SECONDARY SCHOOL IMPROVEMENT PROGRAMME (SSIP) 2019 · SECTION B: NOTES ON CONTENT REVISION OF THE...

© Gauteng Department of Education

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SECONDARY SCHOOL IMPROVEMENT

PROGRAMME (SSIP) 2019

GRADE 12

SUBJECT: MATHEMATICS

LEARNER NOTES (PAGE 1 OF 15)


2

D

E F

G H

NO: 5

TOPIC: EUCLIDEAN GEOMETRY (GRADE 12 TRIANGLE GEOMETRY)

SECTION A: TYPICAL EXAM QUESTIONS

QUESTION 1

1.1 In DEF , GH||EF. DG :GE 5:3 . DE 32 mm and DF 24 mm.

Determine the length of DG, GE, DH and HF. (6)

1.2 In ABC , CD 14 cm, DA 6 cm, CE 21 cm and EB 9 cm .

Prove that DE||BC. (5)

1.3 In ACE , BF||CE, 38

BC AC and AE : ED 4:3 .

Determine DG :GB . (6)

QUESTION 2

PB is a tangent to circle ABC. PA||BC. Prove that:

2.1 PAB||| ABC (5)

2.2 PA:AB AB:BC (2)

2.3 2PA . BC AB (2)

2.4 AP AB

BP AC (2)

A

B

C

D

E

F

G


3

A B

CD

E

12

1 2

1

212

O

C

A

B

1 2

1

212

1 2

D1

2 3

4

QUESTION 3

ABCD is a cyclic quadrilateral. AB and DC produced

meet at E.

Prove that: AE AD

CE BC (7)

QUESTION 4

ABCD is a cyclic quadrilateral, AC and BD intersect at P.

E is a point on BD such that AE||DC.

Prove that:

4.1 AP PE

PC PD (6)

4.2 2AP BP . PE (9)

QUESTION 5

In trapezium ABCD, DC 2BC ,

1ˆ ˆA E and BC EC .

Prove that:

5.1 AD BD

EC DC (6)

5.2 BD 2AD (3)

QUESTION 6

ABOC is a kite in which ˆB C 90 .

6.1 Why is OCD||| OAC ? (2)

6.2 Hence complete:

6.2.1 2OC ..... ..... (1)

6.2.2 2CA ..... ..... (1)

6.2.3 2CD ..... ..... (1)


4

4 parts 3 parts

21cm

A C B

6.3 Prove that:

6.3.1 2

2

BD AD

AOOB (6)

6.3.2 2 2OC OD OD. DA (2)

6.4 If 12

OD DA x , prove that CD 2 . OD (3)

SECTION B: NOTES ON CONTENT

REVISION OF THE CONCEPT OF RATIOS

Consider the line segment AB. If AB 21 cm and C divides AB in the ratio AC:CB 4:3 , it

is possible to find the actual lengths of AC and CB.

It is clear that AC doesn’t equal 4cm and CB doesn’t equal 3 cm because 4 3 21 cm .

However, if we let each part equal k, it will be possible to find the length of AC and CB in

centimetres.

The length of AC is (4 )cmk and the length of CB is (3 )cmk .

4 3 21 cm

7 21 cm

3 cm

k k

k

k

Each part represents 3 cm.

AC 4(3 cm) 12 cm

and CB 3(3 cm) 9 cm

4k 3k

A C B

k k k k k k k

A C B

3 cm 3 cm 3 cm 3 cm 3 cm 3 cm 3 cm

12 cm 9 cm


5

A

B C

D E

hk

Note:

AC 12cm 4AC:CB

CB 9cm 3

4:3 is the ratio of AC:CB.

THEOREM

A line drawn parallel to one side of a

triangle cuts the other two sides so as

to divide them in the same proportion.

If DE||BC then AD AE

DB EC

Proof

In ADE , draw height h relative to base AD and height k relative to base AE.

Join BE and DC to create BDE and CED . 12

12

. AD .Area ADE AD

Area BDE . BD . BD

h

h

12

12

. AE .Area ADE AE

Area CED . EC . EC

k

k

Now it is clear that

Area BDE Area CED

(same base, height and

lying between parallel lines)

Area ADE Area ADE

Area BDE Area CED

AD AE

BD EC

Corollaries

(1) AB AC

AD AE (2)

AB AC

DB EC (3)

BD CE

DA EA (4)

BD CE

BA CA

Whenever you use this theorem the reason you must give is: Line || one side of


6

A

B C

D E

THEOREM CONVERSE

If a line cuts two sides of a triangle proportionally,

then that line is parallel to the third side.

If AD AE

DB EC then DE||BC

Whenever you use this theorem the

reason you must give is:

Line divides sides of proportionally

THEOREM (MIDPOINT THEOREM)

The line passing through the midpoint of

one side of a triangle, parallel to another

side, bisects the third side and is equal to

half the length of the side it is parallel to.

If AD DB and DE||BC , then AE EC

and 12

BC 2DE or DE BC .

Also, if AD DB and AE EC , then

DE||BC and 12

BC 2DE or DE BC .

SIMILARITY OF TRIANGLES

If two triangles are similar, we use the symbol ||| to indicate this.

If ABC is similar to DEF then we write this as follows: ABC||| DEF

If ABC||| DEF then the following conclusions can be made:

(a) The triangles are equiangular which means that:

ˆ ˆA D ˆ ˆB E ˆ ˆC F

(b) The corresponding sides are in the same proportion which means that:

AB BC AC

DE EF DF


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Whenever two triangles are similar we can use the following diagram to match the

corresponding angles and sides: ˆ ˆA D ˆ ˆB E ˆ ˆC F

AB BC AC

DE EF DF

THEOREM

If two triangles are equiangular then the corresponding sides of the two triangles are in

the same proportion and therefore the triangles are similar.

Proof

On AB mark off AG DE .

On AC mark off AH DF .

Join GH.

In AGH and DEF:

(1) AG DE construction

(2) ˆ ˆA D given

(3) AH DF construction

AGH DEF SAS

1ˆ ˆG E

But ˆ ˆB E given

1ˆ ˆG B

GH||BC corr 's equal

AB AC

AG AH

AB AC

DE DF ( AG DE , AH DF )

Similarly, by constructing BG and BH

on AB and BC respectively, it can be

proved that

AB BC

DE EF

AB AC BC

DE DF EF

Therefore the triangles are similar.

A

B C

D

E F

G

H

1

A

B C

D

E F

G H1

A B C D E F


8

THEOREM

If the corresponding sides of two triangles

are in the same proportion, the two triangles

are similar.

Proof

Construct ABC||| GEF (as in diagram)

AB BC AC

GE EF GF

But AB BC AC

DE EF DF

AB AB

GE DE (both equal

BC

EF)

GE DE

Similarly, it can be proved that GF DF

Therefore it can be concluded that DEF GEF (SSS).

DEF||| GEF

But ABC||| GEF

ABC||| DEF

THEOREM

The perpendicular drawn from the vertex of the right angle of a right-angled triangle to

the hypotenuse, divides the triangle into two triangles that are similar to each other and

similar to the original triangle.

Proof

In ABC and DBA:

(a) 1ˆ ˆA D 90 given

(b) ˆ ˆB B common

(c) 1ˆ ˆC A sum of the 's of

ABC||| DBA

In ABC and DAC:

(a) 2ˆ ˆA D 90 given

(b) ˆ ˆC C common

(c) 2ˆB A sum of the 's of

ABC||| DAC

ABC||| DBA||| DAC

A

B C

D

E F

G

1 1


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Corollaries

2

ABC||| DBA

AB BC AC

DB BA DA

AB BD . BC

2

ABC||| DAC

AB BC AC

DA AC DC

AC CD . CB

2

DBA||| DAC

DB BA DA

DA AC DC

AD BD . DC

THEOREM (THE THEOREM OF PYTHAGORAS)

2 2 2BC AB AC

Proof

From the corollaries: 2AB BD. BC and 2AC CD. CB

2 2

2 2

2 2

2 2 2

2 2 2

AB AC BD . BC CD . CB

AB AC BC (BD CD)

AB AC BC (BC)

AB AC BC

BC AB AC

SECTION C: HOMEWORK QUESTIONS

QUESTION 1

1.1 In ABC , DE||AB, CE: EB 4:3 and

AC 28cm . Determine the length of AD. (4)


10

P

Q S V R

T

U

1.2 In ACE , BG||CF and AF FE . AB 2cm, BC 3cm, CD 1cm

and DE 4cm.

1.2.1 Determine AG:GF (1)

1.2.2 Determine EG

GA (1)

1.2.3 Prove that DG||CA (4)

QUESTION 2 In PQR , PS||VT and QS:SR 2:3 .

T is a point on PR such that PT:TR 2:7 .

Prove that: QU : UT 3:1 (5)

QUESTION 3

AB is a diameter of circle ABC.

DC is a tangent at C and BD CD .

Prove that:

3.1 BDC||| BCA (5)

3.2 BD BC

CD AC (1)

QUESTION 4


11

A

B CD

2

3 x

PT is a tangent to circle BAT. BA is

produced to P. TB is joined. AT is a chord.

Prove that:

BT AT

BP PT (7)

QUESTION 5

AOB is a diameter of circle centre O.

Chord AC produced meets tangent

BD at D. BC is joined.

Prove that: 2AB AC . AD (8)

QUESTION 6

TSQR is a cyclic quadrilateral.

SR||PQ and TQ bisects ˆPTR.

Prove that:

6.1 PQ is a tangent to the circle. (6)

6.2 SQ QR (5)

6.3 2QS TR .SP (10)

QUESTION 7

In ABC , A 90 , AD BC ,

BD 3, DC and AC 2x .

Calculate the length of:

7.1 DC (7)

7.2 AB (simplest surd form) (4)


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A

B

C

D

E

F

G

3BC AC

8

BC 3

AC 8

3k

8k 5k

AE :ED 4:3

4p

3p

SECTION D: SOLUTIONS FOR SECTION A

QUESTION 1

1.1 5DG 32 20

8mm mm

3GE 32 12

8mm mm

DH 5

HF 3 Line || one side of triangle

5DH 24 15

8mm mm

3HF 24 9

8mm mm

DG 20mm

GE 12mm

DH 5

HF 3

DH 15mm

HF 9mm

reason (6)

1.2 CD 14 7

DA 6 3

cm

cm

CE 21 7

EB 9 3

cm

cm

CD CE

DA EB

DE||AB Line divides sides of prop

CD 14 7

DA 6 3

cm

cm

CE 21 7

EB 9 3

cm

cm

CD CE

DA EB

DE||AB

reason (5)

1.3 DG 3

GB EF

p Line || one side of triangle; BF||GE

Now EF 3

4 8

k

p k

3

2

3EF 4

8

3EF

2

DG 3

GB

DG2

GB

DG:GB 2 :1

p

kp

k

p

p

DG 3

GB EF

p

EF 3

4 8

k

p k

3

EF2

p

3

2

DG 3

GB p

p

DG:GB 2:1 reason (6)


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QUESTION 2

2.1 In PAB and ABC:

(1) 1 2ˆ Â B Alt angles equal

(2) 1ˆB C Tan-chord

(3) 2ˆP A Sum of the angles of a triangle

PAB||| ABC

1 2ˆ Â B

1ˆB C

2ˆP A

PAB||| ABC

reason (5)

2.2 PA AB PB

AB BC AC

PA AB

AB BC

PA:AB AB:BC

PA AB PB

AB BC AC

PA:AB AB:BC (2)

2.3 PA AB

AB BC

2PA . BC AB

PA AB

AB BC

2PA . BC AB (2)

2.4 PA PB

AB AC

AP AB

BP AC

PA PB

AB AC

AP AB

BP AC (2)

QUESTION 3

In ADE and CBE:

(a) ˆ Ê E Common

(b) 2ˆ Â C Ext angle of a cyclic quad

(c) 2ˆ ˆD B Sum of the angles of a triangle

ADE||| CBE

AD DE AE

CB BE CE

AD AE

CB CE

AE AD

CE BC

ˆ Ê E

2ˆ Â C

2ˆ ˆD B

ADE||| CBE

AD DE AE

CB BE CE

AE AD

CE BC

reasons (7)


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QUESTION 4

4.1 In APE and CPD:

(1) 1 3ˆ ˆP P Vertically opp angles

(2) 2 2ˆ Â C Alt angles equal ; AE||DC

(3) 1 2ˆ Ê D Sum of angles of

APE||| CPD

AP PE AE

CP PD CD

AP PE

PC PD

1 3ˆ ˆP P

2 2ˆ Â C

1 2ˆ Ê D

APE||| CPD

AP PE

PC PD

reasons (6)

4.2 In ABP and EAP:

(1) 1 1ˆ ˆP P Common

(2) 1 2ˆB C Arc AD subtends equal angles

2 2ˆ ˆC A Alt angles equal ; AE||DC

1 2ˆB A

(3) 1 2 1ˆ ˆ Â A E Sum of angles of

2

2

ABP||| EAP

AB BP AP

EA AP EP

BP AP

AP EP

BP . EP AP

AP BP . PE

1 1ˆ ˆP P

1 2ˆB C

2 2ˆ ˆC A

1 2ˆB A

1 2 1ˆ ˆ Â A E

ABP||| EAP

AB BP AP

EA AP EP

2AP BP . PE

reasons (9)

QUESTION 5

5.1 In ABD and EDC:

(1) 1ˆ Â E Given

(2) 1 2ˆ ˆB D Corr angles equal

(3) 1 1ˆD C Sum of angles of

ABD||| EDC

AB BD AD

ED DC EC

AD BD

EC DC

1ˆ Â E

1 2ˆ ˆB D

1 1ˆD C

ABD||| EDC

AD BD

EC DC

reasons (6)

5.2 AD.DC BD.EC AD.DC BD.EC


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AD.2BC BD.BC [ DC 2BC ; BC EC ]

2AD BD

2AD BD reasons (3)

QUESTION 6

6.1 OCD||| OAC In OAC :

ÔCA 90 , 1D 90 (diags of kite)

Perpendicular from right-angled

vertex to hypotenuse

ÔCA 90 , 1D 90

reason (2)

6.2.1 2OC OD OA 2OC OD OA (1)

6.2.2 2CA AD AO 2CA AD AO (1)

6.2.3 2CD OD DA 2CD OD DA (1)

6.3.1 2BD OD.DA In OAB :

2OB OD.OA ÔBA 90 , 2D 90 (diags of kite)

Perpendicular from right-angled

vertex to hypotenuse 2

2

2

2

BD OD.DA

OD.OAOB

BD AD

AOOB

2BD OD.DA

2OB OD.OA

ÔBA 90 , 2D 90

reason

2

2

BD OD.DA

OD.OAOB

2

2

BD AD

AOOB (6)

6.3.2 2 2 2OC OD CD Pythagoras

But 2CD OD DA

2 2OC OD OD.DA

2 2 2OC OD CD

2 2OC OD OD.DA (2)

6.4 2CD OD DA 2

2 2

CD ( ).(2 )

CD 2

CD 2

CD 2.OD

x x

x

x

2CD OD DA

2CD ( ).(2 )x x

CD 2x (3)


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SECONDARY SCHOOL IMPROVEMENT PROGRAMME (SSIP) 2019 · SECTION B: NOTES ON CONTENT REVISION OF THE...

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