Chapter 10 Circle Geometry Section 10.1 Exploring Angles ...oneillmath9.wikispaces.com/file/view/ML9...

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°.


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°.


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


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.


a) FCG is a central angle subtended by the same

arc as inscribed FEG.

FCG = 2FEG

FCG = 2(45)


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.


Section 10.1 83 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.


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°



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°.


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°.



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°.


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°.



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.



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°.



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



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



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°.


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.



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°.



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°.


Section 10.2 Exploring Chord Properties


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.


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.



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.


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.



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.



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.



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.


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



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.


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.


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.



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.



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°.



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



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.



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



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°.


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.


Section 10.3 Tangents to a Circle


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




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°.



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.



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.



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



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.



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




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.



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.


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



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.


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°.



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°



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.




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.




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).


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.



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.




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.



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.


Chapter 10 Review 4 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.


A CHORD is a line segment that has both endpoints on the same circle.


A PERPENDICULAR BISECTOR is a line or line segment that passes through the midpoint of a

line segment at 90°.


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°.


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°.



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.


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°.


EFG is an inscribed angle subtended by the diameter of the circle.

Therefore, the measure of EFG is 90°.


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°.


Example: The line is a perpendicular bisector of the chord, and

therefore passes through the centre of the circle.



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.


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.


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.


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.



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



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.




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



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


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°.



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°.


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.


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.


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.



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°.


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.


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°.



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°.


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.



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.

x

20 cm 20 cm

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