Geometry

Exam Paper 3

Page

Circle Geometry (Part 1)

Assessment Standard: 12.3.2 Circle Geometry (Part 1)

Accept as axioms all results established in earlier grades and the fact that a tangent is perpendicular to the radius drawn at the point of contact with the circle and then investigate and prove the theorems of the geometry of circles.

Over the next few weeks we will lead you through the Grade 2 theory.To be successful in the Geometry section of Paper 3 (which counts 40%) you must be absolutely familiar with the basics from way back in Grade 8.

Do you know that: angles on a straight line add up to 80

the angles of a triangle add up to 80

the exterior angle of a triangle is equal to the sum of the interior opposite angles

Collect your Paper 3 Lessons every week!!Guys, both NSC and IEB examinations

candidates have the option of writing Paper 3 at the end of the year! Paper 3 covers additional

mathematics material and is out of 00 marks. Maths Paper 3 will really set you apart in the job market, and make studying technical subjects at tertiary level easier.

We have hooked you up with these lessons - written by IEB Maths Paper 3 examiner, Heather Frankiskos.

The Examination Guidelines for the DOE stipulate that the proofs of all Grade and the Grade 2 theorems of the

geometry of circles are required for examination purposes. This bookwork has a suggested maximum mark allocation of 5 out of 00. The IEB Subject Assessment Guidelines

suggest that no bookwork i.e. NO PROOFS be examined. We can expect Complete the statement type questions

but no complete proofs. These Geometry modules will not present proofs. You must get these from your teacher.

a b

a + b = 80

c

a

b a + b = c

a

b

c a + b + c = 80

Exam Paper 3

Page 2


When we have parallel lines

a) Alternate Angles are equal

^ d = ^ f ( )

^ c = ^ e ( )

b) Corresponding Angles are equal

^ c = ^ g ( ); ^ b = ^ f ( )

^ d = ^ h ( ); ^ a = ^ e ( )

c) Cointerior Angles are Supplementary

^ d + ^ e = 80

^ c + ^ f = 80

The Geometry Section of the Paper contains some questions which cover only Grade or only Grade 2 Theory. However, some questions may blend the two. We will revise the Grade theory for you when we are ready to present examination type questions.

Grade 12 Theory

Before we launch into the theory let us define some terms:

An Axiom: is an accepted body of knowledge or principle for example: The Theorem of Pythagoras

A Theorem: is a statement that can be proved by logical reasoning

An Arc: is part of the circumference of a circle

A Chord: is a straight line joining the ends of an arc

A Diameter: is a special chord passing through a circle centre. A diameter cuts the circle into two semi-circles

A Radius: is the line from the circle centre to any point on the circumference

A Segment: is part of a circle cut off by a chord. Every chord cuts a circle into two segments

A Tangent: is a line that makes contact with the circle at only one point on the circumference

FFFF

FF

FF

A B

C D

a b

cd

e fgh

minor segment

major segment

chord AC

minor arc AC

major arc AC

A

CDiameter

Radius

Tangent APT

A

P

T

Exam Paper 3

Page 3


Theorem 1

a) If a line is drawn from the centre of a circle so that it is perpendicular to a chord, it will bisect the chord

If OM ^ AB then AM = MB (perp. line from centre to chord)

b) If a line is drawn from the centre of a circle to the midpoint of a chord, it will be perpendicular to the chord

If AM = MB then OM ^ AB (midpoint - chord)Application. Given BD = 8 and OD = 5 and OC ^ BD Find OC

OC ^ BD given\ BC = CD (perp. line from centre to chord)\ CD = 4In D OCD OD2 = OC2 +CD2 (Pythagoras) 25 = OC2 + 6 9 = OC2

3 = OC

A M B A M B

B C D

A

M

B B

M

A

Exam Paper 3

Page 4


R

A B

C

2.

Given : O is the centre PT = TV = 6 OT = x TR = 2

a) Write down the length of PO in terms of x

b) Determine the value of x

a) PO =x + 2 (equal radii)

b) OT ^ PV (midpoint - chord) \ PO2 = PT2 + OT2 (Pythagorus) (x + 2)2 = 62 + x2 x2 + 4x + 4 = 36 +x2 4x = 32 x = 8

For the next theorem we must clarify some vocabulary:

a) In the picture we say:

arc AB subtends A ^ O B to the centre

arc AB subtends A ^ C B to the circumference

b) In the picture we say:

chord AB subtends A ^ O B to the centre

chord AB subtends A ^ C B to the circumference So an arc or a chord can subtend (hang) an angle off of it to the centre or the circumference

Theorem 2

2. The angle subtended by an arc/ chord at the centre of a circle is twice the size of the angle subtended by the same arc/ chord at the circumference ( at the centre = 2 at the circumference)

This theorem can present itself in a number of ways.

Look at each of these.

Can you see that each time the same theory is involved?

AB

C

P VT 2

x

Exam Paper 3

Page 5


Case 1 (Normal Case)Angle is subtended above the one at the centre

Case 2 (Skew Case)Angle is subtended off to the right/left of the one at the centre

Case 3 (Diameter Case)The Angle at the centre is a straight angle (80) so the angle subtended into the semi-circle segment will always be 90. We use this case so often that we refer to it as in semi-circle.

Case 4 (Reflex Case)

The major arc subtends the reflex angle at the centre.

2x

2x

x

Right

x

Left

80

80

x2x

2x

Turn me upside down!

x

2x

x

2yy

Exam Paper 3

Page 6


Application Find x and or y:

Findx:

C ^ B O = 40 (equal radii, Isos D) B ^ O C = 00 (s of D) B ^ D C = 50 ( at centre = 2 at circumference)

x = 50

x = 80 ( at centre = 2 at circumference)

(The angle is subtended skewly to the circumference)

2x - 30 = 30 ( at centre = 2 at circumference)

(The reflex angle is subtended by major arc AC)

2x = 60

x = 80

2x = 90 ( at centre = 2 at circumference)

OR ( in a semi-circle)

x = 45

^ O = 40 ( of Isos D) x = 70 ( at centre = 2 at circumference)

y = 90 ( in a semi-circle)

c)

20

I

y

e)

d)

b)

D

B C40

a)

x

x

A

B

C

40 x

260

A

D

C2x - 30

80

2x

Geometry

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