module 3: geometry and trigonometry - open school...

Module 3: Geometry and Trigonometry

Section 1: Circle Geometry

Introduction 3

Lesson 1 Vocabulary of Circles 5

Lesson 2 Geometry of Circles and Chords 19

Lesson 3 Angles and Arcs of a Circle 31

Lesson 4 Properties of Tangents 47

Lesson 5 Exploring Polygons 59

Review 69

Section Assignment 3.1 73

Section 2: Trigonometry

Introduction 83

Lesson 1 The Trigonometric Ratios of Angles 0° ≤ θ ≤ 360° 85

Lesson 2 Related Angles and Solving TrigonometricEquations of Linear Form 95

Lesson 3 Solving Triangles 103

Lesson 4 The Ambiguous Case 115

Review 123


Principles of Mathematics 11 Contents 1

Module 3

2 Contents Principles of Mathematics 11

Module 3

Section 3: Analytic Geometry I

Introduction 135

Lesson 1 Circle 137

Lesson 2 Distance Between Points and Lines 147

Lesson 3 Systems of Linear Equations 157

Lesson 4 Solving Systems of Equations Algebraically 169

Lesson 5 Systems of Equations ContainingThree Variables 181

Review 189


Section 4: Analytic Geometry II

Introduction 201

Lesson 1 Applications of Systems 203

Lesson 2 Non-Linear Systems 209

Lesson 3 Graphing Linear Inequalitiesin Two Variables 221

Lesson 4 Quadratic, Rational, and AbsoluteValue Inequalities 231

Lesson 5 Verifying and Proving Assertions inCoordinate Geometry 245

Review 253


Module 3 Answer Key 263

Module 3, Section 1

Geometry

IntroductionIn this section, you will study the circle. You will becomefamiliar with the vocabulary of terms pertaining to the circle,and investigate properties of chords, angles, arcs, and tangentsof a circle. Once you have made conjectures based on yourinvestigations, you will use your knowledge and proofs to verifyyour conjectures. You will use these properties to solve othergeometric problems. You will also establish relationships amonginterior and exterior angles of polygons.

Section 1 — Outline

Lesson 1 Vocabulary of Circles

Lesson 2 Geometry of Circles and Chords

Lesson 3 Angles and Arcs of a Circle

Lesson 4 Properties of Tangents

Lesson 5 Exploring Polygons

Review

To do the exercises in this section, you will need to drawaccurate circles. Use a compass (preferred) or a circle templateup to 2” or 50 mm. You will also need a protractor.

Principles of Mathematics 11 Section 1, Introduction 3

Module 3

Notes

4 Section 1, Introduction Principles of Mathematics 11

Module 3

Module 3

Lesson 1

Vocabulary of Circles

OutcomeWhen you complete this lesson, you will be able to

• use the mathematical vocabulary associated with circles

OverviewThe circle is a geometric shape. Circles can be found aroundyou. The invention of the wheel was the forerunner of theindustrial era. Clocks and windmills were great advances incivilization based on the application of wheels. In the newinformation era that is now evolving, note that the circularcomputer disks and compact disks are circular in shape.

In studying geometry, you must know the various definitionsand properties that you have learned about triangles andquadrilaterals to prove deductions. To do the mathematicsinvolving circles, you need to know the vocabulary of the circleand understand the significance of each word.

The word circle is derived from the Latin word “circus” whichmeans “ring” or “racetrack.”

Definition

A circle is the set of all points in a plane equidistant from agiven fixed point.

The centre is the given fixed point. You name a circleby its centre. If the centreis P, then the circle can bedenoted by OP. In thiscourse, when you see adot at the centre of acircle, assume that itrepresents the centrepoint.

l

diameter

radius

centre

P

Principles of Mathematics 11 Section 1, Lesson 1 5

•

Module 3

The points inside the circle form the circle’s interior. Thepoints outside the circle form its exterior. The circle is only thecurved line. The radius of a circle is a segment that has thecentre as one endpoint and a point on the circle as the otherendpoint. The radius can be referred to as a line segment or as alength of a line segment. By the definition of a circle, all radii ofa circle are congruent. If two circles have the same radius, theyare congruent circles.

If two or more circles share the same centre, then they areconcentric circles.

You are going to do some investigations that allow you todevelop some more definitions. You will be given two diagrams.The first diagram will show examples of geometrical terms andthe second diagram, will show you counterexamples.

l

C o n c e n t r i c C i r c l e s

l

2l

2

C o n g r u e n t C i r c l e s

6 Section 1, Lesson 1 Principles of Mathematics 11

1. Investigate a chord.

Chord Not a Chord

How would you define a chord?

A chord of a circle is a segment whose endpoints are on thecircle.

2. Investigate a diameter.

Diameter Not a Diameter

How would you define a diameter?

A diameter, d, of a circle is a chord that passes through thecentre.

UV, ST, and QR are not

diametersAB and CD are diameters

lU

l

l

l

l

lV

S

Q

T

R

l

D

B

C

A

l

l

l

l

WY, QR, and ST are

not chords

AB, CD, and EF are

chords

l

l

l

l

l

l

S

T

R

Q Y

X

W

l

E

F

D

C

B

A

l

l

ll

ll


Module 3

The diameter of a circle is twice the radius. That is, d = 2r.

The radius of a circle is half its diameter. That is

AC is the diameter,and AB and CB areradii.

3. Investigate a secant.

Secant Not a Secant

How would you define a secant?

A secant is a line that intersects a circle at two points.

ST, PR, and UV are not

secants

AB, CD, and EF are

secants

lS

l

l

l

lT

Q

V

U

R

lP

l

D

B

C

A

l

l

l

l

l

l

E

F

l

C

A

l

l

diameter

radi

us

radiu

s

B

r d=12

.


Module 3

4. Investigate a tangent.

Tangent Not a Tangent

How would you define a tangent?

A tangent is a line that intersects the circle at only onepoint. The point where the tangent touches the circle iscalled the point of tangency.

Notice in the diagram that no part of the tangent line lies inthe interior of the circle. Notice that part of the secant line isin the interior of the circle.

l

secant

l

l

lpoint o f tangency

tange n t

PQ, ST, and UV are not tangents

CD, EF, and AB are tangents

S

l

l

l

l

T

Q

UV

P

l

l

l

D

B

C

A

l

l

l

l

lE

F


Module 3

You have learned the definitions of a chord, radius, diameter,secant, and tangent. These deal with lines and line segments.You will now examine some facts about the curve that is calledthe circle.

An arc of a circle is two points on the circle and the part of thecircle between the two points. The two points are called theendpoints of the arc. The symbol for an arc is . The symbol isplaced above the letters that name the endpoints of the arc. ArcAB is symbolized as (Note: Arc AB =

Just as we classify triangles and angles into different types, wedivide arcs into three different classifications.

1. A semicircle is an arc of a circle whose endpoints are theendpoints of the diameter. It issymbolized as

2. A minor arc is an arc of a circle that is smaller than asemicircle. In the diagram,

represent minorarcs.

minor arcs AP and PB

3. A major arc is an arc of a circle that is larger than asemicircle. In the diagram, dashes representmajor arcs.

major arc PBA major arc PAB

B

P

OA l

l

l ll l B

P

OA

l

l

PBA and PAB

AP and PBl l BA

P

O

l

l

semicircle APB

APB.l ll

l

BA

P

O

AB .)AB or BA .


Module 3

arc. You name semicircles and major arcs with the letters ofthree points — the first and last letters are the endpoints andthe middle letter is any other point on the arc.

Circles can be divided into various regions by radii and/orchords.

1. A sector of a circle is a region bounded by two radii of a circle and the intercepted arc. This is theshaded region in the figure.

Sectors can be minor, major, or halfcircles as determined by the minor,major, or semicircle arcs that theyintercept.

2. A segment of a circle is theregion bounded by a chord andits intercepted arc (the arc thatshares its endpoints).

Segments can be classified asminor, major, or half circles asdetermined by minor, major, orsemicircle arcs.

There are two types of angles that can be drawn in a circle.

1. Central angle — an angle which has its vertex at the centre of acircle. ∠ AOB and ∠ COD arecentral angles. ∠ AOB is“subtended” by and ∠ CODis “subtended” by .

l

C

D

B

A

O

Ol

B

Asegment

sector

l


Module 3

AB»

CD¼

2. Inscribed angle — an angle which has its vertex on the circleitself. ∠ ASB and ∠ APB areinscribed angles. The inscribed

angles are subtended by

Self-Marking Activity 1. Use the diagram on the right to

answer the followingquestions:a) Name one diameter.

b) Name four central angles

c) Name two semicircles.

d) Name a secant line.

e) Name five chords.f) Name five minor arcs.

g) Name seven inscribed angles

h) Name eight major arcs.

2. Use the diagram on the right to answer the followingquestions.a) Shade a major sector of

the circle.

b) Shade a minor segment ofthe circle.

c) Name four central angles.

d) Name four inscribedangles.

e) Name two points oftangency.

f) Describe two minorsegments in the circle.

g) Name two tangent lines.

Check your answers in the Module 3 Answer Key.

l

C

l

ll

l

l

A

D

F

O

E

G

l

l

B

l C

l

l

l

l

l

A

DE

O

B

AB . l

S

P

BA

O


Module 3

In the next lessons you will encounter geometry terms andrelationships that you have studied in previous mathematicscourses. Here is a brief summary for reference.

Special Types of Angles

1. Acute Angles

An angle whose measure is greater than 0º but less than 90ºis called acute.

2. Obtuse Angles

An angle whose measure is greater than 90º but less than180º is called obtuse.

3. Right Angles

An angle whose measure is 90º is called a right angle.


Module 3

65º 35º

These are acute angles.

170º 130º

These are obtuse angles.

These are right angles.Note: We use the symbol to denote a rightangle.

4. Complementary Angles

Two angles are called complementary if the sum of theirmeasures is 90º.

5. Supplementary Angles

Two angles are called supplementary if the sum of theirmeasures is 180º.

Two angles which are supplementary can consist of a pair of rightangles or can consist of one acute angle and one obtuse angle.

6. Vertical Angles

1

2

3

4

Angles 1 and 3 are called vertical angles.Angles 2 and 4 are called vertical angles.

Vertical angles are congruent. Here,congruent means that the angles havethe same measure.

70º 110º

These two angles aresupplementarybecause 70º + 110º = 180º

a b

a + b = 180º a + b = 180º

ab

a

ba + b = 90º

50º40º

These two angles arealso complementary.


Module 3

Consider ∠ 3 and ∠ 6.

So are ∠ 4 and ∠ 5.

When a transversal crosses two lines, it forms 2 pairs of interiorangles. If the lines are parallel, then these interior angles aresupplementary.

Congruent Triangles

1. If three sides of a triangle are equal in length to three sides ofanother triangle, the triangles are congruent (SSS).

Example:

AB EF

BC FG

AC EG

ABC EFG SSS

=

=

=∴∆ ≅ ∆ ( )

A

B C F

E

G

4

5

3

6

These are called interior angles.


Module 3

2. If two sides and a contained angle of a triangle are equal to twosides and the contained angle of another triangle, the trianglesare congruent (SAS).

Example:

3. If two angles and the contained side of a triangle are equal totwo angles and the contained side of another triangle, thetriangles are congruent (ASA).

Example:

4. If two angles and one side of a triangle are equal to two anglesand one side of another triangle, the triangles are congruent(AAS).

Example:

∠ = ∠∠ = ∠

=∴∆ ≅ ∆ ( )

A DG F

BG EF

ABG DEF AASA

B

G D F

E

∠ = ∠

=∠ = ∠

∴∆ ≅ ∆ ( )

BAG TEN

AG ENAGB ENT

BAG TEN ASAA

B

G E N

T

HE TOHEN TOP

EN OP

HEN TOP SAS

=∠ = ∠

=∴∆ ≅ ∆ ( )

H

E N O

T

P


Module 3

5. If the hypotenuse and one leg of a right triangle are equal to thehypotenuse and one leg of another right triangle, then the righttriangles are congruent (HL).

Example:

6. If the hypotenuse and one angle of a right triangle are equal to the hypotenuse and one angle of another right triangle, then theright triangles are congruent (HA).

Example:

∠ ∠∠ = ∠

=∆ ≅ ∆ ( )

B and E are right anglesA D

AC DF

ABC DEF HA

A B

C F

ED

∠ ∠

=

=∆ ≅ ∆ ( )

A and X are right angles

AB XY

BC YZ

ABC XYZ HL

A

B

C X

Y

Z


Module 3

Module 3

Lesson 2

Geometry of Circles and Chords


• identify the properties of chords and other features of a circle

OverviewYou will investigate some of the properties of chords in a circleby doing some construction work and then proving the situation.If you have access to technology and the use of the softwarecalled Cabri or Geometry Sketchpad, you may use it.

The first investigation you will do pertains to the relationshipbetween a perpendicular line from the centre of the circle to thechord.

Investigation 1 1. Using a compass, construct

a circle with centre O.

2. Draw two points on thecircle and label them Aand B. Join the twopoints.

3. Using a protractor or aset square, construct aperpendicular to thechord that passesthrough O andintersects the chord at D.

4. Measure AD and BD. What can you conclude?

The lengths of AD and BD should be equal.

l

D BA

O

ll


Module 3

In Principles of Mathematics 10 you learned how to provetheorems and work geometry problems.

If you take your diagram and join , you willform two triangles, ∆ OADand ∆ OBD which you canprove to be congruent.

Because arethe radii of a circle, ∆ OABis isosceles with ∠ OAD and∠ OBD being congruentbase angles and ∠ ODA and∠ ODB are 90° because OD was the perpendicular. ∆ OAD ≅∆ OBD by SAA or AAS and AD = BD because of correspondingsides of congruent ∆s. This means that have thesame measure.

Therefore, you have proven that a line from the centre,perpendicular to a chord bisects the chord.

A second investigation involves the line segment drawn fromthe centre to the midpoint of the chord.

Investigation 21. Using a compass, construct

a circle with centre O.

2. Construct two points onthe circle and labelthem A and B.

3. Bisect to find thecentre of the chord at D.

4. Use a protractor to findthe measure of theangles formed at D.

What is the measure of the angles?

(The measure should be 90° for both angles.)

AB

l

DBA

O

ll

AD and BD

OA and OB

OA and OB

l

D BA

O

ll


If you take our diagram and join you willform two triangles OADand OBD, which you canprove to be congruent.

Because OA OB (equalradii of the same circle),AD BD (D wasconstructed to be themidpoint), and is theside common to bothtriangles, OAD OBD by SSS. Therefore, ODA OBDby corresponding angles and because they are supplementary,each angle is 90°. Therefore, the line segment from the centreto the midpoint of the chord is perpendicular to thechord.

It can be shown that the perpendicular bisector of a chordpasses through the centre of the circle.

Investigation 3

1. Using a compass, draw a circle with centre O.

2. Draw two non-parallel chords thatare not the samelength (AB and CD.)

3. Draw theperpendicularbisector of each chordand extend until theyintersect.

What is special about the point of intersection?

(The two perpendicular bisectors intersect at the centre.)

OD

OA and OB,

D BA

O


Module 3

D

BA

O

F

E

C

=

=

∠ = ∠ ≅ ∆ ∆

The perpendicular bisector of a line segment contains all pointsthat are equidistant from the endpoints of the chord. Regardlessof the length of the chord, the perpendicular bisector will passthrough the centre of the circle.

Therefore, the perpendicular bisector of a chord alwayspasses through the centre of the circle.

Another property of chords of a circle that you can investigate isthat congruent or equal chords are equidistant from the centreof the circle.

Investigation 4

1. Construct a circle with centreO.

2. Mark off two arcs of equallength and join theendpoints.

3. Label the equal chords asAB and CD.

4. Draw the perpendicularbisectors of

5. Label the midpoint of as E and the midpoint of as F.

6. Measure using your ruler.

What did you find about the measures? (They are equal inmeasure.)

It appears that congruent chords are equally distant from thecentre.

OE and OF

CDAB

AB and CD.


Module 3

D

B

A

OE

C

F

You can prove this geometrically. Because

(equal radii of the same circle)and are congruentchords, ∆ OAB and ∆ OCD arecongruent by SSS.

Because the triangles arecongruent, ∠ OAE and ∠ OCFare corresponding parts,∠ OEA and ∠ OFC are 90°(perpendicular bisector) andOA = OC (equal radii).

∴ ∆ OAE ≅ ∆ OCF (AAS)

∴ OE = OF are corresponding parts verifying that congruent

chords are equidistant from the centre of the circle.

The converse of the last statement is that chords equidistantfrom the centre of a circle are congruent.

Investigation 51. Construct a circle with

centre O.

2. From centre O, draw twoequal (congruent) lengthsless than the radius ofthe circle (OX, OY).

3. At the endpoints X and Yof the two segments,construct perpendicularsand label the intersectionpoints on the circle at A and B and C and D.

4. Measure the lengths of What conjecture can youmake?

(The chords that are equidistant from the centre arecongruent.)

AB and CD.

l

D

A

O

B

C

Y

X

AB and CD

= = =AO OC OB OD

l D

B

A

OE

C

F


Module 3

Because OX = OY (chords are equidistant from the centre,∠ OXB =∠ OYD (both 90°)and OB = OD (equal radii),∆ OXB ≅ ∆ OYD hypotenuseleg (H.L.) and BX = DY(corresponding parts ofcongruent triangles).Similarly ∆ OXA and ∆ OYCare congruent (H.L.) and AX= CY (corresponding parts).BX + AX = CY + DY (addition). This is equivalent to AB = CD.

This proves that chords equidistant from the centre of acircle are congruent.

Summary of Properties of Chords of a Circle

• A line from the centre of a circle, perpendicular to a chord,bisects the chord.

• The line segment drawn from the centre to the midpoint ofthe chord is perpendicular to the chord.

• The perpendicular bisector of a chord passes through thecentre of the circle.

• Congruent chords are equidistant from the centre.

• Chords equidistant from the centre of a circle are equal.

Example 1Finda) AB

b) OA

c) BC

d) OC

e) perimeter of ∆ OACf) area of ∆ OAC

g) area of O

h) circumference of O

Note: OO means “circle withcenter labelled O.”

Given: OO, OB ACOB 3AC 8

⊥==

l

A

O

B C

l

D

A

O

B

C

Y

X


Module 3

•

•

•

O is the shorthandsymbol for a circle.

•

•

Solution

a) The line through the centre perpendicular to the chordbisects the chord. ∴ AB = 4.

b) ∆ OAB is a right triangle because OB ⊥ AC

c) BC = 4;

d) is a radius having the same length as ∴ OC = 5 units.

e) Perimeter of ∆ OAC = OA + OC + AC= 5 + 5 + 8= 18 units

f) Area of ∆ OAC

g) Area of O = πr2

= π • 52

= 25π units2

h) Circumference of O: = 2πr= 2π • 5= 10π units

= ( )( )

= ( )( )

=

1212

3 8

12 2

OB AC

units

OA.OC

OB bisects AC

OA OB AB

OA units

2 2 2

2 23 4

25

5

= +

= +==


Module 3

•

•

Self-Marking Activity1. For each of the following, state the property of the chords

that are being illustrated.

a) b)

c) d)

e)

l

B

O

A

C

D

l

B

OA

l

B

OA

C

D

A B = C D

l

B

O

A

C

DlX O

Y


Module 3

2. Find the indicated values for the following and supply thereason(s). These diagrams are not drawn to scale.

a)

b)

c) Find OE.

lO

B

A

C

D

5

A B = C DE

Find AB.

l

A

O

B

C

D

E

4.2

FindCD.

l

A

O

B

C D

12


Module 3

3. Given: Circle with centre O, , OC = 4, and AC is 3.

Find the length of

c) the radius of the circle

d) the diameter of the circle

Give reason(s) for your answers.

4. Given: Circle with centre O, CD = 16, EF = 12, and

Find the length of


5. Given: Circle with centre O, , DE = 20, OF = DF.

Find the length of


a) OF

b) EO

c) DF

d) CDl

E

O

F

D

C

OF CD⊥

a) OF

b) EX

c) OXl

E

O

F

X

D

C

OX EF.⊥

a) BC

b) AB

l

A

O

B

C

OC AB⊥


Module 3

6. A circle with centre O has a radius of 24 cm. is theperpendicular bisector of If CD is 36 cm, find thelength of SB.

7. In a circle, a chord is 24 cm in length and is 8 cm from thecentre. Calculate the length of the diameter.

8. a) Draw a sketch for the following situation. Given a circlewith centre O, two parallel chords are drawn on oppositesides of the diameter. The two chords are 16 cm and20 cm in length. The diameter of the circle is 24 cm.

b) Find the shortest distance between the two chords.

9.

You are to verify using geometrical reasons why bisects∠ APD.

OP

l

A

O

E

D

C

B

F

P

AB DC

OE PA

OF PD

≅

⊥

⊥

CD.AB

l

A

O

SDC

B


Module 3

10.

JI = 6 units

HK = 8 unitsHI = 6 units

PB = 4 units

PA = 3 units

Find the perimeter of theresulting pentagon AHIBP.


lA

P

I

K

J

B

H


Module 3

Lesson 3

Angles and Arcs of a Circle


• identify the relationships that exist between the angles and arclengths in a circle

OverviewIn Lesson 1, you learned the definition of a central angle and aninscribed angle.

In the diagram to the left thereare actually two central anglesAOB, one subtended by majorarc ACB and the othersubtended by minor arc AB.The sum of these two angles is360°, so ∠ AOB (major) = 360° – 60° or 300°.

l

A

O

B

°60

lC

If the endpoints of an arc arethe endpoints of a diameter,then the arc is a semi-circleand the central ∠ ACB = 180°.

lAO

B

lC

180°

The measure of the minor arc isdetermined by the measure of thecentral angle (and, of course, thecircle's radius). You will learn aformula in Principles ofMathematics 12 which relatesthem. Our diagram shows central∠ AOB = 60°.

l

A

O

B

°60


Note: To avoidconfusion between

the two centralangles AOB, we will

refer to the onesubtended by a

major arc as ∠AOB(major) and the one

subtended by theminor arc as simply

∠AOB.

Module 3

Investigation 1You will now investigate the relationship between the measureof the central angle and the measure of the inscribed angle.

1. Construct a circle with centre O.

2. Construct three points onthe circle A, B, and C.

3. Construct chords AC and BCand radii AO and BO.

4. Measure ∠ ACB and ∠ AOB.

5. What do you notice aboutthe measure of these twoangles? (∠ AOB = 2 ∠ ACB)

It appears that the measure of the central angle is twice the sizeof the measure of the inscribed angle subtended by the samearc.

l

A

O

B

C

∠ ACB is an inscribed angle if ACand BC are chords of a circle. ABis called the subtending arc of theangle.

C

B

A

If you have concentric circles,notice that the measurements ofthe central angle remain the sameregardless of the radius. Theradius of a circle does not affectthe measure of the central angle.

If ∠AOB = 60°, then ∠COD = 60°.Similarly, if ∠AOB (major) = 300°,then ∠COD (major) =300°.

l

C

O

B

°6 0

A

D


You can now verify this. You have to join and extend it tomeet the circle at D.

∆ ACO is isosceles because of the sides , being radiiof the same circle. Thatmakes ∠ OCA and ∠ OACcongruent base angles.∠ AOD will be equal inmeasure to ∠ OCA + ∠ OACbecause the measure of theexterior angle of a triangle isequal to the sum of themeasures of the two interioropposite angles. This isequivalent to saying ∠ AOD = 2 ∠ OCA.

Similarly, ∆ BCO is isosceles with ∠ OCB and ∠ OBC beingcongruent base angles and ∠ BOD = ∠ OCB + ∠ OBC.

∠ BOD = 2 ∠ OCB

We can then formulate the following:

∠ AOD = 2 ∠ OCA (1)∠ BOD = 2 ∠ OCB (2)

Adding (1) and (2) you get:

∠ AOD + ∠ BOD = 2 ∠ OCA + 2 ∠ OCB∠ AOB = 2(∠ OCA + ∠ OCB)∠ AOB = 2 ∠ ACB

This property can be stated as the measure of the centralangle is equal to twice the measure of the inscribed angleintercepted by the same arc.

OC and OA

l

A

O

B

C

D

CO


Module 3

∠ BCD is an exterior angle of ∆ABC∠ BAC + ∠ ΑBC = ∠ BCDWhy?

∠ 1 + ∠ 2 + ∠ 3 = 180° (triangle sum)

∠ 3 + ∠ 4 = 180°(angles on a line)

so ∠ 1 + ∠ 2 = ∠ 4

A

B

C D1

2

3 4

Example 1What is the measure of ∠ AOB?

Solution

Because ∠ACB is the inscribedangle intercepted by AB, ∠ AOBis the central angle intercepted byAB and will have a measure oftwice the measure of theinscribed angle.

Example 2Given: OB, || , ∠ ACB = 15°

Find: ∠ DBC

Solution

If || to , ∠ DAC and∠ ACB are the same becausethey are alternate interiorangles. So ∠ DAC is also equalto 15º. ∠ DBC is the centralangle which has a measure oftwice the measure of inscribedangle ∠ DAC.

∴ ∠ DBC = 2 • 15° or 30°.

BCAD l

A

B

D

C

BCAD

∴ ∠ ∠

⋅

=

AOB = 2 ACB

= 2 20o

o40

l

A

O

B

C

2 0°


Module 3

• ||= “parallel to”

Next investigate the case where you have two or more inscribedangles subtended by the same arc.

Investigation 2

1. Construct a circle with centre Oand two points A and B onthe circle.

2. Construct points R, S, and Tin the major arc.

3. Construct line segmentsfrom A and B to point Rforming ARB, to point Sforming ASB, and to pointT forming ATB.

4. Measure each angle.

5. What do you notice? (All angles appear to have the samemeasure.)

Verify this:

The property restated is that the inscribed angles subtendedby the same arc are congruent or have the same measure.

congruent because they allhave the same measure∴ ∠ = ∠ = ∠ARB ASB ATB

the measure of an inscribed angle ishalf the measure of the central angle.

1ARB AOB21ASB AOB21ATB AOB2

∠ = ∠

∠ = ∠

∠ = ∠


Module 3

∠∠

∠

Example 3

Find the size of a° and b°.

Because a° and b° aresubtended by the same arc, AB,they both have the samemeasure, so b° = 55°.

Investigate the property of angles inscribed in a semicircle.

1. Construct a circle with centre 0.

2. Construct a diameter AB.

3. Inscribe three angles in thesame semicircle using pointsR, S, and T on the circle.

4. Measure each angle withyour protractor.

5. What do you notice?

(The measures of ARB,ASB, and ATB are

equal to 90°.)

You can verify this because all three inscribed angles aresubtended by semicircle ACB, as is central angle AOB = 180º.

So the measure of each angle is

The property can be stated an angle incribed in a semicirlceis a right angle (90º).

( ) ( )1 1AOB 180º 90 .

2 2o∠ = =

A

OB

T

C

S

R

a

a

o

o

= ∠

= =

1212

110 55

AOB

o o

Solution

A

O

B

C

D

1 10 °

°b

°a


Module 3

∠

∠∠∠

( )

Example 4Given: Diameter AB = 10, AC = 8

Find ∠ CAB.

Solution∠ C is inscribed in a semicircle by diameter . ∴ ∠ C = 90°.∆ ACB is a right triangle so recall the basic definitions.

Points on the same circle are said to be concyclic.

Points A, B, C, and D areconcyclic.

l

A

D

C

B

l

l

l

l

−

∠ =

=

=

∠ =

1 o

o

ACcos CAB

AB8

or .810

cos 0.8 36.9

Therefore, CAB 36.9 .

AB

lAO

B

C


Module 3

θ =adjacent

coshypotenuse

Investigation 3Investigate the following property of a cyclic quadrilateral.

1. Construct a circle with centre O. Construct four points A, B,C, and D on the circle.

2. Construct chords forming quadrilateralABCD.

3. Measure each angle of thequadrilateral.

4. Find the sum of

a) angles A and C

b) angles B and D

5. What do you notice?(They add up to 180°or are supplementary.)

6. Move the verticesaround thecircumference forming different shapes of quadrilaterals.Find the sum of the pairs of opposite angles. What do younotice? (The answer is always 180°.)

You have to verify the fact that opposite angles of a cyclicquadrilateral are supplementary. Construct line segments(radii) OB and OD from the circle centre O. Label ∠ BOD(major) as ∠ 1 and ∠ BOD as ∠ 2.

l

A

D

C

Bl

ll

l

AB, BC, CD, and DA

A quadrilateral whose verticesare concyclic is called a cyclicquadrilateral.

This feature may be describedas a quadrilateral inscribed ina circle or a circle iscircumscribed about aquadrilateral.

l

A

D

C

B

l

l

l

l


Module 3

You have to show that

A + C = 180°

and B + D = 180°

Because A and C are inscribed angles,

In a similar way, you can show that B + D = 180°.

The opposite angles of a cyclic quadrilateral aresupplementary.

Example 5

Circle with centre O, find x°.

Solution

∠ = ∠

∴ ∠ =

DAB DOB

The inscribed angle is half the

measure of the central angle

DAB =12

o o

12

130 65

A

D

C

B

13 0°

x °

O

( )

( )

∴ ∠ + ∠ = ∠ + ∠

=

=

o

o

1A C 2 1

21

3602180

∠ = ∠

∠ = ∠

1A 2

21

and C 12


Module 3

∠ ∠

A

D

C

B

21

∠ ∠

∠ ∠∠ ∠

( )

But ∠ DAB + ∠ DCB = 180°

Opposite angles of a cyclic quadrilateral are supplementary

∴ x = 180° – ∠ DAB= 180° – 65°

= 115°

Example 6Given: Circle with centre E.

Solutiona) Because are equal radii, ∆ BEC is isosceles. If

the central angle BEC = 100°, then each of the base angles is

b) Because inscribed ∠ BDC and the central ∠ BEC aresubtended by BC,

c) ∠ DAB and ∠ DCB are the opposite angles of quadrilateralABCD. Because opposite angles are supplementary and∠ DCB = 30° + 40° or 70°, then ∠ DAB = 180° – 70° or 110°.

d) The sum of the angles of a triangle is 180°. In ∆ DBC, wenow know the size of two of the three angles. ∠ BDC = 50°and ∠ DCB = ∠ ECB + ∠ BCE or 30° + 40° = 70°; ∠ DBC is180° – (50° + 70°) or 60°.

∠ = ∠

∠ = ⋅

=

o

o

1BDC BEC

21

BDC 100250

180 1002

40 40−

= ∴ ∠ =o oECB. .

BE and EC

Find the measure of

a) ∠ ECBb) ∠ BDCc) ∠ BADd) ∠ DBCl

A

D

C

B

1 00 °

3 0°

lF

E


Module 3

Exterior Angles of Cyclic Quadrilaterals

∠ BCF in Example 6 is called an exterior angle of cyclicquadrilateral ABCD. Its measure is equal to ∠ ABC. This angle∠ ABC is referred to as the interior opposite angle.

This can be verified for any cyclic quadrilateral.

∠ 2 + ∠ 3 = 180° because opposite angles of a cyclicquadrilateral are supplementary∠ 1 + ∠ 3 = 180° because is a straight line

∴ ∠ 1 = ∠ 2 because supplements of the same angle have thesame measure

Sometimes angles are inscribed on the same chord. In thislesson, relationships between arcs and angles were investigated.Inscribed angles subtended by the same chord are congruent aswell.

Summary of Properties of Angles and Arcs of an Angle:

1. If a central angle and an inscribed angle are subtended bythe same chord or arc, the central angle is twice theinscribed angle.

2. Inscribed angles subtended by the same arc or chord arecongruent.

3. An angle inscribed in a semicircle is a right angle (90°).

4. Opposite angles of a cyclic quadrilateral are supplementary.

5. An exterior angle of a cyclic quadrilateral is equal to theinterior opposite angle.

AE

B

A D

C

E

2

31


Module 3

Self-Marking Activity 1. Given: Circle with centre O, ∠ AOB = 75°, ∠ DOC = 40°.

2. For the diagram, ∠ DBA = 60°, ∠ EBO = 35°.

a) Is ACB a semicircle?Why or why not?

b) If AE is the interceptedarc, what two inscribedangles are intercepted?

c) What is the measure of∠ AOE?

d) What two angles have ameasure of 45°?

e) What is the measure of∠ BAD?

f) What is the measure of∠ BOD?

l

E

O

B

A

C

D

a) Name four central angles thathave acute angle measures.

b) Name four central angles thathave obtuse angle measures.

c) Name four arcs which aresemicircles.

d) What is the measure of ∠ BOC?e) What is the measure of

∠ EOD?f) Name seven minor arcs.g) What is the measure of the

largest central angle?

l

E

O

B

A

C

D

40°


Module 3

3. Given: Circle with centre O, AC = 5, and OC = 6.5.

4. Given that is the diameterof the circle, prove that thearea of the circle is given by

5. Find the values of the angles as indicated.

a) Find:i) z°ii) y°iii) x°

Aa b

=+

π2 2

4.

O

B

C

A

ab

AB

a) Find the length of AB.b) Find the length of BC.c) Find the area of ABC.d) Find the area of the circle

to one place of decimal.e) Find the exact value of the

circumference of the circle.

O

B

A

C


Module 3

B

C

A

Dy

x

z

°

°

°

5088

°°O

∆

( )

b)

c)

d) Find:

i) xii) ∠ DOBiii) ∠ AOC (major)

l

B

C

A

D

O 3x °

(5x +8)°(5x +8 ) °

(4x +4)°

Find:

i) xii) ∠ DOBiii) ∠ AODiv) ∠ BOC (major)l

B

CA

D

O(2x 1 ) °

(6 x +5)°

Find:i) ∠ Cii) ∠ Aiii) ∠ D

B

C

A

Dx°

10 0°(5x )°


Module 3

e)

f)

6. The hydro meter in British Columbia has four dials.

a) What is the measure of the central angle betweenconsecutive numbers?

b) How large of an angle is formed when the dial moves from3 to 8?

0 1

2

3

456

7

8

90 9

8

7

654

3

2

1

0 1

2

3

456

7

8

90 9

8

7

654

3

2

1

Find:i) xii) ∠ Ciii) ∠ B

∠ = +

∠ = +∠ =

A

CD

4 20

2 1013

2

2

x

xx

l

B

C

A

D

Find:

i) ∠ Rii) ∠ Siii) ∠ Q

∠ = ∠Q S12

l

Q

R

P

S

11 4°


Module 3

7. a) Through how many degrees does the minute hand movein 55 minutes?

b) Through how many degrees does the hour hand move in2.5 hours?

8. In 1998, a high school class surveyed 1600 people todetermine what type of menu item they chose when diningout at certain areas in the city. The survey found thefollowing: 560 preferred beef items, 240 chose pork, 80 saidlamb, 400 asked for chicken, and 320 liked fish.

a) Find the percentage of people for each category.

b) Determine the central angle measure for each categoryand construct a circle graph.


1

2

3

4567

8

9

1011

12

l


Module 3

Module 3

Lesson 4

Properties of Tangents

OutcomesWhen you complete this lesson, you will be able to

• investigate, verify, and apply properties of tangent lines withrespect to circles

• investigate relationships concerning the relationship betweenchords in a circle and tangents to the circle

OverviewA line can intersect a circle at one point, two points, or nopoints. Remember that a line which intersects the circle at onepoint is a tangent, two points is a secant, and at no points is justa line outside the circle.

To construct a tangent to a circle, you need to know theproperties of tangent lines.

Investigation 1Investigate the relation that exists between the tangent to thecircle and the radius where the tangent makes contact with thecircle.

1. Construct a circle with centre 0.

2. Using a straight edge,draw a line whichappears to touch thecircle at only onepoint. Label the pointX. Join

3. Use your protractor tomeasure the angles atX. What do you notice?(Both angles are 90°.)

Next verify that is tangent to the circle at the point ofcontact where Choose any point X' on AB. Assume X'is on AB and on the circle.

In ∆ OXX', ∠ OXX' is given to be 90°.

OX AB⊥ .AB

OX.l

A

O

B

l

X

Xl


Note: This is calledan “indirect”proof—we provesomething isimpossible in orderto prove that theopposite is true.

Module 3

Thus, ∠ OXX' > ∠ OX' X — In a right triangle, there can only beone right angle that is greater than either acute angle.

OX' > OX — The side opposite the larger angle is longer.

For all positions of X', will always be longer than the radiusOX and X', therefore, it does not lie on the circle.

∴ X is the only point of intersection, which means that isthe only tangent.

∴ The tangent to a circle is perpendicular to the radiusat the point of tangency or the point of contact.

Investigation 2Next, investigate the situation where two tangents are drawn toa circle from an external point outside the circle.

1. Construct a circle with centre O and choose a point P outsidethe circle.

2. Join Bisect and mark that as point R. With centre Rand radius , draw a circle cutting the given circle atpoints S and T. Join

Diameter subtends angles OSP and OTP giving themmeasures of 90°. Thus are tangents to theoriginal circle.

3. Measure What do you notice? (They have thesame measures.)

Next, you will verify that tangents drawn from a point outsidethe circle are equal in length.

SP and TP.

SP and PTOP

OS and OT.OR

OPOP.

lOR

l l

S

T

P

AB

OX'


Investigation 3Because are equal radii, angles OSP and OTP are90° angles because the radius is perpendicular to the tangentline and is the common side, ∆ OSP ≅ ∆ OTP byhypotenuse-leg (H.L.).

This makes the tangents SP and TP to be equal correspondingsides.

∴ Tangent segments to a circle from any external pointare congruent.

Example 1are tangents. OA = 5. OB = 13. Find the length of.

Solution

∆ OAB is a right triangle because the radius is perpendicular tothe tangent at the point of contact.

If AB is 12 units, then so is BC because they are tangentsdrawn from the same external point.

∴ + =

+ =

= −

=

AB OA OB

AB

AB

AB or units

2 2 2

2 2 2

2 2 2

5 13

13 5

144 12

Ol l

A

C

B13

5

AB and BCAB and BC

Ol l

S

T

P

OP

OS and OT


Module 3

Example 2AP and BP are tangents from an external point P.

a) Find the size of ∠ APB.b) Find the size of ∠ PAB.

c) What specific type of triangle is ∆ PAB?

Solution

a) ∠ PAO = ∠ PBO = 90° because tangent lines areperpendicular to the radius at the point of contact

∴ o AOBP is cyclic ∠ PAO + ∠ PBO = 180°

∴ ∠ AOB + ∠ APB = 180° cyclic quadrilateral AOBP120° + ∠ APB = 180°

∠ APB = 180° – 120°= 60°

b) OAB is a base angle of isosceles ∆ AOBIf the vertical angle = 120°, than each base angle is 30°∴ ∠ OAB = 30°Because ∠ PAO = 90°Then, ∠ PAB = ∠ PAO – ∠ OAB

= 90° – 30°= 60°

c) Because all angles of ∆ PAB are 60°, ∆ PAB is equilateral.

l

A

O

B

P 12 0°


Module 3

Next investigate some properties of inscribed angles, arcs,chords, and their relationship to tangents of a circle.

Investigation 41. Construct a circle with

centre O and radiusOA..

2. Draw a perpendicularto O at A. This is atangent line.

3. Draw chord andjoin

4. Measure angle AOCand angle CAB. Whatappears to be the relationship between those two angles?

You can verify the conjecture that the measure of the angle between the tangent and the chord is half the measure of thecentral angle.

x + y = 90° because the radius ⊥ tangent

.

2y + z = 180° becausethe sum of the anglesin a triangle is 180°.

Because x = 90 – y andz = 180 – 2y

z = 2(90 – y)

∴ z = 2x or

x z=12

DBOA

l

A

O

Bl ll

D

C

x °

z°

y°

y°

( )

∠ = ∠

∠ = ∠

1(The measure of CAB AOC.

21

Similarly CAD AOC major .)2

OC.AC

l

A

O

Bl ll

D

ClP


Module 3

The measure of the angle between the tangent and thechord is half the central angle subtended by that chord.

The next investigation involves the relation between a tangentand a chord and the inscribed angle on the opposite side of thechord.

Investigation 51. Draw a circle with

centre O andradius OA.

2. Draw aperpendicularto O at A. Thisis a tangentline.

3. Draw a point Bon the circleand join it to Ato form a chord.

4. Add a point D on the tangent line.

5. Construct another point P on the circle on the exterior of∠ DAB.

6. Construct chords .

7. Measure ∠ DAB and ∠ APB. What do you notice? (Themeasures of the two angles are equal.)

Now verify the statement that the angle between the tangentand the chord is equal to the inscribed angle on the oppositeside of the chord.

Join .

y = 90 – x because and 2y + z = 180° since thethree angles of ∆ OBA = 180°z = 180 – 2yz x

xz x

x z

= − −

= − +=

=

180 2 90

180 180 2212

b g

or rewritten

OA AD⊥

l

A

O B

l lD

l

lP

v

zy

xy

OA and OB

AP and BP

l

A

O B

l lD

l

lP


Module 3

But v = that is, the inscribed angle being half the centralangle

∴ x = v, which translates into the angle between a tangentand chord is equal to the inscribed angle on the oppositeside of the chord.

Note: The above result can be proven if ∠ DAB is obtuse(between 90º and 180º).

Example 3PA and PD are tangents

Find:

∠ BCA∠ BDA∠ ABD∠ ACD∠ BAC∠ BDC∠ AOB∠ CDS∠ CBD∠ CAD∠ DAP∠ ADP∠ COD

Solution

∠ BCA = ∠ TAB = 40°, because the angle between a tangent anda chord is equal to the inscribed angle on the oppositeside of the chord.

∠ BDA: ∠ BCA = ∠ BDA, because they are inscribed angles subtended by the same arc .∴∠ BDA = 40°.

∠ ABD = ∠ ACD because they are inscribed angles subtendedby .

∠ AOD = 70°. ∴∠ ABD = ∠ ACD = 35° because the measure ofthe inscribed angle is half the measure of the centralangle.

AD»

AB

A

B

C

l

Dl

S

Q

T

P

O

40º

170º

70º

∠ = ° ∠ = ° ∠ = °AOD BOC TAB70 170 40, ,

12

z,


Module 3

∠ BAC and ∠ BDC: ∠ BAC = ∠ BDC as inscribed angles subtended by the same arc/chord.

∠ BOC = 170° so ∠ BAC = ∠ BDC = 85° because the measure of theinscribed angle is half the measure of the central angle.

∠ AOB = 80° because the central angle is twice the inscribed ∠BCA = 40°

∠ CDS: ∠ COD = 360° – ∠ BOC – ∠ AOB – ∠AOD = 360° – 170° – 80° – 70°

The angle between the tangent and the chord is half the central anglesubtended by that chord.

∠ CBD and ∠ CAD: ∠ CDS = ∠ CBD = ∠ CAD because the anglebetween the tangent and chord equals the inscribed angle onthe opposite side of the chord

∴ ∠ CBD = ∠ CAD = 20°

∠ DAP: ∠ DAP = ∠ ABD — the angle between tangent and chord equals the inscribed angle on the opposite side of the chord

∴ ∠ DAP = 35°

∠ ADP: Similarly, ∠ ADP = 35°

Summary of the Property of Tangents

1. A tangent to a circle is perpendicular to the radius at the point oftangency.

2. Tangent segments to a circle from any external point are congruent.

3. The angle between the tangent and a chord is half the measure ofthe central angle subtended by that chord.

4. The angle between the tangent and chord is equal to the inscribedangle on the opposite side of the chord.

( )

∠ = °

∠ = ∠

= ° = °

COD 401

CDS COD21 40 202


Module 3

Self-Marking Activity

1.

2. , , and are tangents to the circle. Determine the measureof ∠ BAC, if ∠ DEF = 60° and ∠ EFC = 70°. Provide a reason foreach step in your solution strategy.

A

B

C

E

D

F

BCACAB

Given:

ED is tangent tocircle at C∠ BCA = 60°∠ ACD = (2x)°∠ BCE = (3x – 5)°

Find: ∠ B

l

A

O

B

CE D

6 0°


Module 3

3. PQR is a triangle whose sides are tangent to OO at A, B, andC.

a) If AP = 5 units, BR = 5 units, and QC = 9 units, find theperimeter of ∆PQR.

b) If ∠ AOC = 140° and ∠ BOC = 110°, find the number ofdegrees in each of the angles of ∆ PQR.

For this question only, do not use the results in part (a) to dopart (b).

4. AC is a tangent.∠ FOB (major) is 248°∠ BFD = 42°

Find the measure of

a) ∠ 1b) ∠ 2c) ∠ 3

21

3

O

F

B

A

C

D

E42°

Q

P

R

B

A

C

lO


Module 3

•

5. is a tangent and is a secant in the OO. ∠ APC =27º and ∠ ABP = 35º. Find the measure of:

a) ∠ PACb) ∠ ACBc) ∠ BAQd) ∠ COB

6. Describe how you can construct a tangent to a circle at agiven point on the circle when you are given the centre.

7. is a tangent line. Verify that || .

8. are external tangents to OO at Q and R,respectively. Prove that bisects ∠ QPR.


OPPQ and PR

C

D

B

A

P

Q

R

RQABRQ

27°

O

C

A

B

Q

l

P

35°

PCBPAQ


Module 3

•

•


Module 3

Notes

Lesson 5

Exploring Polygons


• identify, name, and classify polygons, and find the measures ofthe angles of the polygon

OverviewThe word “polygon” originated in ancient Greece where it wasinterpreted as meaning “many angles.” Today, we tend to thinkof it as meaning “many sides.”

A polygon is a plane figure that is formed by three or moresegments called sides. Two conditions must be met:

1. Each side intersects exactly two other sides, once at eachendpoint

2. No two sides with a common endpoint are collinear

For example,

Notice that these are not polygons:


Module 3

A polygon is convex if no two points of it lie on opposite sides ofa line containing a side of a polygon. A polygon that is notconvex is called nonconvex or concave.

This section will deal primarily with convex polygons.

Polygons are classified according to the number of sides.Because a polygon must have a minimum of three sides, thesimplest polygon is the triangle. Each type of polygon has aprefix from which you can tell the number of sides. You willalready be familiar with some of these.

There are points on oppositesides of a line containing aside of the polygon.

No two points lie on oppositesides of a line containing theside of the polygon.

convex nonconvex or concave


Number of Sides Type of Polygon

3 triangle4 quadrilateral5 pentagon6 hexagon7 heptagon8 octagon9 nonagon10 decagon11 undecagon12 dodecagon

If two vertices of a polygon are connected by a side, they arecalled consecutive vertices. If two sides share a commonvertex, they are consecutive sides. If two angles share acommon side, they are consecutive angles.

When naming a polygon, identify each vertex by a letter. Whenyou name the polygon, list the lettered vertices consecutively.

The polygon is named ABCDEFG.

A B

C

D

E

F

G

c o n s e c u t i v e

v e r t i c e s


s i d e s


a n g l e s


Module 3

Example 1Complete the following chart. Take each of the various types ofconvex polygons and in each draw diagonals from one vertex todivide the polygon into triangular regions. Find the totalnumber of degrees in all the triangles.

t r i a n g l e q u a d r i l a t e r a l

A regular polygon is a polygon that isequilateral and equiangular.

An equiangular polygon is a polygonwith all its interior angles equal inmeasure.

An equilateral polygon is a polygon withall sides equal in length.

A diagonal of a polygon is a segment thatjoins two nonconsecutive vertices. Everysegment that joins two vertices of apolygon must be a side or a diagonal.


Module 3

Solution

Study the pattern that results and come up with a formula thatwill give you the sum of the measures of the interior angles.

If n = number of sides, what must be done to n to get thenumber of triangles? (Subtract 2 from n to get the number oftriangles.)

To get the number of degrees in the interior, what do you dowith (n – 2)? (Multiply by 180°.)

The sum of the measures of the interior angles of a convex n-gonis (n – 2)(180°).

In a regular polygon, all its interior angles are congruent. If (n – 2)(180°) gives you the sum of the measures of the interiorangle, what would you do to find the measure of each angle?You are right if you said divide by n.

Therefore, the measure of each interior angle of a regular

polygon is .

h e x a g o np e n ta g o n


Module 3

PolygonNumber of

SidesNumber ofTriangles

Sum of Measuresof Interior Angles

TriangleQuadrilateralPentagonHexagonHeptagonOctagonNonagonDecagon

345678910

123

1(180°) = 180°2(180°) = 360°3(180°) = 540°

Investigate the size of each exterior angle of a convex polygon.

1. Draw a convex quadrilateral, pentagon, and hexagon.

2. Extend each side to form one exterior angle at the vertex.

3. Measure each exterior angle formed and find the sum of theexterior angles for each polygon.

What sum(s) did you get? (360°)

The sum of the measures of the exterior angles, one from eachvertex of a convex polygon, is 360°.

The measure of each exterior angle in a regular n-gon is

Example 2

For a regular polygon with 14 sides, find

a) the sum of the interior angles

b) the measure of each interior anglec) the measure of each exterior angle

Solution

a) The sum of the interior angles for a 14-gon = (n – 2)180°= (14 – 2)180°= 2160°

b) The measure of each interior angle of a 14-gon =−

=

n 2)(180)

15427

o

o

14

o360 .n


Module 3

(

c) The exterior angles of a 14-gon = 360°

∴ The measure of each exterior angle of a 14-gon

Self-Marking Activity

1. Which of the following figures are polygons? If it is not,explain why not. If it is a polygon, is it convex?

a)

b)

c)

=

=

=

360

36014

2557

o

o

o

n


Module 3

2. Draw the following figures:

a) a hexagon that is equilateral but not equiangular

b) an equiangular polygon that is not equilateralc) a regular hexagon

3. How many diagonals are in the following polygons?

a) quadrilateral

b) trianglec) hexagon

d) heptagon

4. Find the measure of y:

5. a) Find the sum of the interior angles of a 15-gon.

b) What is the sum of the exterior angles of a 15-gon?

6. What is the number of sides of a convex polygon if the sum ofthe measures of the interior angles is 2700?

7. The measure of each interior angle of a regular polygon isgiven. Find the number of sides of the regular polygon.

a) 120°

b) 160°

15 0°

85 °

12 0°

42°

60 °

y


Module 3

8. The measure of each exterior angle of a regular polygon isgiven. Find the number of sides.

a) 45°

b) 72°

9. The number of sides of a regular polygon is given. Find themeasure of each interior and exterior angle.

a) 9

b) 15

10. Find the measure of each interior angle in the figure.

11. a) How many diagonals does ABCDE have?

b) Find the length of each diagonal.


Review Section 1 before attempting the review questionsbeginning on the next page. These questions should help youconsolidate your knowledge as you prepare for the SectionAssignment 3.1.

l

l

l

l

l

x

y

B(2, 6)

C(4, 5 )

D(5, 2)

E(3, 0)

A(0, 2 )

x

x + 20°

x + 40° x + 60°

x + 80°

x + 100°


Module 3


Module 3

Notes

Review

1. Review the definitions in Lesson 1.

2. Circle O. OP ⊥ AB

AC = 16

AP = OPFind the length of

a) OP

b) OC

c) AB

d) AP

3. Circle O. AC = 5

OC = 6.5Find:

a) the length of AB

b) the length of BC

c) the area of ∆ ABC

d) the area of the circle

4. Circle O.

Find:

a) the measure of ∠ BOD in terms of x

b) the measure of ∠ CODin terms of y

c) the measure of ∠ BACin terms of x and y

d) the measure of ∠ BOCin terms of x and y

lO

B

A

C

D

x y

lO

B

A C

l

A

O

P

B

C

Principles of Mathematics 11 Section 1, Review 69

Module 3

•

•

•

5. On a circular ferris wheel, thereare 8 chairs, equally spaced.Braces between the chairs areplaced as shown in thediagram. Find the size ofangles 1, 2, and 3.

6. Given circle with centre O, CE tangent to D, AOB = 164°, BAD = 48°, find:

a) ODE

b) ADBc) BOD

d) OBDe) AODf) ADE

g) BDCh) ODA

i) BOD (major)

7. Given that AB is tangent at B, AD BD, and 1 = 70°, findthe measures of 2, 3, 4, and 5.

BA

D

C

1

3

2

54

ll ll

1 3

2

70 Section 1, Review Principles of Mathematics 11

Module 3

•

C

D

EO

A

B

∠∠

∠∠∠∠∠∠

∠∠∠

∠∠ ∠ ∠ ∠

≅

8. Circle O, ∠ COD = 100°. If ∠ B = 130° and ∠ A = 80°, find:

a) ∠ ODC

b) ∠ OCBc) ∠ ADC

d) ∠ OCD

9. BC is tangent

BE bisects ∠ ABD.

Verify that BC = CE

10. What is the number of sides of a convex polygon if the sum ofthe measures of the interior angles is 3960°?

11. The measures of each interior angle of a regular polygon is168°. Find the number of sides of the regular polygon.

12. Find the sum of the interior angles of a 19-sided polygon.

13. Find the number of sides of a regular polygon if the measureof each exterior angle is 30°.

B

A

D

C

E

32

14

5678

9

B

A

D

C

lO

Principles of Mathematics 11 Section 1, Review 71

Module 3

•

14. Find the values of:

a ) x

b ) COB

c ) AOC


Now do the Section Assignment which follows this section.

B

A

D

C

(2x + 25)°(4x - 13)°O

7 11 scitamehtaM fo selpicnirPweiveR ,1 noitceS2

Module 3

∠∠

PRINCIPLES OF MATHEMATICS 11

Section Assignment 3.1

Module 3

Principles of Mathematics 11 Section Assignment 3.1 73

General Instructions for Assignments

These instructions apply to all the Assignments, but will not be reprinted eachtime. Remember them for future sections.

(1) Treat this assignment as a test, so do not refer to your Module or notes orother materials. A scientific calculator is permitted.

(2) Where questions require computations or have several steps, show your work.

(3) Always read the question carefully to ensure you answer what is asked. Oftenunnecessary work is done because a question has not been read correctly.

74 Section Assignment 3.1 Principles of Mathematics 11

Module 3

Module 3

Section Assignment 3.1

Geometry

1. Use the diagram below to answer the following questions.

a) Name a diameter ________________

b) Name a central angle

_____________________________________________________

c) Name a semi-circle ________________

d) Name a secant line ________________

e) Name a tangent line ________________

f) Name a chord ________________

g) Name a minor arc ________________

h) Name an inscribed angle subtended by the minor arc named in (g)

_____________________________________________________

i) Name a point of tangency ________________

j) Name a right angle ________________

l

l

ll

l

l

D

X

Y

OAl C

B

E


Total Value: 50 marks(Mark values in brackets)

(10 x 1 = 10)

Module 3

2. List three properties of chords of a circle

3. Circle with centre at O. OD ⊥ XY

OD = 3 XY = 8

Find the following,explaining your reasoning.

a) perimeter of ∆ OXY

l

X

Y

O D


(3)

(4 x 2 = 8)

b) area of ∆ OXY

c) circumference of circle

d) area of circle O


Module 3

4. ∠ ACB = (x + 17)°

∠ AOB = (4x – 22)°

Find ∠ ADB indegrees. Explain yourreasoning.

l

C

B

O

A

D


Module 3

(3)

5. AB and CD are diameters.

CX and BP are tangents.

∠ BPX = 100°

Give reasons for:

a) ∠ ACB = 90°

b) OB ⊥ BP

c) ∆ CPB is isosceles

d) Quadrilateral OBPC is cyclic

e) ∠ COB = 100°

f) ∠ CAB = 50°

g) ∠ CBP = 50°

h) ∠ ARC = ∠ ABC

i) AC = DB

l

D

BO

C

R

P X

A


Module 3

(9 x 1 = 9)

6. PA and PB are tangents to circle with centre O. Verify thatOP bisects ∠ AOB.

7. a) Find the sum of the interior angles of a regular 20-gon.

b) Find the measure of each interior angle of a regular 20-gon.

c) Find the sum of the exterior angles of a regular 20-gon.

d) Find the size of each exterior angle of a regular 20-gon.

l

A

B

O P


Module 3

(4)

(4)

8. What is the number of sides of a convex polygon if the sum ofthe measures of the interior angles is 2880°?

9. The measure of each interior angle of a regular polygon is144°. Find the number of sides of the regular polygon.

10. The measure of each exterior angle of a regular polygon is18°. Find the number of sides of the regular polygon.


Module 3

(1)

(1)

(1)

11. Given that the circle is inscribed in the quadrilateral andthat R, P, S, and Q are points of tangency with AP = 5,PD = 9, CQ = 4, and QB = 3, what is the perimeter ofquadrilateral ABCD.

12. AB // CDProve OD = OC. BA

O

D C

D

S

P

C

Q

B

R

A


Module 3

(3)

(3)

(Total: 50)

1

23

module 3: geometry and trigonometry - open school...

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