2 ∠KNL = 96°
∠KNL = 48°
∠M = ∠KNL (buitehoek van kv. LNOM / ext. ∠ of Cyc. quad. LNOM)
= 48°
Antwoord/Answer 10.
10.1
In ΔBDE en / andΔBDC
i). DE = DC (gegee / given)
ii). DE = DC (gegee / given)
iii). DE = DC (gegee / given)
∴ ΔBDE ≡ ΔBDC (SHS / SAS)
∴ ∠DBE = ∠DBC (ΔBDE ≡ ΔBDC)
10.2
∠ABD = ∠BDC (verwis. ∠'e, / alt. ∠'s, AB ∥ DC)
∠BDC = ∠BDE (gegee / given)
∴ ∠ABD = ∠BDE (gegee / given)
Maar / But ∠DBE = ∠DBC (10.1)
∠DBC = ∠ADB (verwis. ∠'e, / alt. ∠'s, AD&nbap;∥ BC)
∴ ∠EBD = ∠ADB (gegee / given)
∠ABE + ∠EBD = ∠ADE + ∠ADB
∴ ∠ABE = ∠ADE
∴ AE onderspan gelyke hoeke / subtands equal angles)
∴ ABDE is 'n koordevierhoek. / a cyclic quadrilateral)
10.3
∠ABD = ∠EDB (10.2)
ΔFBD is gelykbenige Δ / ΔFBD is an equilateral Δ (basishoeke = / base ∠s =)
FB = FD
FA + AB = FE + ED
AB = CD (oorst. sye van / opp. sides of ∥m ABCD)
= ED (gegee / given)
FA = FE
Antwoord/Answer 11.
11.1
∠PQR = 90° (∠ in halwe sirkel / ∠ in semi-circle)
∠PQR + ∠PQB = 180° (RQB 'n rt. lyn / RQB a str. line)
∠PQB = 90°
In ΔBDE en / and ΔBDC
∠PQR = ∠PQB (elk / each = 90°)
∠PRQ = ∠QPB (raaklyn / tangent PAB, koord / chord PQ)
∴ ∠RPQ = ∠B (oorblywende ∠'e van, / remaining ∠'s of ΔPQR en / and ΔPQB)
11.2
Laat RC en PQ mekaar in T sny. / Let RC and PQ intersect at T
In ΔCTQ en / and ΔPTR
∠CTQ = ∠PTR (regoorst. ∠'e / vert. opp. ∠'s)
∠TQC = ∠PRT (onderspan deur PC / subtended by PC)
∴ ∠TCQ = ∠RPT (oorblywende ∠'e van, / remaining ∠'s of ΔCTQ en / and ΔPTR)
∴ ∠TCQ = ∠B (∠RPT = ∠B (10.1))
∴ ABQC is 'n koordevierhoek (buithoek = teenoorst. binnehoek) /
ABQC is a cyclic quadrilateral, (ext. ∠ = opp. int. ∠)