710 MHR ● MathLinks 9 Solutions Chapter 10 Circle Geometry Section 10.1 Exploring Angles in a Circle Section 10.1 Page 382 Question 3 ADB and AEB are inscribed angles that are subtended by the same arc as central ACB. The measure of ACB is 82°. The measures of ADB and AEB are one-half the measure of ACB. 82 41 2 So, the measures of ADB and AEB are 41°. Section 10.1 Page 382 Question 4 a) FJG is an inscribed angle subtended by the same arc as inscribed FHG. Since the measure of FHG is 23°, the measure of FJG is also 23°. b) FCG is a central angle subtended by the same arc as inscribed FHG. FCG = 2FHG FCG = 2(23) The measure of FCG is 46°. Section 10.1 Page 383 Question 5 The inscribed angles in the diagram subtend the same arc as the central angle, which measures 60. The inscribed angles each measure 30°. 60° 30° 30°
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710 MHR ● MathLinks 9 Solutions
Chapter 10 Circle Geometry
Section 10.1 Exploring Angles in a Circle
Section 10.1 Page 382 Question 3
ADB and AEB are inscribed angles that are subtended by the
same arc as central ACB. The measure of ACB is 82°. The
measures of ADB and AEB are one-half the measure of ACB.
8241
2
So, the measures of ADB and AEB are 41°.
Section 10.1 Page 382 Question 4
a) FJG is an inscribed angle subtended by the same arc as
inscribed FHG. Since the measure of FHG is 23°, the measure
of FJG is also 23°.
b) FCG is a central angle subtended by the same arc as
inscribed FHG.
FCG = 2FHG
FCG = 2(23)
The measure of FCG is 46°.
Section 10.1 Page 383 Question 5
The inscribed angles in the diagram
subtend the same arc as the central
angle, which measures 60. The
inscribed angles each measure 30°.
60°
30°
30°
MHR ● MathLinks 9 Solutions 711
Section 10.1 Page 383 Question 6
a) ABD is an inscribed angle, which is subtended by the diameter
of the circle. The measure of ABD is 90°.
b) Since the measure of ABD is 90°, ABD is a right triangle.
Use the Pythagorean relationship to find the length of AB.
AB2 + BD
2 = AD
2
AB2 + 15
2 = 17
2
AB2 + 225 = 289
AB2 = 64
AB = 64
AB = 8
Therefore, AB = 8 cm.
Section 10.1 Page 383 Question 7
a) FCG is a central angle subtended by the same
arc as inscribed FEG.
FCG = 2FEG
FCG = 2(45)
The measure of FCG is 90°.
b) Since FCG = 90°, FCG is a right triangle.
Use the Pythagorean relationship to find the length of FG.
FC2 + CG
2 = FG
2
82 + 8
2 = FG
2
64 + 64 = FG2
128 = FG2
128 = FG
11.3 FG
Therefore, the length of FG is about 11.3 cm.
712 MHR ● MathLinks 9 Solutions
Section 10.1 Page 383 Question 8
In the diagram, a circle with a central angle of 30° is shown. Use one arm of the central angle as
the radius of the circle. Construct any number of different inscribed angles that represent Jacob’s
flashlight beam projecting light through the same arc as the beam projected by his mother’s
flashlight. The measure of each of these inscribed angles will be one-half the measure of the
central angle. Each inscribed angle will measure 15°, which corresponds to the beam of Jacob’s
flashlight. So, Jacob could place his flashlight anywhere on the major arc MN.
Section 10.1 Page 383 Question 9
In the diagram, a circle with the centre, X, is
shown. Construct an inscribed angle of 22°
that represents the spotlight’s beam
projecting light to the stage. Draw a central
angle subtended by the same arc as the
inscribed angle. This angle would be 44°. So,
the centre of the circle, X, would be the ideal
location.
X 44° 22°
M
N
30°
MHR ● MathLinks 9 Solutions 713
Section 10.1 Page 383 Question 10
a) The measure of ACD is 76°. Example: ACD is a central angle
subtended by the same arc as inscribed ABD. The measure of ABD
is 38°, so the measure of ACD is 2 38°, or 76°.
b) ACD is an isosceles triangle. Sides AC and DC are radii of the
same circle and therefore have equal length.
c) To find the measure of CAD, subtract the measure of ACD,
76°, from 180° and divide by 2.
(180 – 76) ÷ 2 = 104 ÷ 2
= 52
The measure of CAD is 52°.
Section 10.1 Page 383 Question 11
a) FCE is an isosceles triangle since sides FC and EC are radii of the
same circle. To find the measure of FCE, subtract the measures of the
two equal angles, EFC and CEF, from 180°.
180 – 2(25) = 130°
The measure of FCE is 130°.
b) EGF is an inscribed angle subtended by the same arc as the central
FCE. The measure of EGF is one-half the measure of FCE.
130 ÷ 2 = 65°
The measure of EGF is 65°.
714 MHR ● MathLinks 9 Solutions
Section 10.1 Page 384 Question 12
a) KLM and KJM are inscribed angles subtended by
the same arc. Since the measure of KJM is 15°, the
measure of KLM is also 15°.
b) JKL and JML are inscribed angles subtended by
the same arc. Since the measure of JML is 24°, the
measure of JKL is also 24°.
c) JCL is a central angle subtended by the same arc as
inscribed JML.
JCL = 2 JML
= 2 24
= 48
The measure of JCL is 48°.
d) KCM is a central angle subtended by the same arc as KJM.
KCM = 2 KJM
= 2 15°
= 30
The measure of KCM is 30°.
Section 10.1 Page 384 Question 13
a) ABE and ADE are inscribed angles subtended by
the same arc. Since the measure of ADE is 56°, the
measure of ABE is also 56°.
b) To find the measure of AGB in AGB, subtract the
measures of GAB and ABE from 180°.
180 – (34 + 56) = 90
The measure of AGB is 90°.
c) Since the measure of AGB is 90°, ABG is a right
triangle.
d) To find the measure of DGE in DGE, subtract the
measures of GDE and DEG from 180°.
180 – (34 + 56) = 90
The measure of DGE is 90°.
MHR ● MathLinks 9 Solutions 715
Section 10.1 Page 384 Question 14
Amanda cannot use the Pythagorean relationship to
calculate the length of chord AB. Example: Neither
ADB nor ACB is a right triangle. The angles in
triangle ADB are all less than 90°. In ACB, ACB is a
central angle subtended by the same arc as inscribed
ADB. The measure of ADB is 55° so, the measure of
ACB is 2 55, or 110°. The Pythagorean relationship
can only be used with right triangles.
716 MHR ● MathLinks 9 Solutions
Section 10.1 Page 384 Question 15
a) The angle labelled 135° and the angle labelled x form a
straight angle.
180 – 135 = 45
The measure of the angle labelled x is 45°.
The angle labelled x and the angle labelled y are inscribed
angles subtended by the same arc. The measure of the angle
labelled y is 45°.
b) The measure of the angle labelled x is 60°, since the inscribed
triangle is an equilateral triangle.
The angle labelled y is a central angle subtended by the same arc
as the inscribed angle labelled x, which is 60°. The measure of the
angle labelled y is 2 60, or 120°.
c) The angle labelled x is an inscribed angle subtended by the
same arc as the inscribed angle labelled 15°, so x = 15.
The measure of angle labelled y is 30°, since it is a central angle
subtended by the same arc as the inscribed angle labelled x, which is
15°.
d) Both triangles are right triangles. Each of the inscribed angles is
subtended by a diameter of the circle, so its measure is 90°.
Since the measures of two angles in the first triangle are known, the
measure of the angle labelled x can be determined.
x = 180 – (55 + 90)
= 35
The measure of the angle labelled x is 35°.
The second right triangle is an isosceles right triangle. To find the
measure of the angle labelled y, subtract 90 from 180 and divide the difference by 2.
(180 – 90) ÷ 2 = 45
The measure of the angle labelled y is 45°.
MHR ● MathLinks 9 Solutions 717
Section 10.1 Page 384 Question 16
Example: In the diagram, point O is the centre of the circle and OQ = OP = PQ. Find the measure
of PMQ.
In the diagram, POQ is an equilateral triangle, so POQ is a central angle that measures 60°.
Since POQ is subtended by the same arc as inscribed angle PMQ, the measure of PMQ is
60 2, or 30°.
O
P
Q
M
718 MHR ● MathLinks 9 Solutions
Section 10.1 Page 384 Question 17
In the diagram, inscribed ABE is subtended by the same arc as inscribe ADE.
ABE = 14°
ACE is a central angle subtended by the same arc as ADE.
ACE = 2 ABE
= 2 14
= 28
The measure of ACE is 28° and the measure of ABE is 14°.
A
D
E
C
14°
B
MHR ● MathLinks 9 Solutions 719
Section 10.1 Page 384 Question 18
a) The inscribed angle that has a measure labelled 25° subtends an arc
that has the same measure as the inscribed angle labelled x. Therefore,
the measure of the angle labelled x is 25°.
The angle labelled y is a central angle subtended by an arc that has a
measure that is the same as the arc subtended by the inscribed angle of
25°. Therefore, the measure of the angle labelled y is 2 25°, or 50°.
b) The inscribed angle labelled x is subtended by the same arc as the
central angle labelled 190°. The measure of the angle labelled x is
190° 2, or 95°.
There are two central angles in the diagram. One is labelled 190° and
the second angle is 360 – 190, or 170°. This second central angle, the
40° angle, and the 95° angle form three angles of a quadrilateral. The
angle sum of a quadrilateral is 360°. To find the measure of the angle
labelled y, find the sum of the three angles of the quadrilateral and
subtract the sum from 360°.
360 – (95 + 40 + 170) = 55
The measure of the angle labelled y is 55°.
Section 10.1 Page 385 Question 19
The diameter of the circle divides the square into two right isosceles
triangles.
Let x represent the side length of the square. The diameter is the
hypotenuse of the right triangle. Use the Pythagorean relationship to
find x.
x2 + x
2 = 20
2
2x2 = 400
x2 = 200
x = 200
x 14.14
The side length of the square is about 14.14 cm.
720 MHR ● MathLinks 9 Solutions
Section 10.1 Page 385 Question 20
a) Inscribed ABD, is subtended by the same arc as inscribed AED.
ABD = AED
3x = x + 18 Solve for x.
2x = 18
x = 9
Therefore, x = 9°.
b) Inscribed EFD is subtended by the same arc as central ECD.
ECD = 2 EFD
4x + 2 = 2(3x – 16) Solve for x.
4x + 2 = 6x – 32
4x + 34 = 6x
34 = 2x
17 = x
Therefore, x = 17°.
MHR ● MathLinks 9 Solutions 721
Section 10.1 Page 385 Question 21
a) Inscribed BHA is subtended by the diameter so
its measure is 90°.
b) Inscribed BHE is subtended by the diameter so
its measure is 90°. Therefore BHE is a right
triangle. In BEH, find the measure of BEH.
180 – (90 + 27) = 63
The measure of BEH is 63°.
c) AEG is opposite BEH. Opposite angles have the same measures. So, AEG is 63°.
d) All three sides of ACG are equal in length to the radius of the circle. Therefore, ACG is
an equilateral triangle, so ACG is 60°.
e) BCG and ACG are supplementary.
BCG = 180 – ACG
= 180 – 60
= 120
The measure of BCG is 120°.
722 MHR ● MathLinks 9 Solutions
Section 10.2 Exploring Chord Properties
Section 10.2 Page 389 Question 4
Since CD is a radius that bisects the chord AB, then CD
is perpendicular to AB and AEC = 90°.
The length of AE is 12 cm because CD bisects the chord
AB, which has length 24 cm. The radius AC is 15 cm.
Use the Pythagorean relationship in ACE.
AE2 + CE
2 = AC
2
122 + CE
2 = 15
2
144 + CE2 = 225
CE2 = 81
CE = 81
CE = 9
The length of CE is 9 cm.
Section 10.2 Page 389 Question 5
Since CF is a radius that bisects the chord HJ, then CF is
perpendicular to HJ and CGH = 90°.
The length of HG is 7 mm because CF bisects the
chord HJ, which measures 14 mm. The segment CG
measures 4 mm.
Use the Pythagorean relationship in CGH.
HG2 + CG
2 = HC
2
72 + 4
2 = HC
2
49 + 16 = HC2
65 = HC2
65 = HC
8.1 HC
The length of HE is about 8.1 mm.
MHR ● MathLinks 9 Solutions 723
Section 10.2 Page 390 Question 6
Example: Hannah could draw any two chords on the circle. She could then locate the midpoint of
each chord and draw the perpendicular bisectors. The intersection of the perpendicular bisectors
is the centre of the trampoline.
Section 10.2 Page 390 Question 7
Use the Pythagorean relationship to find the length of EB.
EC2 + EB
2 = BC
2
82 + EB
2 = 17
2
64 + EB2 = 289
EB2 = 225
EB = 225
EB = 15
Since CD is perpendicular to AB, CB bisects AB.
AB = 2(EB)
= 2(15)
= 30
The length of AB is 30 m.
O
O is the centre of the trampoline.
724 MHR ● MathLinks 9 Solutions
Section 10.2 Page 390 Question 8
In the diagram, draw radius CK to form right CQK.
QC = MC – MQ
= 11.1 – 3.4
= 7.7
The length of segment QC is 7.7 cm.
Use the Pythagorean relationship to find the length of QK.
QC2 + QK
2 = CK
2
7.72 + QK
2 = 11.1
2
59.29 + QK2 = 123.21
QK2 = 63.92
QK = 63.92
QK 8.0
Since MC is perpendicular to LK, MC bisects LK.
LK = 2(QK)
= 2(8.0)
= 16.0
The length of LK is about 16.0 m.
MHR ● MathLinks 9 Solutions 725
Section 10.2 Page 390 Question 9
a) In the diagram, assume that C is the centre of the circle. If the
bisector of a chord in a circle passes through the centre, then the
bisector is perpendicular to the chord. Therefore, the triangle formed
in the diagram is a right triangle. Use the Pythagorean relationship to
find the length of x.
The legs of the right triangle are x and 6 – 3, or 3. The hypotenuse has
length 6.
32 + x
2 = 6
2
9 + x2 = 36
x2 = 27
x = 27
x 5.2
The length of the side labelled x is about 5.2 units.
b) In the diagram, assume that C is the centre of the circle. The
diameter is perpendicular to the chord, so the triangle formed in the
diagram is a right triangle. Use the Pythagorean relationship to find the
length of x.
The legs of the right triangle are y = 10 – x and 9. The hypotenuse has
length 10, which is the radius of the circle.
92 + y
2 = 10
2
81 + y2 = 100
y2 = 19
y = 19
y 4.4
y = 10 – x
10 – 4.4
= 5.6
The length of the side labelled x is about 5.6 units.
726 MHR ● MathLinks 9 Solutions
Section 10.2 Page 390 Question 10
In the diagram, C is the centre of the circle
and CA and CF are radii of length 50 2,
or 25 cm. CF bisects chord EA. Therefore,
the length of DA is 34 2, or 17 cm. The
distance DF represents the depth of the
water.
Use the Pythagorean relationship to find the length of CD.
CD2 + DA
2 = CA
2
CD2 + 17
2 = 25
2
CD2 + 289 = 625
CD2 = 336
CD = 336
CD 18.3
The depth of the water is about 25 – 18.3, or 6.7 cm. To the nearest centimetre, the depth is 7 cm.
Section 10.2 Page 390 Question 11
Draw segment AC in the diagram to form ACB, which is a right
triangle. Use the Pythagorean relationship to find the length of
segment AB.
AB2 + CB
2 = CA
2
AB2 + 3
2 = 5
2
AB2 + 9 = 25
AB2 = 16
AB = 16
AB = 4
The length of AB is 4 cm.
Determine the area of ABD.
Area = height × base ÷ 2
= 8 × 4 ÷ 2
= 16
The area of ABD is 16 cm2.
25 cm
17 cm
C
D A E
F
MHR ● MathLinks 9 Solutions 727
Section 10.2 Page 390 Question 12
Refer to the diagram. The shortest distance to the centre of the circle
would be the perpendicular distance to the midpoint of the chord,
which is D. Use the Pythagorean relationship to find the length of
segment CD.
CD2 + DF
2 = CF
2
CD2 + 7
2 = 25
2
CD2 + 49 = 625
CD2 = 576
CD = 576
CD = 24
The shortest distance from the centre of the circle to the chord is 24 mm.
Section 10.2 Page 391 Question 13
Example: Locate and draw the perpendicular bisectors of any two sides of
the octagon. The sides of the octagon represent the chords of a circle. The
point where the two perpendicular bisectors intersect is the centre of the
octagon.
Section 10.2 Page 391 Question 14
Example: Draw any two chords on the piece. Locate and draw the perpendicular bisectors of the
two chords. The point of intersection of the two perpendicular bisectors is the centre of the circle.
Measure the distance from the centre of the circle to the endpoint of any chord. If the
measurement is 8 cm, the diagram was accurate.
728 MHR ● MathLinks 9 Solutions
Section 10.2 Page 391 Question 15
a) ADE is an inscribed angle subtended by the diameter AE.
The measure of ADE is 90°. An inscribed angle subtended by a
diameter has a measure of 90.
b) ADE is a right triangle. Use the Pythagorean relationship to
determine AD.
DE2 + AD
2 = AE
2
162 + AD
2 = 20
2
256 + AD2 = 400
AD2 = 144
AD = 144
AD = 12
The length of AD is 12 cm.
c) DFE is a right triangle. Determine the length of FE.
FE = AE – AF
= 20 – 5
= 15
The length of FE is 15 cm.
Use the Pythagorean relationship to determine DF.
FE2 + DF
2 = DE
2
152 + DF
2 = 16
2
225 + DF2 = 256
DF2 = 31
DF = 31
DF 5.6
The length of DF is about 5.6 cm.
d) BD = 2DF
2(5.6)
= 11.2
The length of BD is about 11.2 cm.
MHR ● MathLinks 9 Solutions 729
Section 10.2 Page 391 Question 16
a) HMP is an inscribed angle subtended by the same arc as central
angle HCP.
HMP = HCP 2
= 130 ÷ 2
= 65
The measure of HMP is 65°.
b) Segment CJ, which is a radius of the circle, is the perpendicular
bisector of chord MP. Therefore, HEM is 90°.
c) Since HEM is 90°, HME is a right triangle.
Use the angle sum of a triangle to determine MHJ.
MHJ = 180 – (HEM + HMP)
= 180 – (90 + 65)
= 25°
The measure of MHJ is 25°.
d) MPJ is an inscribed angle subtended by the same arc as the inscribed angle MHJ.
So MPJ = MHJ.
The measure of MPJ is 25°.
e) HCP and PCE are supplementary angles. The sum of supplementary angles is 180°.
Therefore, to find the measure of PCE, subtract 130 from 180.
PCE = 180 – HCP
= 180 – 130
= 50°
The measure of PCE is 50°.
f) CEP is a right angle, so CEP is a right triangle. Use the angle sum of a triangle to
determine CPE.
CPE = 180 – (CEP + PCE)
= 180 – (90 + 50)
= 40°
The measure of CPE is 40°.
730 MHR ● MathLinks 9 Solutions
Section 10.2 Page 391 Question 17
Refer to the diagram.
Segments CF and CD are radii of the circle.
Since the diameter is 12 m, each radius has a
length of 6 m. Radius CF is the perpendicular
bisector of chord DG, so DEC is a right
triangle. Chord DG represents the horizontal
width of the water.
DE = DG 2
= 7.3 2
= 3.65
Use the Pythagorean relationship to find the
length of segment CE.
DE2 + CE
2 = DC
2
3.652 + CE
2 = 6
2
13.3225 + CE2 = 36
CE2 = 22.6775
CE = 22.6775
CE 4.76
Segment EF represents the depth of the water.
EF = CF – CE
= 6 – 4.76
= 1.24
The depth of the water is about 1.2 m.
C A B
D E
F
G
MHR ● MathLinks 9 Solutions 731
Section 10.2 Page 392 Question 18
Example: Gavyn made two mistakes. The first mistake is that the
diagram is labelled incorrectly. Segment AC is a radius and should be
labelled as 13 cm. The second mistake is that segment AC is the
hypotenuse of right AEC, not a leg. So, by the Pythagorean
relationship,
EC2 + AE
2 = AC
2
52 + AE
2 = 13
2
25 + AE2 = 169
AE2 = 144
AE = 144
AE = 12
The length of segment AE is 12 cm.
Since CD is a radius and it is perpendicular to AB, then CD bisects the chord AB.
AB = 2 12
= 24
The length of chord AB is 24 cm.
Section 10.2 Page 392 Question 19
Refer to the diagram.
In the diagram, the length of segment AC is 6 mm. This
is half the perpendicular distance between the two
parallel chords. The length of segment AB is 5 mm.
This is one-half the length of one of the parallel chords.
ACB is a right triangle. Use the Pythagorean
relationship to find the radius, BC.
AC2 + AB
2 = BC
2
62 + 5
2 = BC
2
36+ 25 = BC2
61 = BC2
61 = BC
7.8 BC
The radius of the circle is about 7.8 mm, so the diameter is about 2 7.8, or 15.6 mm.
A
C
B
732 MHR ● MathLinks 9 Solutions
Section 10.2 Page 392 Question 20
a) If a bisector of a chord passes through the centre of a circle, then the bisector is perpendicular
to the chord. So, FGH = 90.
b) Solve the following equation for x.
3x + 2 + x + 90 = 180
4x + 92 = 180
4x = 88
x = 22
The acute angle, HFG, is labelled with x; therefore, its measure is 22°.
The acute angle, GHF, is labelled with 3x + 2. Replace x with 22.
3x + 2 = 3(22) + 2
= 66 + 2
= 68
Therefore, the measure of the acute angle, GHF, is 68°.
Section 10.2 Page 392 Question 21
Since the bisector of chords AB and DE passes
through the centres of both circles, it is
perpendicular to chords AB and DE. Two line
segments perpendicular to the same segment
are parallel.
MHR ● MathLinks 9 Solutions 733
Section 10.3 Tangents to a Circle
Section 10.3 Page 399 Question 3
a) Since AB is tangent to the circle at point D, then radius CD is
perpendicular to line segment AB. The measure of BDC is 90°.
b) The sum of the angles in a triangle is 180°.
In BCD, DCB = 180 – (90 + 60).
The measure of DCB is 30°.
Since DCE and DCB are supplementary angles,
DCE = 180 – 30 = 150
The measure of DEC is 150°.
c) CDE is an isosceles triangle since two sides are radii of the circle and are therefore equal.
d) DEC is an inscribed angle subtended by the same arc as central DCB, which is 30°.
DEC = DCB 2
= 30 ÷ 2
= 15
The measure of DEC is 15°.
734 MHR ● MathLinks 9 Solutions
Section 10.3 Page 399 Question 4
a) CGL is an isosceles triangle since two of the sides are radii of
the circle and are therefore equal.
b) The sum of the angles in a triangle is 180°. Since CGL is
isosceles, LGC = GLC.
GCL = 180 – (LGC + GLC)
= 180 – (10 + 10)
= 160°
The measure of GCL is 160°.
c) Since JCH and GCL form a straight line, they are supplementary.
JCH = 180 – GCL
= 180 – 160
= 20
The measure of JCH is 20°.
d) Since segment JH is tangent to the circle at point H, it is perpendicular to the radius CH.
Therefore, the measure of JHG is 90°.
e) CJH is a right triangle so CHJ = 90.
CJK = 180 – (JCH + CHJ)
= 180 – (20 + 90)
= 70°
The measure of CJK is 70°.
MHR ● MathLinks 9 Solutions 735
Section 10.3 Page 400 Question 5
a) Diameter BD is perpendicular to tangent AB because B is
the point of tangency on the circle. Therefore, ABD is 90°
and ABD is a right triangle. Use the Pythagorean
relationship to find BD.
AB2 + BD
2 = AD
2
62 + BD
2 = 10
2
36 + BD2 = 100
BD2 = 64
BD = 64
BD = 8
The length of diameter BD is 8 m.
b) BCE is an equilateral triangle, so all sides are equal.
Since BE = BC and BC is a radius, BE = 8 2, or 4 m.
c) BED is an inscribed angle subtended by a diameter. Therefore, the measure of BED is 90°.
d) Since BED is a right triangle, use the Pythagorean relationship to find the length of DE.
BE2 + DE
2 = DB
2
42 + DE
2 = 8
2
16 + DE2 = 64
DE2 = 48
DE = 48
DE 6.928
The length of chord DE is about 7 m.
736 MHR ● MathLinks 9 Solutions
Section 10.3 Page 400 Question 6
a) The diameter is twice the radius. The radius is 5 mm, so
the diameter is 10 mm.
b) Since the inscribed angle, GJH, is subtended by the
diameter, GH, it is a right angle. Therefore, GHJ is a right
triangle.
c) Since GHJ is a right triangle, use the Pythagorean
relationship to find the length of chord HJ.
GJ2 + HJ
2 = GH
2
52 + HJ
2 = 10
2
25 + HJ2 = 100
HJ2 = 75
HJ = 75
HJ 8.7
Therefore, the length of chord HJ is about 8.7 mm.
d) Since segment FG is tangent to the circle at point G and segment GH is a diameter of the
circle, the measure of FGH is 90°.
e) Since FGH is a right triangle, use the Pythagorean relationship to find the length of FH.
FG2 + GH
2 = FH
2
72 + 10
2 = FH
2
49 + 100 = FH2
149 = FH2
149 = FH
12.2 FH
The length of segment FH is about 12.2 mm.
MHR ● MathLinks 9 Solutions 737
Section 10.3 Page 400 Question 7
Draw a segment from the pole to the cat door. Refer to the diagram.
The distance from the pole to the cat door, x, is the hypotenuse of a right triangle. Use the
Pythagorean relationship to find the distance.
52 + 16
2 = x
2
25 + 256 = x2
281 = x2
281 = x
16.8 x
The distance from the pole to the cat door is about 16.8 m.
To determine how close the dog can get to the cat door, subtract 5 from 16.8.
16.8 – 5 = 11.8
The closest the dog can get to the cat door is about 11.8 m.
pole
cat door 16 m
5 m x
738 MHR ● MathLinks 9 Solutions
Section 10.3 Page 400 Question 8
a) Since line l is tangent to the circle at one endpoint of the
diameter, the diameter is perpendicular to line l. Therefore, the
triangle with sides labelled 15 m, 8 m, and x is a right triangle,
with the side labelled x being the hypotenuse.
Use the Pythagorean relationship to find the length of x.
152 + 8
2 = x
2
225 + 64 = x2
289 = x2
289 = x
17 = x
The length of x is 17 m.
b) Since line l is tangent to the circle at one endpoint of the
radius, the radius is perpendicular to line l. Therefore, the
triangle is a right triangle with hypotenuse 20 cm long and
one leg 16 cm long. The third side of the right triangle is a
radius, which is x.
Use the Pythagorean relationship to find x.
162 + x
2 = 20
2
256 + x2 = 400
x2 = 144
x = 144
x = 12
The length of x is 12 cm.
MHR ● MathLinks 9 Solutions 739
Section 10.3 Page 401 Question 9
a) Since line l is tangent to the circle at one endpoint of the
diameter, line l is perpendicular to the diameter. Thus, the
triangle containing is a right triangle. One angle of the
right triangle is a central angle, which is subtended by an
arc the same size as the arc that subtends a central angle of
55°. Find the measure of angle .
= 180 – (90 + 55)
= 180 – 145
= 35
The measure of angle is 35°.
b) Refer to the diagram.
Since line l is tangent to the circle at point D, line l is
perpendicular to the radius CD. Thus, CDB is a right
triangle. Find the measure of BCD.
BCD = 180 – (90 + 74)
= 180 – 164
= 16
The measure of BCD is 16°.
BCD and the angle labelled are supplementary.
Therefore, to find the measure of angle , subtract 16 from 180.
= 180 – BCD
= 180 – 16
= 164
The measure of angle is 164°.
740 MHR ● MathLinks 9 Solutions
Section 10.3 Page 401 Question 10
a) Since the circles are identical in size, the radii OR and CK have equal length. The lengths of
OC and RK represent the lengths of the diameters of the circles, so OC = RK. The line l is
tangent to the circles at points R and K, so l is perpendicular to OR and CK. The angles of the
quadrilateral are 90°. Quadrilateral ROCK is a rectangle.
b) Segments OC and RK have the same measurements as the diameter of each circle. Since the
radius of each circle is 5 cm, the diameter is 10 cm. To find the perimeter of ROCK, add the
sides.
5 + 10 + 5 + 10 = 30
The perimeter of ROCK is 30 cm.
MHR ● MathLinks 9 Solutions 741
Section 10.3 Page 401 Question 11
Refer to the diagram. Since AB is tangent to the circle at point A, AB is perpendicular to AD.
Therefore, ABD is a right triangle. Use
the Pythagorean relationship to find DB.
AB2 + AD
2 = DB
2
4.22 + 7.3
2 = DB
2
17.64 + 53.29 = DB2
70.93 = DB2
70.93 = DB
8.4 cm DB
The length of segment DB is about
8.4 cm.
Section 10.3 Page 401 Question 12
a) Since AD is tangent to the circle at D and DB is
a diameter, ADB is 90°.
b) ADB is a right isosceles triangle.
DBE = (180 – 90) ÷ 2
= 45°
The measure of DBE is 45°.
c) DFE is an inscribed angle subtended by the
same arc as inscribed DBE. Therefore, the two
angles have equal measurements. The measure of
DFE is 45°.
D A
7.3 cm
4.2 cm
B
742 MHR ● MathLinks 9 Solutions
Section 10.3 Page 401 Question 13
a) Since line l is tangent to the circle at point H and CH is a
radius, line l is perpendicular to CH. Therefore, the measure of
CHM is 90°.
b) Since chord JK is parallel to line l and line l is
perpendicular to segment CH, segment JK is perpendicular to
segment CH. Therefore, the measure of CGJ is 90°.
c) CH passes through the centre of a circle and is
perpendicular to JK. Therefore, JK is bisected by CH.
JG = JK 2
= 17 2
= 8.5
The measure of segment JG is 8.5 cm.
d) Since JK is perpendicular to CH, CGJ is a right triangle. Use the Pythagorean relationship.
JG2 + CG
2 = JC
2
8.52 + CG
2 = 9.1
2
72.25 + CG2 = 82.81
CG2 = 10.56
CG = 10.56
CG 3.2
The length of segment CG is about 3.2 cm.
Section 10.3 Page 402 Question 14
Since segment JG is tangent to the circle at point J,
segment GJ is perpendicular to diameter JH.
Therefore, JHG is a right triangle. Use the angle
sum of a triangle.
90 + 5x – 2 + 2x + 15 = 180
7x + 103 = 180
7x = 77
x = 11
To find the measure of JGH, replace x with 11 in
the expression 5x – 2.
5(11) – 2 = 55 – 2
= 53
The measure of JGH is 53°.
MHR ● MathLinks 9 Solutions 743
Section 10.3 Page 402 Question 15
Example: The inscribed angle that measures 85° is
subtended by the same arc as the central angle.
The measure of the central angle is 2 85, or 170°.
One of the angles in the right triangle has a measure
of 170 – 140, or 30°.
To find the measure of angle , which is the third angle
of the right triangle, use the angle sum of a triangle.
= 180 – (90 + 30)
= 60°
The measure of angle is 60°.
Section 10.3 Page 402 Question 16
Example: Refer to the diagram. The four congruent circles represent watered
regions of the square field. The circles are tangent to the sides of the square. If
the area of the field is 400 m2, what is the area of the field that is not watered?
Answer: The side length of the square field is 400 , or 20 m. The radius of
each circle is 20 ÷ 4, or 5 m. To find the area that is not watered, subtract the
total area of the circles from the area of the square field.
400 – 4r2 400 – 314.16
= 85.8
The area of the field that is not watered is about 85.8 m2.
744 MHR ● MathLinks 9 Solutions
Section 10.3 Page 402 Question 17
Refer to the diagram.
Label the point of tangency to the small circle D.
Draw segments AC and CD. Since AB is tangent to the small circle
at D, AB is perpendicular to CD.
Therefore, ADC is a right triangle.
CD is the radius of the small circle, so the length of CD is 8 cm.
CD bisects AB, so AD is 26 2, or 13 cm.
Use the Pythagorean relationship to find AC, which is the radius of
the large circle.
CD2 + AD
2 = AC
2
82 + 13
2 = AC
2
64 + 169 = AC2
233 = AC2
233 = AC
15.26 AC
The radius of the large circle is about 15.26 cm.
Find the circumference of the large circle.
C = 2r
C 2(3.14)(15.26)
C = 95.83
The circumference of the large circle, rounded to the nearest centimetre, is 96 cm.
MHR ● MathLinks 9 Solutions 745
Section 10.3 Page 402 Question 18
Refer to the diagram.
Since the circles are congruent and the centre of Circle A is at
(2, 2), the radius of each circle is 2 units.
Point D is the point of tangency for Circles A and B.
So, CD is perpendicular to AB.
AB is twice the radius, AD, so AB is 4 units long.
The centre of Circle B is four units right of (2, 2). The centre
of Circle B is at the point (6, 2).
The length of CD is equal to the length of AB. So, the length
of CD is 4 units.
The coordinates of the centre of Circle C are 2 units right and 4 units up from point (2, 2). The
centre of Circle C is at the point (4, 6).
Section 10.3 Page 402 Question 19
In the diagram, the segments touching the circle at the dotted
lines resemble tangents to the circle. Perpendicular lines
drawn through these points of tangency would intersect at the
centre of the circle.
746 MHR ● MathLinks 9 Solutions
Section 10.3 Page 403 Question 20
Refer to the diagram. Draw a line segment from the centre of the small circle, O, to point D on
the radius BC that is perpendicular to BC.
DCO is a right triangle, with hypotenuse OC and legs DC and OD.
The length of segment DC is 18 – 7, or 11 cm.
The length of segment OD is 60 cm because quadrilateral ABDO is a rectangle.
Use the Pythagorean relationship to find the length of OC.
DC2 + OD
2 = OC
2
112 + 60
2 = OC
2
121 + 3600 = OC2
3721 = OC2
3721 = OC
61 = OC
The length of OC is 61 cm.
Find the total length of the metal band.
Total length = OA + AB + BC + OC
= 7 + 60 + 18 + 61
= 146
The length of the metal band needed is 146 cm.
MHR ● MathLinks 9 Solutions 747
Section 10.3 Page 403 Question 21
Refer to the diagram.
In the diagram, r represents the radius of the circle where the ball touches the ice surface. Use the
Pythagorean relationship to find the length of the radius, r.
12 + r
2 = 3
2
1 + r2 = 9
r2 = 8
r = 8
r 2.8
The radius of the circle where the ball touches the ice surface is about 2.8 cm.
Substitute 2.8 into the formula, C = 2r.
C = 2r
C 2(3.14)(2.8)
C = 17.6
The circumference of the circle is about 17.6 cm.
3 cm
2 cm
1 cm
r
This is a side-view. The red dotted line is the
edge marking where the ball is above the ice
surface.
r
This is a top-view. The green dotted
line represents the circumference of
the circle where the ball touches the
ice surface.
748 MHR ● MathLinks 9 Solutions
Section 10.3 Page 403 Question 22
Since segments BD and ED are tangent to the circle at points
B and E respectively, segment BD is perpendicular to radius
BC and segment DE is perpendicular to radius EC. Therefore,
DEC and DCB are right triangles.
EC is a radius, which has length 20 2, or 10 cm.
Use the Pythagorean relationship to find the length of DE.
DE2 + EC
2 = DC
2
DE2 + 10
2 = 24
2
DE2 + 100 = 576
DE2 = 476
DE = 476
DE 21.8
Since DC = DC, EC = BC, and DEC = DBC,
DEC and DCB are congruent. So, DE = DB.
The total length of chain needed is about 2 21.8, or 43.6 cm.
MHR ● MathLinks 9 Solutions 749
Chapter 10 Review Page 404 Question 1
The RADIUS is the distance from the centre to any point on the circle.
Chapter 10 Review Page 404 Question 2
An INSCRIBED ANGLE is an angle formed by two chords that share a common endpoint.
Chapter 10 Review Page 404 Question 3
A CHORD is a line segment that has both endpoints on the same circle.
Chapter 10 Review Page 404 Question 4
A PERPENDICULAR BISECTOR is a line or line segment that passes through the midpoint of a
line segment at 90°.
Chapter 10 Review Page 404 Question 5
a) The inscribed angle, ABD, is subtended by the same arc
as inscribed angle, AED. Therefore, the measure of ABD is
equal to the measure of AED, which is 24°.
b) The central angle, ACD, is subtended by the same arc as
the inscribed angle, AED. Therefore, the measure of ACD
is twice the measure of AED. The measure of ACD is 48°.
Chapter 10 Review Page 404 Question 6
The angle labelled x is an inscribed angle subtended by the same arc
as the central angle that measures 96°. Therefore, the measure of the
inscribed angle labelled x is one-half of 96°, or 48°.
The angle labelled y is an inscribed angle subtended by the same arc
as the central angle that measures 96°. Therefore, the measure of the
inscribed angle labelled y is one-half of 96° or 48°.
750 MHR ● MathLinks 9 Solutions
Chapter 10 Review Page 404 Question 7
Parmjeet’s thinking is not correct. The measure of a central angle subtended by the same arc as
an inscribed angle is double the measure of the inscribed angle.
Chapter 10 Review Page 404 Question 8
The dartboard is divided into twenty congruent central angles. To find the measure of each angle,
divide the entire measure of a circle, 360°, by 20.
360 ÷ 20 = 18
The measure of each central angle on the dartboard is 18°.
Chapter 10 Review Page 404 Question 9
EFG is an inscribed angle subtended by the diameter of the circle.
Therefore, the measure of EFG is 90°.
Chapter 10 Review Page 404 Question 10
ADB is an inscribed angle subtended by the diameter of the circle.
The measure of ADB is 90°. Use the angle sum of a triangle to
find the measure of BAD.
BAD = 180 – (ADB + ABD)
= 180 – (90 + 62)
= 28
The measure of BAD is 28°.
Chapter 10 Review Page 404 Question 11
Example: The line is a perpendicular bisector of the chord, and
therefore passes through the centre of the circle.
MHR ● MathLinks 9 Solutions 751
Chapter 10 Review Page 405 Question 12
Example: She should have found the perpendicular bisector of her string and the perpendicular
bisector of a second string. The intersection of the two perpendicular bisectors gives the location
of the centre.
Chapter 10 Review Page 405 Question 13
Since the bisector of AE passes through the centre of the circle,
the bisector, CD, is perpendicular to AE. Therefore, ACB is a
right triangle. Use the Pythagorean relationship to find AB.
CB2 + AB
2 = AC
2
102 + AB
2 = 26
2
100 + AB2 = 676
AB2 = 576
AB = 576
AB = 24
AE = 2AB
= 2(24)
= 48
The length of AE is 48 m.
Chapter 10 Review Page 405 Question 14
Example: The archaeologists could draw two chords and the perpendicular bisectors of those
chords. The intersection of the two perpendicular bisectors is the centre of the circle. Next, they
can measure the distance from the centre to a point on the circle, which is the radius. Finally, they
can use the radius to calculate the circumference.
Chapter 10 Review Page 405 Question 15
Since the radius is perpendicular to FG, it also bisects FG.
Name the intersection of chord FG and the radius H.
HCG is a right triangle.
The length of segment HC is the shortest distance between FG and the
centre of the circle. To find the length of HC, use the Pythagorean
relationship. The length of HG is 9 cm, since FG is bisected at H. The
length of the radius, GC, is 11 cm, since the diameter is 22 cm.
HG2 + HC
2 = GC
2
92 + HC
2 = 11
2
81 + HC2 = 121
HC2 = 40
HC = 40
HC 6.3
The shortest distance between FG and the centre of the circle is about 6.3 cm.
752 MHR ● MathLinks 9 Solutions
Chapter 10 Review Page 405 Question 16
Since DE is tangent to the circle at G, the radius CG is
perpendicular to DE. Therefore, CGE is a right triangle.
Find the measure of ECG.
ECG = 180 – (EGC + GEC)
= 180 – (90 + 43)
= 47
The measure of ECG is 47°.
ECG and FCG form a straight line, so they are supplementary.
180 – 47 = 133
The measure of FCG is 133°.
Chapter 10 Review Page 405 Question 17
Since AB is tangent to the circle at B, AB is perpendicular to
radius CB. Therefore, the triangle formed is a right triangle.
Find the length of the diameter. Label the other endpoint of the
diameter D. Find the length of BD by using the Pythagorean
relationship.
AB2 + BD
2 = AD
2
92 + BD
2 = 15
2
81 + BD2 = 225
BD2 = 144
BD = 144
BD = 12
The length of the diameter is 12 mm. Therefore, the length of the radius is 6 mm.
MHR ● MathLinks 9 Solutions 753
Chapter 10 Review Page 405 Question 18
Refer to the diagram.
The segment d represents the horizontal distance. Since the segment labelled d is tangent to the
circle at the radius, a right triangle is formed. Use the Pythagorean relationship to find the length
of d.
502 + d
2 = 140
2
2500 + d2 = 19 600
d2 = 17 100
d = 17 100
d 130.76696
The horizontal distance is 131 m rounded to the nearest metre.
50 m 140 m
d
754 MHR ● MathLinks 9 Solutions
Chapter 10 Review Page 405 Question 19
a) Since AF is tangent to the circle at E, AF is perpendicular to EC.
Therefore, the measure of CEF is 90°.
b) CEF is a right triangle. Use the angle sum of a triangle.
ECF = 180 – (CEF + EFC)
= 180 – (90 + 48)
= 42
The measure of ECF is 42°.
c) ECD and ECF are supplementary angles.
180 – 42 = 138
The measure of ECD is 138°.
d) DEC is an isosceles triangle, since two of its sides are radii of the circle. To find the
measure of DEC, subtract 138 from 180 and divide by 2.
DEC = (180 – 138) ÷ 2
= 21
The measure of DEC is 21°.
e) AED, DEC, and CEF are supplementary.
AED = 180 – (90 + 21)
= 69
The measure of AED is 69°.
f) EDB is an inscribed angle subtended by the same arc as central ECF, which is 42°.
EDB = ECF 2
= 42 2
= 21
Therefore, the measure of EDB is 21°.
MHR ● MathLinks 9 Solutions 755
Chapter 10 Review Page 405 Question 20
a) ACE is a central angle subtended by the same arc as
inscribed FDE.
ACE = 2 FDE
= 2 21
= 42.
The measure of ACE is 42°.
b) Since AB is tangent to the circle at E, AB is
perpendicular to CE. Therefore, CEA is a right triangle.
Use the angle sum of a triangle to find the measure of
CAB.
CAB = 180 – (CEA +ACE)
= 180 – (90 + 42)
= 48
The measure of CAB is 48°.
756 MHR ● MathLinks 9 Solutions
Chapter 10 Practice Test
Chapter 10 Practice Test Page 406 Question 1
Statement A is false. A central angle is twice the measure of an inscribed angle subtended by the
same arc.
Statement B is false. Two inscribed angles are equal if they are subtended by equal arcs.
Statement C is true.
Statement D is false. The largest inscribed angle in a circle is 90°.
The correct choice is C.
Chapter 10 Practice Test Page 406 Question 2
The measure of an inscribed angle is one-half the measure of a
central angle subtended by the same arc. Since the measure of the
central angle is 100°, the measure of the inscribed angle is 50°.
The correct choice is B.
Chapter 10 Practice Test Page 406 Question 3
Refer to the diagram. Label the intersection of radius CA and
chord EF point D. Since CA and EF are perpendicular, CA
bisects EF. To find the length of chord EF, find the length of
segment ED and double it. Since EDC is a right triangle, use
the Pythagorean relationship to find the length of segment ED.
The length of segment DC is 13 – 8, or 5 mm.
DC2 + ED
2 = EC
2
52 + ED
2 = 13
2
25 + ED2 = 169
ED2 = 144
ED = 144
ED = 12
Since the length of ED is 12 mm, the length of EF is 24 mm.
MHR ● MathLinks 9 Solutions 757
Chapter 10 Practice Test Page 406 Question 4
Since AB is tangent to the circle at D, DB is perpendicular to DC.
Therefore, BCD is a right triangle. To find the measure of
BCD, use the angle sum of a triangle.
BCD = 180 – (CDB + DBC)
= 180 – (90 + 54)
= 36
The measure of BCD is 36°.
Chapter 10 Practice Test Page 406 Question 5
In the diagram, the segment labelled, x, is the radius of the
circle. The right triangle has a radius as one of its legs. Use the
Pythagorean relationship to find the radius.
92 + x
2 = 13
2
81 + x2 = 169
x2 = 88
x = 88
x 9.4
The length of the radius is about 9.4 cm.
Chapter 10 Practice Test Page 406 Question 6
Since line l is tangent to the circle, it is perpendicular to the
radius. Therefore, the triangle is a right triangle. The central
angle, which is one of the angles of the right triangle,
measures 39° because it is opposite the angle that is labelled
39°. To find the measure of , use the angle sum of a
triangle.
= 180 – (90 + 39)
= 51
The measure of is 51°.
758 MHR ● MathLinks 9 Solutions
Chapter 10 Practice Test Page 407 Question 7
ADB and AEB are inscribed angles subtended by the same
arc, so they are congruent. Therefore, the measure of ADB is
41°. ACB is a central angle subtended by the same arc as
inscribed AEB, which is 41°. Therefore, the measure of ACB
is 82°.
Chapter 10 Practice Test Page 407 Question 8
Refer to the diagram. If a radius is drawn to one of the endpoints of
the 20-mm chord, and the shortest distance is perpendicular to the
water level, a right triangle is formed. Use the Pythagorean
relationship where d represents the shortest distance and the radius
is 17 mm.
102 + d
2 = 17
2
100 + d2 = 289
d2 = 189
d = 189
d 13.75
The shortest distance, d, is about 14 mm.
MHR ● MathLinks 9 Solutions 759
Chapter 10 Practice Test Page 407 Question 9
In the diagram, x represents the side length of the square.
The four triangles formed by the two intersecting diameters are congruent right triangles. Use the
Pythagorean relationship to find the measure of x.
202 +20
2 = x
2
400 + 400 = x2
800 = x2
800 = x
28.28 x
The largest dimension of a square that can be cut from the log is about 28.28 cm.