MATHEMATICS - Western Cape Curriculum Planner for Mathematics E-mail: [email protected]...

Western Cape Education Department

Examination Preparation Learning Resource 2016

GEOMETRY MEMORANDUM

MATHEMATICS

Razzia Ebrahim

Senior Curriculum Planner for Mathematics

E-mail: [email protected]

Website: http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835-

grade-12?Itemid=-1

Website: http://wcedeportal.co.za

Tel: 021 467 2617

Cell: 083 708 0448

Grade Theorems 11 Grade

on Questions Geometry 12

mailto:[email protected]

http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835-grade-12?Itemid=-1

http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835-grade-12?Itemid=-1

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2

Index Page

1. 2016 Feb-March Paper 2 3 – 4

2. 2015 November Paper 2 5 – 8

3. 2015 June Paper 2 9 – 12

4. 2015 Feb-March Paper 2 13 – 16

5. 2014 November Paper 2 21

6. 2014 Exemplar Paper 2 22 – 23




10. 2010 November Paper 3 35 – 38

11. 2009 November Paper 3 39 – 41

12. 2008 November Paper 3 43 – 45

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

QUESTION/VRAAG 8

8.1

8.1.1 == 22 MK 40° [tan chord theorem/raakl-kdst] S R (2)

8.1.2 11 KN = [∠s in the same seg/∠e in dies segm]

1K = 84° – 40° = 44° ∴ °= 44N1

S R S

(3) 8.1.3

1NT = = 44° [alt/verw ∠s/e; KT | | NM] S R (2)

8.1.4 TKL 22 += [ext ∠ of ∆/buite ∠ v ∆] = 40° + 44° = 84°

R

S (2)

8.1.5 In ∆KLM: 44° + 84° + 40° + 1L = 180° [∠s sum in ∆/∠e som in ∆] ∴ 1L = 12°

S

(1)

M

K 1 2

84°

T L 2

1

1 2 40°

N 1 3

3

Mathematics/P2/Wiskunde V2 DBE/ Feb.–Mar./Feb.–Mrt. 2016 NSC/NSC – Memorandum


8.2

8.2 =C 108° [opp∠s of ||m/tos ∠e v ||m] °=°+°+ 180108402x [opp∠s of cyc quad/tos∠e v kdvh]

°= 322x °= 16x

OR/OF

=C 180° – (2x + 40°) [opp∠s of cyc quad/tos∠e v kdvh]

180° – (2x + 40°) = 108° [opp∠s of ||m/tos ∠e v ||m] °= 322x

°= 16x

S R S R answ/antw

(5)

S R S R answ/antw

(5) [15]

B

C A

D

E 108°

2x + 40°

4

Mathematics/P2/Wiskunde/V2 DBE/November 2015 NSC/NSS – Memorandum


QUESTION/VRAAG 8

8.1.1 twice/twee keer of dubbel R

(1) 8.1.2 A2O1 = [∠ at centre = 2×∠ at circ/midpts∠ =2×omtreks∠]

C2O2 = [∠ at centre = 2×∠ at circ/midpts∠ =2×omtreks∠] °=+ 360OO 21 [∠s in a rev/∠e in omw of om 'n pt] °=+ 360C2A2

∴ °=+ 081CA

S S S

(3)

O

A

B C

D 1

2

5



8.2

2C = A [ext ∠ of cyclic quad/buite∠ v kdvh]

2C180E −°= [opp ∠s of cyclic quad/tos∠e v kdvh] ∴ A180E −°= ∴ EF | | AB [co-interior ∠s 180°/ko-binne∠e 180°] OR/OF

1D = B [ext ∠ of cyclic quad/buite∠ v kdvh]

1D180F −°= [opp ∠s of cyclic quad/tos∠e v kdvh] ∴ B180F −°= ∴ EF | | AB [co-interior ∠s 180°/ko-binne∠e 180°]

S R S R R

(5) S R S R R

(5) [9]

A

D

E

B

C

F

1 2

1 2

6



QUESTION/VRAAG 9

9.1 CK3 = [corresp ∠s/ooreenk ∠e ; CA| |KT]

x=C [tan-chord th/raakl-koordst] x=3K

S R S R

(4)

9.2 33 AK == x [proved/bewys in 9.1]

∴ AKBT is cyc quad [line (BT) subtends equal ∠s/ lyn (BT) onderspan gelyke ∠e]

S R

(2) 9.3 x== CB2 [tan-chord th/raakl-koordst]

x== KB 22 [∠s in the same segm/∠e in dies segm] ∴ x== KK 23 ∴ TK bisects/halveer BKA OR/OF

x== AB 32 [tans for same pt; ∠s opp equal sides/ rkle v dies pt; ∠e to gelyke sye]

x== KB 22 [∠s in the same seg/∠e in dies segm] ∴ x== KK 23 ∴ TK bisects/halveer BKA

S R

S R

(4)

S R

S R

(4)

1 2

x

1

2 3

2

3

1

2 1

A

C

B

•S

T

K H

7



OR/OF TA = TB [tans for same pt/rkle v dies pt] KATB is cyc quad ∴ x== KK 23 [equal chords subtend equal angles/ gelyke koorde onderspan gelyke hoeke] ∴ TK bisects/halveer BKA

S R

S R

(4) 9.4 x== KA 23 [proven/bewys]

∴ TA tangent [converse tan chord theorem/ omgekeerde raakl-kdst]

S R

(2)

9.5 The circle passing through points A, K and B contains the point S on the circumference (A, S, K and B concyclic)./Die sirkel deur punt A, K en B bevat die punt S op die omtrek (A, S, K en B konsiklies). The circle passing through A, K and B contains the point T on the circumference (proven in 9.2)./Die sirkel deur punt A, K en B bevat die punt T op die omtrek (bewys in 9.2). ∴ points A, S, B and T are also concyclic/punte A, S, B en T is konsiklies OR/OF

AKBASB = = 2x [A,S,K & B concyclic/konsiklies] x2180BTA −°= [A,T,B & K concyclic/konsiklies]

∴ points A, S, B and T are also concyclic/punte A, S, B en T is ook konsiklies [opp ∠s of quad = 180°/tos ∠e van vierhoek=180°]

S S

(2)

S (both/beide statements/ bewerings) R

(2) [14]

8

Mathematics/P2/Wiskunde V2 DBE/2015 NSCNSS – Memorandum


QUESTION/VRAAG 8

8.1.1 °= 65P2 (∠s opp equal sides/∠e teenoor gelyke sye) S R

(2) 8.1.2 °= 40D (ext∠ of ∆CDP/buite∠ v ∆CDP )

OR/OF (∠s on a str line; sum of ∠s in ∆/ ∠e op regt lyn; som v ∠e in ∆)

S R

(2) 8.1.3 °= 40 A1 (ext∠ of ∆CAT/buite∠ v ∆ CAT )

OR/OF (∠s on a str line; sum of ∠s in ∆/ ∠e op regt lyn; som v ∠e in ∆)

S R

(2)

8.2 °== 40D A1 ∴CA is a tangent to the circle (∠ between line and chord)/ CA is 'n raaklyn aan die sirkel (∠ tussen lyn en koord)

S R

(2) [8]

P

1 2

2 2

A

B

C

D

T 25°

65° 1 1

2 1

3

9



QUESTION/VRAAG 9

9.1.1 ext∠ of cyclic quad/buite ∠ van koordevh R

(1) 9.1.2 ∠ at centre = 2 ×∠ at circumference / midpts∠ = 2 × omtreks∠ R

(1) 9.2.1 x== 1EADC

(corresp∠s/ooreenk ∠e; EB || DC)

∴ x== CADC ∴ AC = AD (sides opp equal ∠s/sye teenoor gelyke ∠e)

S R S (justification)

(4) 9.2.2 x2180A −°= (sum of ∠s in ∆/som van ∠e in ∆)

x21O = OR °=+−°=+ 18022180OA 1 xx ∴ ABOD is a cyclic quad/koordevh (opp∠s quad supp/ teenoorst ∠e van vh suppl)

S linking the 2 ∠s R

(3) [9]

O

x

A

C

D E

B

1

1

1

1 2

2

2

3

10


Copyright reserved Please turn over/Blaai om asseblief

QUESTION/VRAAG 11 11.1 Draw diameter AD and join DC.

Trek middellyn AD en verbind DC.

Proof/Bewys:

°=+ 90DABPAB (tangent/raaklyn ⊥ radius) °=+ 90BCABCD (∠ in semi circle/halfsirkel)

but BCDDAB = (∠s in same segment/∠e in dies segm)

∴ BCAPAB = OR/OF Draw diameter AD and join DB. Trek middellyn AD en verbind DB.

Proof/Bewys: °=+ 90DABBAP (tangent/raaklyn ⊥ radius)

°= 90ABD (∠ in semi circle/halfsirkel)°=+ 90BDADAB (sum of ∠s in ∆/som van ∠e in ∆)

BCABDA = (∠s in same segment/∠e in dies segm) ∴ BCAPAB =

construction/ konstruksie S R S R

S/R (6)

construction/ konstruksie

S R S R S/R

(6)

D

P

A

B

C

O

D

A

B

C

O

P

11



OR/OF

Draw radii OA and OB. Trek radii OA en OB.

Proof/Bewys:

°=+ 90PABBAO (tangent/raaklyn ⊥ radius) ∴ =PAB BAO90 −°

ABOBAO = (∠s opp equal sides/∠e to gelyke sye) BAO2180BOA −°= (sum of ∠s in ∆/som van ∠e in ∆)

∴ BAO90BCA −°= (∠ at centre = 2 ×∠ at circumference/ midpts∠ = 2 × omtreks∠) ∴ BCAPAB =

construction/ konstruksie

S R S S/R S

(6)

P

A

B

C

O

12

Mathematics P2/Wiskunde V2 DBE/Feb.–Mrt. 2015 NSC/NSS – Memorandum


QUESTION/VRAAG 7

7.1 MB = 10 cm answer/antw

(1) 7.2 line from centre to midpoint of chord is perpendicular to chord/lyn

vanaf midpt na midpt van koord is loodreg op koord

OR/OF line from centre bisects chord/lyn vanaf midpt halveer koord

answer/antw (1)

answer/antw (1)

7.3

xxx

x

5OP2225OP25

OMMP

=+

=+

=

23OP x

=

OR/OF

23OP

23

OMOP

x=

=

25OP

=+x

x

2

3OP x=

(2)

23

OMOP

=

2

3OP x=

(2)

x

A

B

M

O•

P

13



7.4 OM² + MB² = OB²

80 or54or 94,8

80400594004

2310

2

2

22

222

=

=

=

=+

=+

x

xx

xx

xx

subst into/subst Pythagoras 22 94004 xx =+ answer/antw

(3) [7]

14



QUESTION/VRAAG 8

8.1.1

°== 55O 21 D 1 (∠ at centre=2×∠at circ/∠ by midpt=2×∠by omt)

S R (2)

8.1.2 °== 55O

21A 1 (∠ at centre=2×∠at circ/∠ by midpt=2×∠by omt)

OR/OF

°== 55DA (∠s in same segment/∠e in dieselfde segment)

S R

(2) S R

(2) 8.1.3 DB1 = = 55° (alternate ∠s/verwiss∠e; AB | | DC)

ABE 12 += (ext∠ of∆ = sum of opp∠s/buite∠ v∆=som v tos∠e) = 55° + 55°

°= 110E 2

S R

R answer/antw

(4) 8.2 °== 110OE 12 (proven in/bewys in 8.1.3)

BEOC is a cyclic quadrilateral (equal ∠s subtended by line/ gelyke ∠e onderspan deur lyn)

S

R (2)

[10]

A

B

D

E

O

C

1 2 3

1 2 3

1 110°

1

2

3

15



QUESTION/VRAAG 9 9.1 the interior opposite angle/die teenoorstaande binnehoek. answer/antw

(1)

9.2

Construction: Draw diameter CT and join CV. Konstruksie: Trek middellyn CT en verbind CV.

=+ 21 VV 90° ∠ in semi-circle/∠ in halfsirkel S R

x−°= 90T2 Tangent ⊥ diameter/radius/raaklyn ⊥ middellyn/radius R

∴ =C x Sum of the angles of triangle/Som van die hoeke van 'n driehoek S

∴ x=S ∠'s same segment/∠e in dieselfde segment R

∴ SRTV = (5)

C

V

S

O

T P R 1

2

1 2

x

16

Mathematics P2/Wiskunde V2 DBE/November 2014 NSC/NSS – Memorandum


QUESTION/VRAAG 8 8.1

8.1.1 °96=x (∠ at centre = 2∠ at circumference/

∠ by midpt = 2∠ by omtrek) S R

(2) 8.1.2 °=°−°=+ 8496801BC 22 (sum of ∠s in ∆/ som v∠e in ∆ )

42°B2 ==y (∠s opp = sides/∠e teenoor = sye)

S

S (2)

8.2

8.2.1 °= 90F1 (line from centre to midpt chord/

lyn vanaf midpt na midpt kd) S R

(2)

8.2.2 150° CBA = (opposite ∠s of cyclic quad/ tos ∠e v koordevh )

S R (2)

48°

x

2

1

1

y

A

O

B

C 2

30°

A

B

C D

O

F 1

17



8.3

8.3.1 (a) tangent ⊥ radius/diameter / raaklyn ⊥ radius/middellyn

R

(1) 8.3.1 (b) tangents from common pt OR tangents from same pt /

raaklyne v gemeensk pt OF raaklyne vanaf dies pt R (1)

8.3.2 AB² + BC² = AC²

222 13)7( =++ xx (Theorem of/Stelling vanPythagoras)

169491422 =+++ xxx 0120142 2 =−+ xx 06072 =−+ xx 0)12)(5( =+− xx 5=x ( 12−≠x )

AB² + BC² = AC²

222 13)7( =++ xx standard form answer

(4) [14]

A E

B

C

x

13

x + 7

18



QUESTION/VRAAG 10

10.1.1 Tangent chord theorem/Raaklyn-koordstelling

R (1)

10.1.2 Tangent chord theorem/Raaklyn-koordstelling

R (1)

10.1.3 Corresponding angles equal/Ooreenkomstige ∠e gelyk R (1)

10.1.4 ∠s subtended by chord PQ OR ∠s in same segment ∠e onderspan deur dieselfde koord OF ∠e in dieselfde segment

R (1)

10.1.5 alternate ∠s/verwisselende ∠e ; WT | | SP R (1)

Y

X

R

W

S

P

Q

T

1 2 3

4

1 2

1 2

3

1 2

1 2

1 2 3 4

X

R

W

S

P

T

y

1 2 3

4

1 2

1 2

3

1 2

1 2

1 2 3 4

x

19

Wiskunde/V2 DBE/2014 NSS – Graad 12 Model – Memorandum

Kopiereg voorbehou Blaai om asseblief

VRAAG 8 8.1 die hoek onderspan in die teenoorstaande sirkelsegment korrekte stelling

(1) 8.2.1 == 11 EB 68° (rkl-koordst) =1E 68°

rede (2)

8.2.2 == 31 BE 68° (verwiss∠e; AE | | BC) =3B 68° (S/R) (1)

8.2.3 31 BD = = 68° (buite∠ v koordevh) 1D = 68°

rede (2)

8.2.4 2E = 20° + 68° (buite∠ v ∆)

= 88°

2E = 88° (S/R)

(1) 8.2.5 =C 180° – 88° (tos ∠e v koordevh)

= 92°

=C 92° rede

(2) [9]

21



2

1 3

B

• O C

F

E

D

G

H

1 2

32°

56° 4 5

3

QUESTION/VRAAG 10 10.1

2E = 90° – 56° (∠s in a semi circle)

2E = 34° (∠e in ‘n halwe sirkel)

2E = 34° ∠s in a semi circle

(2) 10.2 HEC 21 EE += = 66°

∴ 32 BBCBE += = 66° (tan ch th) (raaklyn koord) OR

2B = 32° (tan ch th) (raaklyn koord)

3B = 34° (∠s in the same segment) (∠e in selfde seg)

CBE = 66°

HEC = 66° CBE = 66° tan ch th

(3) tan ch th 3B = 34° answer

(3)

10.3 °=++ 122EEE 321

F = 122° (tan ch th) / (raaklyn koord) OR

°= 58C (sum of int ∠s of ∆) / (som van binne∠e ∆) F = 122° (opp ∠s cyclic quad) / (oorst ∠ koordevierhoek) OR D = 58° (sum of int ∠s of ∆) / (som van binne∠e ∆) F = 122° (opp ∠s cyclic quad) / (oorst ∠ koordevierhoek)

F = 122° reason

C = 58° ∠ sum ∆ F = 122° opp ∠s cyclic quad D = 58° ∠ sum ∆ F = 122° opp ∠s cyclic quad

(4) [9]

22



QUESTION/VRAAG 11

Construct radii OA and OC. In ∆OAB and ∆OCB

i. OB is common ii. OA = OC (radii) iii. °== 90CBOABO (given)

∆OAB ≡ ∆OCB (90°HS) AB = BC (≡ ∆s) OR Construct radii OA and OC. In ∆OAB and ∆OCB

i. OB is common OR OA = OC (radii) ii. CA = (∠s opp = radii) iii. °== 90CBOABO (given)

∆OAB ≡ ∆OCB (SAA) AB = BC (≡ ∆s) OR Construct radii OA and OC.

°== 90CBOABO (given) OA2 = OB2 + AB2 (Pythagoras) OC2 = OB2 + BC2 (Pythagoras) OC = OA (radii) OB is common AB2 = BC2 AB = BC

construction OB common radii °== 90CBOABO 90°HS

construction OB common / OA = OC ∠s opp = radii °== 90CBOABO SAA

construction OA2 = OB2 + AB2 OC2 = OB2 + BC2 OA = OC / radii AB2 = BC2

[5]

O

A B C

23

Mathematics/P3 DBE/November 2012 NSC – Memorandum

Copyright reserved Please turn over

76° 29°

63°

29°

34°

117°

76°

117°

41°

M

P

W

T U

Q

R

c b

a

1

29°

75° 34°

1 1

2 3

d

41°

105°

QUESTION 8 8.1 ... equal to the angle subtended by the chord in the alternate

segment. answer

(1) 8.2

°= 29a (tan ch.thm)

°= 34RPQ (∠s in same seg)

°= 41c °= 76b (adj∠s on str. line)

°= 76Q1 (∠s in same seg) °=105d (ext∠ cyclic quad)

OR

°= 29a (tan ch. thm)

c=1T (tan ch. thm) °=°+ 7534c (tan ch. thm)

°= 41c

°= 76b (adj∠s on str. line)

°=105d (adj∠s on str. line) OR An alternative solution for calculating d:

°== 76TPRQ1 (∠s in same seg) TPRRPQTQPQTR ++=+d (ext∠∆)

°+°+°=°+ 76342934d °= 105d

°= 29a tan ch. thm ∠s in same seg °= 41c °= 76b ext∠ cyclic quad

(9)

°= 29a tan ch. thm c=1T tan ch. thm °=°+ 7534c tan ch. thm °= 41c °= 76b

(9)

[10]

°= 34RPQ

°= 76Q1

°=105d

°=105d

24


Copyright reserved

QUESTION 10

10.1 DC = 13x

CD = 13 x (1)

10.2 OD = x

213

OM = x25

OD = x2

13

answer

(2) 10.3 BO = OD (radii)

AM = MB = 12 units (line from circ cent ⊥ch) 22

2

213

2512

=

+ xx (Pythagoras)

)0(22

44

144144

4169

425144

2

2

22

>=±=

=

=

=+

xxx

x

x

xx

The radius = ( )22

13

= 13 units.

MB = 12

22

2

213

2512

=

+ xx

or 222 25,4225,612 xx =+

or 222

4169

42512 xx =+

answer answer

(4) [7]

D

O

B

A

C M 4x

6,5x

2,5x

25



QUESTION 8 8.1 Draw diameter AM and join M to B.

°=+ 90AA 21 (rad ⊥ tangent) °=+ 90BB 21 (∠s in a semi circle)

22 AB = (∠s in same seg)

11 AB = OR

construction S/R °=+ 90BB 21 ∠s in a semi

circle S/R

(5)

Draw radii OC and OA Let x=2A

x=1C (∠ opp = radii) x−°= 90A1 (rad ⊥ tan)

x2180COA −°= (∠ sum Δ) x−°= 90CBA (∠ circ cent = 2 ∠ circumference)

1ACBA = (= x−°90 ) OR Draw QA extend to P. Draw tangent CP at C. PC = PA (tan from comm pt)

12 AC = (∠s opp = sides) CBA2AOC =

(∠ circ cent = 2∠ circumf) °=+ 90AA 21 (tan ⊥ radius)

ABC

AOC21A

A2

AA

CA

)C90A90(180AOC

1

1

11

21

21

=

=

=

+=

+=

−°+−°−°=

OR

construction

x−°= 90A1 rad ⊥ tan

S/R S/R

(5)

construction S/R

S/R

°=+ 90AA 21 tan ⊥ radius

(5)

NOTE: If there is no construction: 0 / 5 marks If candidate changes lettering and states “Similarly”: full marks

A

C

P Q

O

B

1

M

2

2 1

A

C

P Q

O

B

1

1

1

A

C

P Q

O

B

1

2

2

26



Draw diameter AM and Join M and C °= 90ACM (∠s in semi circle)

°=+ 90ACMA 2 (∠ sum Δ) °=+ 90AA 21 (rad ⊥ tangent)

1ACMA = BCMA = (∠s in same seg)

BA1 =

construction S/R

S/R

°=+ 90AA 21 tan ⊥ radius

(5)

8.2.1 °= 90SRW (tan ⊥ radius) statement

(1)8.2.2 °= 50TSR (tan ch th)

°= 40W (∠ sum Δ) OR

°= 90T1 (∠s in semi circle)

11 TRW =+ (ext ∠ Δ) °= 40W

S/R °= 40W

(2)

11 TRW =+ °= 40W

(2)8.2.3 °= 40R 2 (tan ⊥ radius)

°= 40P1 (∠s in same seg) °= 40R 2 °= 40P1 ∠s in same seg

(3)

M

A

C

P Q

O

B

1

2

1

2

O

S

R

T

W

P

V

3

1

1

1

1

1 2

2

2

2

2 50°

50°

40°

40°

40°

27



8.2.4 WP1 = (= 40°) WVPT is a cyclic quadrilateral (ext ∠ = int opp)

STPV1 = (ext ∠ cyclic quad) OR

°= 90T1 (∠s in semi circle)

2T90STP +°=

12 ST = (∠s in same seg)

1S90STP +°=

11 S90V +°= (ext ∠ Δ) STPV1 =

OR

°= 401P2 (∠s on str line) °=+ 180PW 2

WVPT is cyclic quad (opp ∠s suppl) STPV1 = (ext ∠ cyclic quad)

OR

1211 SRRV ++= (ext ∠ Δ)

11 S90V +°=

2T90STP +°= But 12 ST = (∠s in same seg)

STPV1 = OR In ΔPTS and ΔWVS

WP1 = (= 40°)

2S is common STPV1 = (∠ sum Δ)

WP1 = WVPT is a cyclic quadrilateral ext ∠ = in opp ext ∠ cyclic quad

(4)

∠s in semi circle 2T90STP +°= 12 ST = ∠s in same seg

(4)

°=+ 180PW 2 WVPT is a cyclic quadrilateral opp ∠ suppl ext ∠ cyclic quad

(4)

11 S90V +°= 2T90STP +°= 12 ST = ∠s in same seg

(4)

identification of triangles WP1 = 2S is common ∠ sum Δ

(4)[15]

28



QUESTION 9

9. °= 90C (∠s in semi circle) °= 90AEO (corres ∠s; OD || BC)

AE = 8 cm (line from circ cent ⊥ ch bis ch) OE = 6 cm (Pythagoras) ED = 10 – 6 = 4 cm OR

°= 90C (∠s in semi circle) °= 90AEO (corres ∠s; OD || BC)

OE || BC (given) OA = OB (radii) AE = EC = 8cm (midpoint theorem) OE = 6 cm (Pythagoras) ED = 10 – 6 = 4 cm OR

°= 90C (∠s in semi circle) BC2 = (20)2 – (16)2 BC2 = 144 BC = 12

OE = 21 BC (midpoint theorem)

OE = 6 cm OD = 10cm ED = 10 – 6 = 4 cm OR

°= 90C (∠s in semi circle) BC2 = (20)2 – (16)2 BC2 = 144 BC = 12

BC21OE = (midpoint theorem)

OE = 6 cm ED = 4cm

°= 90C °= 90AEO line from circ

cent ⊥ ch bis ch OE = 6 cm ED = 4 cm

°= 90C °= 90AEO midpoint

theorem OE = 6 cm ED = 4 cm

°= 90C BC = 12 reason OE = 6 cm ED = 4 cm

[5]

°= 90C BC = 12 reason

OE = 6 cm ED = 4 cm

[5]

A

C

B

D

E

O

29

Mathematics/P3 DBE/November 2010 NSC - Memorandum QUESTION 7 7.1

1D = 33° (∠ in same segment) DEA = 90° (given)

°= 57A1 (∠ sum Δ) OR

CEB = 90° (given) °= 57B1 (∠ sum Δ) °= 57A1 (∠ in same segment)

OR DE = EB (line from circ cent ⊥ ch bis ch) AE is common

DEA = =1E 90° (given) ΔAED ≡ ΔAEB (SAS)

CBA = 90° (∠s in semi-circle) °== 57AA 21 (∠ sum Δ)

1D = 33° ∠ in same

segment °= 57A1

(3)

°= 57B1 °= 57A1 ∠ in same

segment (3)

DE = EB (S/R)

ΔAED ≡ ΔAEB (SAS)

answer (3)

7.2 °=+ 57DD 12 (OD = OA = radii) °= 24D2

OR

°=114COD (OD = OA = radii) OR ∠ at the centre theorem °= 90E2 °=114D2 °− 90

= °24

°=+ 57DD 12

answer (2)

°=114COD

answer (2)

A

O

C

BD E

2 1

2

1

1 2 3

1 2

33°

30

Mathematics/P3 DBE/November 2010 NSC - Memorandum 7.3 °= 90CBA (∠ in semi-circle)

°= 57A2 (∠ sum Δ) 1A= AE bisects BAD OR DE = EB (line from circ centre bis ch) AE is common

°== 90BEAE1 (given) ΔADE ≡ ΔABE (SAS)

12 AA =

°= 90CBA ∠ in semi-

circle 12 AA = or

AE bisects BAD (3)

DE = EB (S/R)

ΔAED ≡ ΔAEB (SAS)

12 AA = or AE bisects BAD

(3)[8]

31

Mathematics/P3 DBE/November 2010 NSC - Memorandum QUESTION 8 8.1 Draw diameter TP.

Join P to J. °=+ 90TT 21 (tan ⊥ diameter) °=+ 90JJ 21 (∠ in semi-circle)

22 TJ = (∠ in same seg)

1TKJT = OR Draw radii OT and OK Let x=2T

x=1K (∠ opp = radii) x−°= 90T1 (rad ⊥ tan)

x2180KOT −°= (∠ sum Δ) x−°= 90KJT (∠ circ cent)

1TKJT = (= x−°90 )

OR Draw GT extend to H. Draw tangent KH at K. TH = KH (tan from comm pt)

11 TK = (∠s opp = sides) KJT2KOT =

(∠ circ cent = 2∠ circumf) 21 TT + =90° (tan ⊥ radius)

KJT

TOK21T

T2

TT

KT

)K90T90(180KOT

1

1

11

11

11

=

=

=

+=

+=

−°+−°−°=

construction

°=+ 90TT 21 tan ⊥ diameter S/R S/R

(5)

construction

x−°= 90T1 rad ⊥ tan

S/R S/R

(5)

construction S/R

S/R

21 TT + =90° tan ⊥ radius

(5)

T

G

H

•O

J

K

1

P

2

2 1

T

G

H

•O

J

K

1 2

1

T

G

H

•O

J

K

1

2

1

2

NOTE: If there is no construction: 0 / 5 marks If candidate changes lettering and states “Similarly”: max full marks

32

Mathematics/P3 DBE/November 2010 NSC - Memorandum OR

Construct OT, OJ and OK

x== 11 JT (radii) z== 12 KT (radii) y== 22 JK (radii)

°=++ 180222 zyx (∠ sum Δ)

zyxzyx

−°=+°=++

9090

°= 90HTO (rad ⊥ tan))

KJT

90))(90(90

90T3

=

−°=+−°−°=

−°=

zyx

z

construction S/R

S

°=+ 90TT 23

rad ⊥ tan

(5)

T

G

H

•O

J

K

1 2

1 2

1 2

z z

y

y

x

x

3

33

Mathematics/P3 DBE/November 2010 NSC - Memorandum 8.2

8.2.1 x=4B (tan chord theorem)

x== 4BA (corres ∠; BD || AO) x=2B (BO = EO = radii)

x=4B tan chord theorem x== 4BA with

reason x=2B

(4)8.2.2 °= 90EBD (∠ in semi-circle)

x+°= 90EBC OR

°= 90OBC (rad ⊥ tan) x+°= 90EBC

OR

x2O1 = (∠ circ cent) x−°== 90DB 13 (radii)

xxxx

+°=+−°+=

90)90(EBC

°= 90EBD ∠ in semi-circle x+°= 90EBC

(3)

°= 90OBC rad ⊥ tan x+°= 90EBC

(3)

x2O1 = ∠ circ cent x+°= 90EBC

(3)

A

O

E

F

B

D

C

x

1 234

1

2

1 2

32 1

Note: If start with x=A and do not use tan ch th: max 2 marks

34

Mathematics/P3 DoE/November 2009 NSC – Memorandum

• Consistent Accuracy will apply as a general rule. • If a candidate does a question twice and does not delete either, mark the FIRST attempt. • If a candidate does a question, crosses it out and does not re-do it, mark the deleted attempt.


QUESTION 8

8.1 Construct OL and extend to J OMLL2 = (MO = OL)

22 LOMLO += (ext ∠ of ΔOML)

22 L2O = Similarly 11 L2O =

2121 L2L2OO +=+ )LL(2MOK 21 +=

MLK2MOK = OR Join M to K and O to L

x== 12 ML (MO = OL) y== 22 MK (MO = OK)

z== 11 LK (OL = OK) y2180O1 −°= (∠ sum Δ)

2y + 2z + 2x = 180° (∠ sum Δ) 2z + 2x = 180° – 2y 2(z + x) = 180° – 2y

MOK)LL(2 21 =+ MLK2MOK =

construction S/R S/R

11 L2O =

2121 L2L2OO +=+

)LL(2MOK 21 +=(6)

L

O K

M

1 2

1 2

J

If candidate writes: (∠ circ centre = 2∠ at circumference): 0 / 6

MLK2MOK =

Note: Construction can be stated or drawn.

L

O K

M

1

1

1

21

2

2

36




90°- x 180°- 2x P

S

R

O

Q

1 2

3

1 2

1

2

x

T

1 2

x=1R (∠’s opp = radii) S/R 8.2.1 O x2180ˆ

1 −°=ˆ

1

x2180O1 −°= (∠ sum in ΔQRT) S/R P x−°= 90x−°= 90P1 (∠ circle centre = twice ∠ at circumference)

(3)8.2.2 PQ = QR (given)

x−°= 90PRQ (∠ opp = sides in Δ) S/R

Statement

Q x=2

x2RQP = (∠ sum in ΔPQR) xx 2Q2 =+

x=2Q TQ bisects RQP ˆ

ˆ (3)

8.2.3 x2RQP = x2180S −°= (opp ∠’s of cyclic quad are supplementary)

x2180O1 −°= SO1 =

STOR is a cyclic quadrilateral … (converse – ext ∠ of cyclic quad = int opp. ∠)

(ext ∠ quad = int opp ∠)

S/R

Statement

Reason (3)

[15]

37




QUESTION 9

C

O B

M

A

9.1 °= 90ACB (∠’s in a semi-circle) answer (1) 9.2.1 22 810AC −= (Pythagoras)

36= = 6 AM = 3 (line from circle centre ⊥ chord bisects chord OR midpoint theorem)

diameter = 10

AC

AM (3)

9.2.2 22 35OM −= (Pythagoras) = 4 (OR midpoint theorem) Area ΔAOM : Area ΔABC

= 3.4.21 : 6.8.

21

= 6 : 24 = 1 : 4 OR Area ΔAOM : Area ΔABC

= MAOsin.OM.AM.21 : CABsin.AC.AB.

21

= 3.4.21 : 6.8.

21

= 6 : 24

OM

substitution

answer (3)

[7]

38



QUESTION 7

7.1.1 equal to twice the angle subtended by the same chord at the circle.

7.1.2 equal to the angle subtended by the same chord in the alternate segment.

7.1.3 supplementary.

7.2.1 °==∧∧

40BD 11 …(angle between tangent and chord)

∴ °==∧∧

40BD 12 …(CD = CB)

7.2.2 ∴ ∧

C = 180° – ( 40° + 40°) = 100°….( angle sum of triangle)

7.2.3 ∧

A = 180° – 100° = 80° ……… (Opposite angles of a cyclic quad are supp.)

7.2.4 °==∧∧

160A2O1 …. ( angle at the centre is twice…) ALTERNATIVE

From 7.2.1 °==∧∧

40BD 12 Now °=°+°−°= 10)4040(90D3 … (tan ⊥ radius)

∴ °=°+°−°= 160)1010(180O1 …(sum of angles in triangles)

answer (1)

answer (1)

answer (1)

statement & reason statement

(2)

statement °=∧

100C (1)

statement °= 80A

(1)

statement °=∧

160O1 reason (2)

°= 10D3

°= 160O1 (2) [9]

D S R

C

B A

O 1

123

40° 1

234 5

39



QUESTION 8

8.1 x===∧∧∧

213 RRQ …( ext angle of cyclic quad…) and

( RA bisects ∧

R )

x==∧∧

22 QR … ( angles in the same segment)

Now 32 QQ∧∧

=

OR

BQP bisects AQ

QQ

(given) QRRbut

RQ

segment) samein (angles RQbut

quad.) cyclic angle(ext RRQQ

QQ

bisect...)RA segment, samein (angles RRQbut

quad.) cyclic of angle(ext RRQQ

23

121

13

22

2122

23

122

2132

∴

=⇒

==

=⇒

=

+=+

=∴

==

+=+

OR

8.2 x BQ3 ==∧∧

… (angles opp equal sides, AQ = AB)

x==∧∧

BR1 … (from 8.1) ∴ TR = TB …….( sides opp equal angles)

21 RR∧∧

= reason

x==∧∧

22 QR If no valid conclusion 2/3

(3)

x BQ3 ==∧∧

x==∧∧

BR1

(2)

Follow candidates’ argument. To get full marks candidate must reach a valid conclusion

QT

R P

A

B3

1

1

1

2

2 2

40



8.3 1AP∧∧

= (∠ in same segment)

∧∧∧

+= BQA 31 (ext ∠ of ∆ABC = sum into opp ∠’s)

33 Q2BQ =+∧∧

(∧∧

= BQ3 ∠’s opp equal sides)

13 R22Q =∧

(from 8.1) TRPR2 1 = (given) OR x2PRT = ……..(from above)

x2BQA 31 =+=∧∧∧

……( exterior angle of triangle)

And x2AP 1 ==∧∧

….( angles in the same segment) PRT=

x2AP 1 ==∧∧

x2BQA 31 =+=∧∧∧

13 R22Q =∧

x2RR 21 =+∧∧

x2BQA 31 =+=∧∧∧

x2AP 1 ==∧∧

(3) [8]

41

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