Chapter 1
Circle geometry
Exercise 1.1
1 (a) x = 70 (angle at centre on equal chords)
y = 55 (DOCD is isosceles, OC = OD, radii)
(b)∠FOE = 110°(DOFE is isosceles,
(c)∠OLM = 32°
x = 64
y = 16
(d)x = 60
(DOLM is isosceles,
OL = OM, radii)
(exterior angle is sum of
interior opposite angles)
(angles at centre are equal
so chords are equal)
(angle at centre is twice
angle at circumference)
(e) y = 160(angle at centre is twice
angle at circumference)
∠COE = 140°(angle at centre is twice
angle at circumference)
reflex ∠COE = 220° (angles at a point)
z = 110(angle at centre is twice
angle at circumference)
pl
m
Sa
3 (a)
∠DCB = 115° (angle sum of triangle)
reflex ∠DOB = 230° (angle at centre is twice
angle at circumference)
∠DOB = 130°
(angles at a point)
x = 65(angle at centre is twice
angle at circumference)
(b)
5 Draw a diagram:
D
E
40°
F
DODE ≡ DOEF (SSS)
ODE = ∠OED (isosceles
∠
triangle)
\ ∠ODE = ∠OED = 70°
\ ∠DEF = 140°
O
(OC = OB, radii)
7 DOCB is isosceles.
∠OCB = ∠OBC(base angles of isosceles
triangle OCB are equal)
∠OCB = ∠OBC = 38° (angle sum of DOCB is 180°;
[180 - 104] ÷ 2)
\ ∠ACO = 71° - 38° = 33°
DOCA is isosceles. (OC = OA, radii)
∠OCA = ∠OAC
(base angles of isosceles
triangle OCA are equal)
e
(f)
(opposite angles of a rhombus are equal)
∠OHK = z°(DOHK is isosceles)
2z° = 56°
(exterior angle is sum of
interior opposite angles)
z = 28
pa
ge
s
y = 25
OF = OE, radii)
(angles at centre are equal
so chords are equal)
(c) OHPG is a rhombus, so ∠HOG = 56°.
∠COE = 2y°
(angle at centre is twice
angle at circumference)
reflex ∠COE = 4y° (angle at centre is twice
angle at circumference)
6y° = 360° (angles at a point)
y = 60
\ ∠OAC = 33°
9Let ∠OCA = x°.
DAOC is isosceles.
(OC = OA, radii)
\ ∠OCA = ∠OAC = x° (base angles of isosceles
triangle OCA are equal)
\ ∠AOC = (180 - 2x)° (angle sum of triangle)
\ ∠AOB = (180 - 2x)° (equal chords subtend
equal angles at the centre, AC = AB)
But DAOB is isosceles. (OB = OA, radii)
\ ∠OAB = ∠OBA = x° (angle sum of triangle)
\ ∠CAB = 2x°
\ ∠XAC = (180 - 2x)° (XAB is a straight line)
\ ∠XAC = ∠AOC (both equal [180 - 2x]°)
Exercise 1.2
1 The line segment from the centre of the circle to
chord PQ is perpendicular to PQ.
102 - 82 = 100 - 64
= 36
The distance is 6 mm.
3 C is the correct answer.
82 + 152 = 289 = 17
5Let N be the midpoint of ML and join ON.
∠ONL = 90° (line joining midpoint of chord to
centre of circle ^ the chord)
Similarly ∠OAL = 90°.
But ∠ALM = 90°. (given; same as ∠KLM)
\ ∠AON = 90°
(angle sum of a quadrilateral
is 360°)
\ AONL is a rectangle. (each angle 90°)
Chapter 1 Circle geometry
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\ LN = OA (opposite sides of a rectangle)
But ML = 2LN. (N is the midpoint of ML)
\ ML = 2OA
7 XF = FY
(perpendicular from centre of circle to
chord bisects chord)
∠CEF = ∠OFG = ∠DGF = 90°
\ CE  OF  DG (pairs of cointerior angles
supplementary)
But CO = OD. (equal radii)
\ EF = FG (if system of parallel lines cut by equal
intercepts on one transversal, then intercepts
on other transversals are also equal)
EX = EF - XF
and: YG = FG - FY
so: EX = YG
Exercise 1.3
1Join OP and OQ:
B
P In DOAQ and DOAP:
OA is a common side.
OQ = OP (radii)
O
QA
= PA (given)
\ DOAQ ≡ DOAP (SSS)
A
\ ∠OAQ = ∠OAP (corresponding angles in
congruent triangles)
\ BA bisects ∠PAQ.
pa
ge
s
Q
7In DAOB and DAPB:
OA = PA (radii of equal circles)
Similarly OB = PB.
AB is a common side.
\ DAOB ≡ DAPB (SSS)
\ ∠OAB = ∠PAB (corresponding angles of
congruent triangles)
In DAOC and DAPC:
OA = PA (radii of equal circles)
∠OAB = ∠PAB
AC is a common side.
\ DAOC ≡ DAPC (SAS)
(corresponding sides of
\ OC = CP
congruent triangles)
3 O and P are equidistant from KN. (equal chords
in equal circles are equidistant from the centre)
Hence KN  OP.
Sa
m
pl
e
5 ∠OBM = 90° (line joining midpoint of chord
to centre of circle ^ the chord)
Similarly ∠OAK = 90°.
But ∠BOA = 90°. (angle sum of quadrilateral
AOBX is 360°)
\ AOBX is a rectangle.
But OA = OB. (equal chords are equidistant from
the centre of the circle)
AOBX is a square. (rectangle with a pair of
adjacent sides equal)
9Let A be the midpoint of CD and B be the
midpoint of FG.
\ ∠OAD = 90° (line joining midpoint of chord
to centre of circle ^ the chord)
Similarly ∠OBG = 90°.
In DOAE and DOBE:
∠OAE = ∠OBE = 90°
OE is a common side.
OA = OB (radii)
\ DOAE ≡ DOBE (RHS)
\ AE = BE (corresponding sides of
congruent triangles)
But DE = AE - AD and GE = BE - BG.
But AD = BG. (half of equal chords)
\ DE = GE
Exercise 1.4
1 (a) x = 40
(angle at centre is twice
angle at circumference)
(b)x = 100
(angle at centre is twice
angle at circumference)
y = 50 (isosceles triangle)
(c) z + 100 = 130
z = 30
(d)x = 230
(angle at centre is twice
angle at circumference)
(angle at centre is twice
angle at circumference)
(e) x = 60 (angles in the same segment)
y = 40 (angle sum of triangle)
2
z = 40 (angles in the same segment)
(f) x = 90 (angle in a semicircle)
(g)c = 15 (angles in the same segment)
b + 15 = 90
(exterior angle is sum of
interior opposite angles)
b = 75
a = 75 (angles in the same segment)
(h) a = 68 (base angles of isosceles triangle,
equal radii)
b = 22 (angle in a semicircle)
c = 22 (base angles of isosceles triangle,
equal radii)
New Senior Mathematics Extension 1 for Years 11 & 12 Student Worked Solutions
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b = 50 (j) a = 50 b = 25 c = 25 (k) y = 40 a = 28 b = 28 x = 40 (l) a = 42 z = 55 y = 42 x = 35 (m) e = 50 f = 50 a = 40 d = 40 c = 40 b = 50 (n) x = 28 y = 28 z = 28 (o)x = 47 y = 22
(angle in a semicircle,
angle sum of triangle)
(angles in the same segment)
(angles in the same segment)
(angle sum of triangle)
(angles in the same segment)
(angle in a semicircle)
(angle sum of triangle)
(angles in the same segment)
(angles in the same segment)
(angle in a semicircle)
(angle sum of triangle)
(angles in the same segment)
(angles in the same segment)
(angle in a semicircle)
(angles in the same segment)
(base angles of isosceles triangle)
(angles in the same segment)
(angles in the same segment)
(angle in a semicircle)
(alternate angles)
(angles in the same segment)
(angles in the same segment)
(angles in the same segment)
(exterior angle is sum of
interior opposite angles)
3 Draw a diagram:
9 DOCP is isosceles. (OC = CP, given)
\ ∠COP = ∠CPO = a°
∠BCO = 2a°
(exterior angle is sum of
interior opposite angles)
DOCB is isosceles. (OC = OB, radii)
\ ∠OBC = ∠OCB = 2a°
\ ∠BOC = (180 - 4a)° (angle sum of triangle)
But AOP is a straight line so:
∠AOB = 180° - (180 - 4a)° - a°
\ ∠AOB = 3a°
1Let ∠ABC = y°.
1
E
Sa
m
B
pl
C
D
∠CAB = ∠CDB (angles in the same segment)
∠AEC = ∠DEB (vertically opposite angles)
∠ACD = ∠ABD (angles in the same segment)
\ DAEC and DDEB are equiangular.
5 (a) (angle in a semicircle)
(b)(both 90°)
(c) (angles in the same segment)
(d)(sum of the angles of a triangle is 180°)
7 Draw a diagram:
\ ∠ADC = y°
(opposite angles of a
parallelogram are equal)
(angles in the same
\ ∠AEC = ∠ADC = y°
segment standing on AC)
(alternate angles,
and: ∠DAE = ∠AEC = y° DA  EB)
\ ∠AEB = ∠ABE = y°
\ DABE is isosceles. (base angles of triangle equal)
13 (a)In DCDE and DBAE:
∠DCA = ∠DBA
(angles in the
same segment)
∠DEC = ∠AEB
(vertically opposite angles)
DC = AB
(given)
\ DCDE ≡ DBAE (AAS)
e
A
∠DAC = ∠DBC (angles in the same segment)
∠DAC = ∠ACB (alternate angles equal,
DA  CB)
\ ∠DBC = ∠ACB
Hence DEBC is isosceles. (two angles equal)
∠ADB = ∠ACB (angles in the same segment)
\ ∠ADB = ∠DAC
Hence DEDA is isosceles. (two angles equal)
pa
ge
s
(i) a = 50
(b)(continuing …)
\ DE = AE (corresponding sides of
congruent triangles)
\ DEDA is isosceles.
(c) ∠CDE = ∠BAE (corresponding angles of
congruent triangles)
But ∠EDA = ∠EAD (base angles of isosceles
triangle equal)
\ ∠CDA = ∠BAD (sum of equal parts)
In DADC and DDAB:
∠CDA = ∠BAD (from above)
DA is a common side.
DC = AB (given)
\ DADC ≡ DDAB (AAS)
(d)(continuing …)
A
B
\ AC = DB (corresponding sides of
congruent triangles)
E
C
D
Chapter 1 Circle geometry
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15 There is only one circle that passes through A, B
and all the positions of P (or: A, B and all the
positions of P are concyclic).
A
∠APB is the same size
no matter where P is on
the circle.
7 Draw a diagram:
1
A
28°
53°
B
O
P
C
B
D
∠OAC = 25° (DOAB is isosceles; 53° - 28° = 25°)
So ∠AOB = 74°. (angle sum of triangle is 180°;
[180 - 53 - 53]° = 74°)
∠ACB = 37° (angle at circumference is
half the angle at the centre)
∠ADC = 65° (37° stands on AB and
28° stands on BC)
Exercise 1.5
1 (a) x = 82
(opposite angles of a cyclic
quadrilateral are supplementary)
(opposite angles of a cyclic
y = 102
quadrilateral are supplementary)
(b)x = 65
c = 53 (opposite angles of kite are equal)
d = 42
(angles at a point,
sum of angles of a quadrilateral)
( j) x = 109 (cointerior angles supplementary,
reflex angle at centre is twice angle
at circumference)
(k) x = 52 (reflex angle at centre is twice angle
pl
e
(angle in a semicircle,
angle sum of triangle)
y = 115
(opposite angles of a cyclic
quadrilateral are supplementary)
pa
ge
s
P
(c) x = 54
(l) b = c = 109
(base angles of isosceles triangle
are equal, exterior angle of a triangle)
a = d = 71
(opposite angles of a cyclic
quadrilateral are supplementary)
m
(opposite angles of a cyclic
quadrilateral are supplementary, base
angles of an isosceles triangle are equal)
(opposite angles of a cyclic
quadrilateral are supplementary)
(e) x = 76
(opposite angles of a cyclic
quadrilateral are supplementary,
straight angle equals 180°)
Sa
(d)x = 65
3 (a)correct
a = 108 (straight angle)
(b)correct
b = 100
(f) a = 105 (base angles of isosceles triangle equal,
straight angle)
(opposite angles of a cyclic
quadrilateral are supplementary)
c = 55 (angle sum of triangle)
b = 100
(g)a = 155 (straight angle)
b = 25
(opposite angles of a cyclic
quadrilateral are supplementary)
at circumference, straight angle)
(opposite angles of a cyclic
quadrilateral are supplementary)
(c)incorrect
(d)correct
c = 72
(opposite angles of a cyclic
quadrilateral are supplementary)
5 Draw a diagram:
P
(h) a = 122
(opposite angles of a cyclic
quadrilateral are supplementary)
b = 18 (angle sum of triangle)
(i) a = 74 (angle at centre is twice the angle at
the circumference)
b = 106
(opposite angles of a cyclic
quadrilateral are supplementary)
4
A
Q
X
B
Y
New Senior Mathematics Extension 1 for Years 11 & 12 Student Worked Solutions
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∠QAB = ∠PXB
(exterior angle of cyclic
quadrilateral equals interior opposite angle)
∠XBA = ∠QAB (alternate angles, PQ  XY)
(exterior angle of cyclic
∠AQY = ∠XBA
quadrilateral equals interior opposite angle)
The remaining angles in both quadrilaterals must
also be equal. (angle sum of a quadrilateral)
\ ∠PXY = ∠PQY
9 Draw a diagram:
A
B
7 (a) Draw a diagram:
D
D
A
C
Let ∠ABC = x°.
\ ∠ADC = (180 - x)° (opposite angles of a cyclic
quadrilateral are supplementary)
and ∠DAB = x° (cointerior angles supplementary,
AB  DC)
\ ∠DAB = ∠ABC
C
B
Let ∠DAC = x°.
\ ∠DCA = ∠DAC = x°
(base angles of
isosceles triangle are equal)
∠DBC = ∠DAC = x° (angles in the same
segment)
\ ∠BDC = ∠DBC = x° (base angles of isosceles
triangle are equal)
∠BAC = ∠BDC = x° (angles inthe same
segment)
and ∠DBA = ∠DCA = x° (angles inthe same
segment)
\ ∠DAB = 2x° (∠DAC + ∠CAB)
Hence ∠DCB = (180 - 2x°). (opposite angles
of a cyclic quadrilateral
are supplementary)
∠ABC = 2x° (∠ABD + ∠DBC)
Hence ∠ADC = (180 - 2x°). (opposite angles
of a cyclic quadrilateral
are supplementary)
\ ∠ADC = ∠DCB
D
A
B
C
Sa
m
pl
e
pa
ge
s
11 Draw a diagram:
(b)∠ADC + ∠DAB = (180 - 2x°) + 2x° = 180°
These are a pair of supplementary cointerior
angles, so DC  AB.
In DABC and DCDA:
∠BAC = ∠ACD (alternate angles, BA  CD)
∠ACB = ∠CAD (alternate angles, BC  AD)
AC is a common side.
\ DABC ≡ DCDA (AAS)
\ ∠ABC = ∠CDA (matching angles in
congruent triangles)
Because DABC is acute-angled, ∠ABC < 90° and
thus ∠CDA < 90°.
If D is on the circle then ADCB would be a cyclic
quadrilateral and:
∠CDA = 180° - ∠ABC
(opposite angles of a
cyclic quadrilateral are supplementary)
This would only be possible if:
∠ABC = ∠CDA = 90°
\ D is not on the circle if DABC is acute-angled.
Exercise 1.6
1 ∠CBD = 25° (angle sum of DBCD)
\ ∠CAD = ∠CBD (both 25°)
These are a pair of equal angles subtended on the
same side of CD; hence A, B, C and D are concyclic.
\ ∠BAC = ∠BDC = 55°
(angles in the same
segment standing on BC)
∠BEA = 65° (∠CEA is a straight angle)
\ ∠ABD = 60° (angle sum of DABE)
3 (a) correct (a pair of equal angles are subtended
on the same side of BC)
(b)incorrect
(c) correct (a pair of equal angles are subtended
(d)incorrect
on the same side of BC)
Chapter 1 Circle geometry
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