Gr 12 Technical Maths P2 Bili 2025 Possible Answers

PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2025 MARKING GUIDELINES/ NASIENRIGLYNE TECHNICAL MATHEMATICS/TEGNIESE WISKUNDE (PAPER/VRAESTEL 2) (11092) 22 pages/bladsye Marking Codes/Nasienkodes A Accuracy/Akkuraatheid CA Consistent Accuracy/Volgehoue akkuraatheid M Method/Metode R Rounding/Afronding NPR No penalty for rounding/Geen penalisering vir afronding nie NPU No penalty for units omitted/Geen penalisering indien eenhede weggelaat nie SF Substitution into the correct formula/Vervanging in korrekte formule F Correct formula/Korrekte formule S Simplification/Vereenvoudiging ST/RE Statement and reason/Bewering en rede AO Answer only/Slegs antwoord

11092/25 NOTES: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed- out version. • Consistent accuracy applies in all aspects of the marking guidelines, where applicable. LET WEL: • Indien ‘n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. • Indien ‘n kandidaat 'n antwoord deurgetrek het en nie weer beantwoord het nie, merk die deurgetrekte antwoord. • Volgehoue akkuraatheid is deurgaans van toepassing op alle aspekte van die nasienriglyne waar van toepassing. QUESTION/VRAAG 1 CL/DV 1.1 𝐵𝐶 = √(𝑦𝐵 − 𝑦𝐶 )2 + (𝑥𝐵 − 𝑥𝐶 )2 L2 ✓ SF A 𝐵𝐶 = √(5 − 8)2 + (8 − 4)2 𝐵𝐶 = √25 𝐵𝐶 = 5 units/eenhede ✓ answer/antwoord CA AO: Full marks (2) 𝑦𝐵 − 𝑦𝐴 1.2 𝑚𝐴𝐵 = 𝑥 −𝑥 𝐵 𝐴 L1 ✓SF A 5−1 = 8−2 4 2 = ✓ 𝑚𝐴𝐵 = 3 CA 6 2 AO: Full marks (2) =3 2

11092/25 1.3 𝑚𝐴𝐵 = 𝑚𝐶𝑀 (∥ 𝑙𝑖𝑛𝑒𝑠) L3 2 2 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) ; 𝑚 = 3 and/en 𝑝𝑡. (4; 8) ✓ 𝑚𝐶𝑀 = 3 CA 2 ✓ SF CA 𝑦 − 8 = (𝑥 − 4) 3 2 16 ✓ answer/antwoord A 𝑦= 𝑥+ 3 3 (3) OR/OF OR/OF 2 𝑦 = 𝑚𝑥 + 𝑐 ; 𝑚 = and/en 𝑝𝑡. (4; 8) 3 2 ✓ 𝑚𝑀𝐶 = 3 A 2 8= (4) + 𝑐 3 ✓ SF CA 16 𝑐= 3 ✓ value of/waarde van c CA 2 16 (3) ∴ 𝑦= 𝑥+ 3 3 1.4 16 𝑀 (0 ; 3 ) OR/OF 𝑀(0 ; 5, 33) ✓0 A L1 16 ✓ 3 or/of 5, 33 A (2) 3

11092/25 1.5 Using the equation of CM to find 𝑡/Gebruik die CA from/van 1.3 L3 vergelyking van CM om t te verkry 2 16 𝑦= 𝑥+ 3 3 ✓ SF CA 2 16 6 = (𝑡) + 3 3 2 16 𝑡 =6− ✓S CA 3 3 2 2 𝑡= 3 3 ∴𝑡=1 ✓ value of/waarde van 𝑡 CA (3) OR/OF OR/OF Using the equation of CM to find 𝑡/Gebruik die CA from/van 1.3 vergelyking van CM om t te verkry. 3𝑦 = 2𝑥 + 16 ; and/en 𝑝𝑡. (𝑡 ; 6) ✓ equation/vergelyking in 𝑎𝑥 + 𝑏𝑦 = 𝑐 A 3(6) = 2𝑡 + 16 18 = 2𝑡 + 16 ✓S CA ∴ 2𝑡 = 2 𝑡=1 ✓ the value of/waarde van 𝑡 CA (3) [12] 4

11092/25 QUESTION/VRAAG 2 CL 2.1 2.1.1 𝑟 2 = 𝑥 2 + 𝑦 2 L1 𝑟 2 = (0)2 + (5)2 ✓ SF A 𝑟 2 = 25 𝑥 2 + 𝑦 2 = 25 ✓ Equation/Vergelyking CA AO: Full marks (2) 2.1.2 Mid-point/𝑝𝑢𝑛𝑡𝑅𝑥 = 𝑥𝑃 + 𝑥𝑅 L2 2 0 + 𝑥𝑅 ✓ SF (both/beide) A 2= 2 4 = 𝑥𝑅 ✓ 𝑥𝑅 = 4 CA ∴ 𝑥𝑅 = 4 𝑦𝑃 + 𝑦𝑅 Mid-point/punt Ry = 2 5 + 𝑦𝑅 4= 2 8 = 𝑦𝑅 + 5 ✓ 𝑦𝑅 = 3 CA ∴ 𝑦𝑅 = 3 𝑅(4 ; 3) AO: Full marks (3) 5

11092/25 2.1.3 𝑚𝑂𝑀 = 𝑦𝑀 − 𝑦𝑂 𝑥𝑀 −𝑥𝑂 4−0 ✓ SF A L3 = 2−0 4 = 2 =2 ✓ 𝑚𝑂𝑀 = 2 CA 𝑦𝑅 − 𝑦𝑃 𝑚𝑃𝑅 = 𝑥 −𝑥 𝑅 𝑃 3−5 = 4−0 −2 = 4 1 1 = −2 ✓ 𝑚𝑃𝑅 = − 2 CA ∴ 𝑚𝑂𝑀 × 𝑚𝑅𝑃 ✓ 𝑚𝑂𝑀 × 𝑚𝑅𝑃 = −1 A 1 = 2 × − = −1 (4) 2 OR/OF ∴ 𝑂𝑀 ⊥ 𝑅𝑃 OR/OF Theorem of Pythagoras/Stelling van Pythagoras in ∆𝑂𝑃𝑀 𝑂𝑃 = 5 𝑃𝑀 = √(5 − 4)2 + (0 − 2)2 ✓ SF(Pyth. Theorem) A ∴ 𝑃𝑀 = √5 ✓ 𝑃𝑀 = √5 CA 𝑂𝑀 = √(4 − 0)2 + (2 − 0)2 =√16 + 4 ∴ 𝑂𝑀 = √20 ✓ 𝑂𝑀 = √20 CA 𝑂𝑃2 = 𝑃𝑀2 + 𝑀𝑂2 2 2 𝑂𝑃2 = (√5) + (√20) ∴ 𝑂𝑃2 = 25 𝑂𝑀 ⊥ 𝑃𝑅 ∴ ∆𝑂𝑃𝑀 is a right-angled triangle at ✓ Conclusion/Gevolgtrekking CA M/reghoekige driehoek by M (4) 6

11092/25 2.1.4 L2 −1 1 ✓ SF A 𝛼 = 𝑡𝑎𝑛 (− ) 2 ∴ 𝛼 = 26, 6° ✓ answer/antwoord CA (2) 2.2 L2 ✓ 𝑥-intercepts/afsnitte A ✓ 𝑦-intercepts/afsnitte A ✓ shape/vorm CA (3) [14] 7

11092/25 QUESTION/VRAAG 3 CL/DV 3.1.1 L2 𝑟2 = 𝑥2 + 𝑦2 (17)2 = (𝑚)2 + (−4)2 ✓ SF A 289 = 𝑚2 + 16 𝑚2 = 273 ✓ 𝑚2 = 273 CA ∴ 𝑚 = −√273 ✓ m= −√273 CA (3) 3.1.2 sec 2 𝜃 − 1 OR/OF tan2 𝜃 L3 17 17 2 −4 2 ✓ −√273 CA =( ) −1 =( ) −√273 −√273 16 16 = = ✓ answer/antwoord CA 273 273 (2) 8

11092/25 3.1.3 −4 L4 sin 𝜃 = ✓ trig ratio/verhouding A 17 4 𝜃 = 𝑠𝑖𝑛−1 17 ∴ 𝑅𝑒𝑓 ∠ = 13, 61° ✓Ref ∠ = 13, 61° CA ∴ 𝜃 = 180° + 13,61° ✓ correct quadrant/korrekte kwadrant A 𝜃 = 193, 61° ✓ answer/antwoord CA OR/OF (4) −√273 OR/OF 𝑐𝑜𝑠 𝜃 = 17 ✓ trig ratio/verhouding CA √273 𝜃 = 𝑐𝑜𝑠 −1 ( ) 17 ∴ 𝑅𝑒𝑓 ∠ = 13, 61° ✓ 𝑅𝑒𝑓 ∠ = 13, 61° CA ∴ 𝜃 = 180° + 13,61° ✓correct quadrant/korrekte kwadrant A 𝜃 = 193, 61° ✓answer/antwoord CA OR/OF −4 (4) tan  = − 273 Ref  =13, 61   = 180 + 13, 61  = 193, 61 9

11092/25 3.2 5 cosec(2𝜋 − 3 𝜋) L4 1 1 ✓ 3𝜋 A = cosec ( 𝜋) 3 1 180° 1 = cosec ( 𝜋 × ) ✓ sin 60° CA 3 𝜋 1 = sin 60° ✓ answer/antwoord CA = 1, 155 AO: full marks OR/OF (3) 5 OR/OF cosec(2𝜋 − 𝜋) 1 3 ✓ 3𝜋 A 1 = cosec ( 𝜋) 3 1 1 ✓ 1 CA = sin( 𝜋) 3 1 sin (3 𝜋) = 1,155 ✓ answer/antwoord CA [12] 10

11092/25 QUESTION/VRAAG 4 CL 4.1 1 − 𝑐𝑜𝑠 2 𝜃 = 𝑠𝑖𝑛2 𝜃 ✓ 𝑠𝑖𝑛2 𝜃 A L1 (1) 4.2 sin(180° − 𝜃). cos(2𝜋 − 𝜃). tan𝜃 L3 1 + cos𝜃. cos(180° − 𝜃) ✓ sin 𝜃 A (sin 𝜃). (cos 𝜃). (tan 𝜃) = ✓ cos 𝜃 A 1 + cos 𝜃 . (− cos 𝜃) ✓ tan 𝜃 A sin 𝜃 (sin 𝜃). (cos 𝜃). ( = cos 𝜃) sin 𝜃 ✓ cos 𝜃 A 1 + cos 𝜃 . (− cos 𝜃) ✓− cos 𝜃 A 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃 ✓ 1 − 𝑐𝑜𝑠 2 𝜃 CA 𝑠𝑖𝑛2 𝜃 = 𝑠𝑖𝑛2 𝜃 =1 ✓1 CA (7) 4.3 tan2𝑥 = sin41, 3° + sin153, 6° L3 ✓ 0, 66 & 0, 445 A tan2𝑥 = 0, 66 + 0, 445 ✓ 1, 105 CA tan2𝑥 = 1, 105 ✓ ref. angle/verw hoek CA ∴ 2𝑥 = 47, 8° 𝑥 = 23, 9° ✓ 𝑥 = 23, 9° CA (4) [12] 11

11092/25 QUESTION/VRAAG 5 CL 5.1 For graph of/vir L2 grafiek van 𝒇 ✓ shape/vorm A ✓ intercepts/ afsnitte A ✓ turning points/ draaipunte A For graph of/vir grafiek van 𝒈 ✓ shape/vorm A ✓ intercepts/ afsnitte A ✓TP/DP A (6) 5.2 180° ✓ answer/ L1 antwoord A (1) 12

11092/25 5.3 −2 ≤ 𝑦 ≤ 2 ✓ values/ L2 waardes CA OR/OF ✓ notation/ 𝑦 ∈ [−2; 2] notasies A (2) 5.4.1 𝑥 = 90° or/of 𝑥 = 270° CA from L1 graph/vanaf grafiek: ✓ 𝑥 = 90° ✓ 𝑥 = 270° (2) 5.4.2 𝑥 = 0° or/of 𝑥 = 360° ✓ 𝑥 = 0° CA L2 ✓ 𝑥 = 360° CA (2) [13] 13

11092/25 QUESTION/VRAAG 6 CL/DV L1 of 1 6.1 Area ̂ 𝑣𝑎𝑛 ∆ABC = × 𝑎. 𝑐. sinB ✓ sinB A 𝑂𝑝𝑝 2 (1) 6.2 6.2.1 sin𝐸̂ = 𝐸𝐻 sin𝐹̂ (In ∆𝐸𝐹𝐻) ✓F A 𝐻𝐹 L2 sin40° sin60° ✓ SF A = 𝐻𝐹 40 40 𝑚𝑚 ×sin40° ∴ 𝐻𝐹 = sin60° 𝐻𝐹 = 29,69 𝑚𝑚 Accept/Aanvaar 30 𝑚𝑚 ✓ answer/ antwoord CA (3) 6.2.2 𝐹𝐻 ̂ 𝐺 = 𝐻𝐹̂ 𝐺 = 500 ✓ answer/ L1 𝐺̂ = 800 antwoord A (1) 14

11092/25 6.2.3 𝐻𝐹 2 = 𝐹𝐺 2 + 𝐺𝐻 2 − 2. 𝐹𝐺. 𝐺𝐻. cosG ̂ (In ∆𝐺𝐻𝐹) ✓F A L3 𝐻𝐹 2 = 𝑘 2 + 𝑘 2 − 2. 𝑘. 𝑘. cos (180° −100° ) ✓ SF CA 𝐻𝐹 2 = 2𝑘 2 − 2𝑘 2 cos80° (2) 𝐻𝐹 2 = 2𝑘 2 (1 − cos80° ) 6.2.4 𝐻𝐹 2 = 2𝑘 2 (1 − cos80° ) CA HF from/van 6.2.1 L2 (30)2 2 ° = 2𝑘 (1 − cos80 ) ✓F CA 900 2𝑘 2 = ✓ SF CA 1 − cos80° 2𝑘 2 = 1089, 1244 𝑘 2 = 544, 562 … ∴ 𝑘 = 23, 34 𝑚𝑚 ✓ answer/ antwoord CA NPR (3) 6.2.5 Area of/Opp van ∆HFG = 1 . GH. GF. sinG ̂ ✓F A L2 2 1 Area of/Opp van ∆HFG = 2 × 23, 34 × 23, 34 × sin80° ✓ SF CA from 6.2.4 = 268, 24 𝑚𝑚2 ✓ answer/ antwoord CA OR/OF (3) 1 OR/OF ̂ Area of/Opp van ∆HFG = 2 . 𝐺𝐻. 𝐺𝐹. sinG ✓F A 1 2 ° Area of/Opp van ∆HFG = 2 . (23, 34) . sin80 (𝐺𝐻 = 𝐺𝐹) ✓ SF CA = 268, 24 𝑚𝑚2 ✓ answer/ antwoord CA (3) [13] 15

11092/25 QUESTION/VRAAG 7 CL/ DV 7.1 perpendicular/loodreg . . . . ✓ answer/ L1 antwoord A (1) 7.2.1 ̂1 + 𝐾𝐿̂𝑂 + 134° = 180° 𝐾 (𝐼𝑛𝑡 ∠𝑠/𝑏𝑖𝑛𝑛𝑒 ∠𝑒 ∆) ✓ ST/RE A L2 ̂1 + 𝐾𝐿̂𝑂 = 46° 𝐾 ̂1 = 𝐾𝐿̂𝑂 = 23° 𝐾 (∠𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠/ ✓𝐾̂1 = 23° A 𝑒 ∠ 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 𝑔𝑒𝑙𝑦𝑘𝑒 𝑠𝑦𝑒) ✓ RE A (3) 7.2.2 ̂2 = 67° 𝑁 (∠ at centre = 2 × ∠ at circ/ ✓ST A L1 mdpt. ∠ = 2 × 𝑜𝑚𝑡𝑟. ∠ .) ✓ RE A (2) 7.2.3 𝑀𝐿̂𝑂 = 90° (tan ⊥ radius/raaklyn⊥ 𝑟𝑎𝑑𝑖𝑢𝑠) ✓ ST A L2 ✓ RE A 𝐿̂2 = 𝑁 ̂2 = 67° (alt. ∠′ 𝑠/𝑣𝑒𝑟𝑤∠𝑒 , 𝐿𝑀 ∥ 𝐾𝑁) ✓ST/RE CA ∴ 𝐿̂1 = 23° ✓answer/antwoord CA (4) 7.2.4 ̂3 = 180° − (26° + 67°) 𝑁 ✓ ST CA L2 = 87° ✓answer/antwoord CA OR/OF OR/OF ̂1 = 26° 𝑁 (corr. ∠′ 𝑠/𝑜𝑜𝑟𝑒𝑒𝑛𝑘∠𝑒 . , 𝐿𝑀 ∥ 𝐾𝑁) ✓ ST/RE CA ̂3 = 87° 𝑁 ′ 𝑒 (∠ 𝑠 on str line/∠ 𝑜𝑝 𝑟𝑒𝑔𝑢𝑖𝑡𝑙𝑦n) ✓answer /antwoord CA (2) [12] 16

11092/25 QUESTION/VRAAG 8 CL/ DV 8.1 Equal/gelyk OR/OF equal in length/ewe lank ✓A (1) 8.2.1 𝑅𝑆̂𝑃 = 90° (∠ in semi-circle/ ✓ ST A L2 ∠ 𝑖𝑛 ℎ𝑎𝑙𝑤𝑒 𝑠𝑖𝑟𝑘𝑒𝑙/) ✓ RE A 𝑅𝑃̂𝑇 = 90° (tan ⊥ radius/raaklyn ⊥ 𝑟𝑎𝑑𝑖𝑢𝑠 s) ✓ ST A 𝑃̂1 + 𝑃̂2 = 90° (tan ⊥ radius/raaklyn ⊥ 𝑟𝑎𝑑𝑖𝑢𝑠 ∠𝑠 on a ✓ RE A tan-chord theorem/∠ 𝑡𝑢𝑠𝑠𝑒𝑛 ✓ ST A raaklyn 𝑒𝑛 𝑘𝑜𝑜𝑟𝑑 (5) ∠′ 𝑠 on str line/∠𝑒 𝑜𝑝 𝑟𝑒𝑔𝑢𝑖𝑡𝑙𝑦𝑛/) 8.2.2 (a) 𝑅̂ = 68° (∠𝑠 in same segment ✓ ST A L1 ✓ RE A /∠𝑒 𝑖𝑛 𝑑𝑖𝑒𝑠. 𝑠𝑒𝑔𝑚𝑒𝑛𝑡) (2) (b) 𝑃̂3 = 22° (int ∠𝑠 of ∆/∠𝑒 𝑣𝑎𝑛∆) ✓ ST/RE CA L3 𝑃̂4 = 68° ✓ ST CA 𝑃̂4 = 𝑆̂3 = 68° (∠𝑠 opp. = sides/ ✓ RE CA ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 𝑔𝑒𝑙𝑦𝑘𝑒 𝑠𝑦𝑒, ST=TP) ✓ 𝑇̂ = 44° CA 𝑇̂ = 44° (int ∠𝑠 of ∆/∠𝑒 𝑣𝑎𝑛 ∆) OR/OF OR/OF ST = PT (tans from same point/ ✓ ST/RE A raaklyne dieselfde punt gelyk) ✓ ST A 𝑃̂4 = 𝑆̂3 = 68° (∠𝑠 opp. = sides/ ✓ RE A ∠𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 𝑔𝑒𝑙𝑦𝑘𝑒 𝑠𝑦𝑒,) 𝑇̂ = 44° (int ∠𝑠 of ∆/∠𝑒 𝑣𝑎𝑛 ∆) ✓ 𝑇̂ = 44° CA (4) 8.3 exterior/buitehoek ✓ A L1 (1) 17

11092/25 8.4 8.4.1 Equal chords, equal angles/gelyke koorde, gelyke hoeke ✓ RE A L1 (1) 8.4.2 𝐶𝐴̂𝐷 = 20° (= chords = ∠𝑠/(= 𝑘𝑜𝑜𝑟𝑑𝑒 = ∠𝑒 ) ✓ ST/RE A L2 𝐴𝐶̂ 𝐷 = 90° (int ∠𝑠 of ∆/∠𝑒 𝑣𝑎𝑛 ∆) ✓ ST A AD subtends an angle of 90° at the circumference of the circle./AD ✓ conclusion/ onderspan ʼn hoek van 90°by die omtrek van die sirkel) gevolgtrekking A (3) 8.4.3 𝐷𝐸̂ 𝐹 = 90° (ext. ∠ of cyclic quad/ ✓ ST CA L1 𝑏𝑢𝑖𝑡𝑒 ∠ 𝑣𝑎𝑛 𝑘𝑣ℎ) ✓ RE A (2) [19] 18

11092/25 QUESTION/VRAAG 9 CL 9.1 GS = FS = 33 cm (FS and GS are the radii ✓ ST A L1 of the bigger circle/FS en GS is die radii van ✓ RE A die groter sirkel) (2) 9.2 𝐺𝑇̂𝑆 = 90° (tan ⊥ radius/ ✓ ST/RE A L2 raaklyn ⊥ 𝑟𝑎𝑑𝑖𝑢𝑠) ✓ ST/RE A ̂ 𝐹 = 90° 𝐺𝐷 (∠ in semi-circle/ ✓ conclusion/ ∠ 𝑖𝑛 ℎ𝑎𝑙𝑤𝑒 𝑠𝑖𝑟𝑘𝑒𝑙) gevolgtrekking TS || DF (corr. ∠𝑠 equal/ A (3) ooreenk∠𝑒 𝑔𝑒𝑙𝑦𝑘.) 9.3 𝐺̂ ( common/gemeenskaplike ∠) ✓ ST/RE A L1 𝐺𝑇̂𝑆 = 𝐺𝐷̂𝐹 (both/beide 90°) ✓ ST/RE A rd 𝐺𝑆̂𝑇 = 𝐺𝐹̂ 𝐷 (3 ∠ of ∆/3 𝑑𝑒 ∠ 𝑣𝑎𝑛 ∆) ∆OMT||| ∆OPV (∠, ∠, ∠) ✓ conclusion/ gevolgtrekking A (3) [8] 19

11092/25 QUESTION/VRAAG 10 CL 10.1.1 𝐴𝑂̂𝐷 = 315° = 315° × 𝜋 ° ✓ 315° A L1 180 ✓ answer/ 7 antwoord CA = 𝜋 𝑜𝑟/𝑜𝑓 5,50 radians/𝑟𝑎𝑑𝑖𝑎𝑙𝑒 4 (2) 10.1.2 Arc length/Booglengte = 𝑟𝜃 ✓F A L1 7 ✓ SF CA = 2,25 × 𝜋 4 63 = 𝜋 16 ✓ answer/ = 12,37 𝑚 antwoord CA (3) 10.1.3 𝑉 = 𝜋𝐷𝑛 ✓F A L2 12 ✓diameter A = 𝜋(2,25 × 2) ( ) 60 ✓ SF CA 9 𝑚 ✓ answer/ = 𝜋 10 𝑠 antwoord CA = 2,83 𝑚/𝑠 (4) 10.1.4 𝜔 = 2𝜋𝑛 ✓F A L2 12 ✓ SF CA = 2𝜋 ( ) 60 2 = 𝜋 𝑟𝑎𝑑/𝑠 5 ✓answer/ = 1,26 𝑟𝑎𝑑/𝑠 antwoord CA (3) 10.2 Area/Opp = 𝑟2𝜃 ✓F A L2 2 ✓ SF CA 7 2,252 × 4 𝜋 ✓ answer/ = 2 antwoord CA 567 = 𝜋 (3) 128 = 13,92𝑚2 OR/OF OR/OF 𝑟𝑠 Area/Opp = 2 2,25 × 12,37 ✓F A = 2 ✓ SF CA = 13,92 𝑚2 ✓ answer/ antwoord CA 20

11092/25 (3) [15] QUESTION/VRAAG 11 CL/ DV 11.1.1 Area/𝑂𝑝𝑝 = (sum of parallel sides/𝑠𝑜𝑚 𝑣𝑎𝑛 𝑒𝑤𝑒𝑤𝑦𝑑𝑖𝑔𝑒 𝑠𝑦𝑒) ×ℎ L2 2 (2,5 + 1,5) ✓ SF A 8= ×ℎ 2 16 = 4ℎ ✓ answer/ ℎ =4𝑚 antwoord A (2) 11.1.2 AB = √0,52 + 42 ✓pyth theorem/ L3 stelling A = 4,03 𝑚 ✓ 0,5 CA ✓ answer/ antwoord CA (3) 21

11092/25 11.2 Surface area/𝐵𝑢𝑖𝑡𝑒 − 𝑂𝑝𝑝 = (2 × 8) + 2(4 × 4,03) + (1,5 × 4) ✓ SF A L2 = 54,24 𝑚2 ✓𝑅70 × 54,24 Cost of painting interior/Koste om binne te verf = 𝑅70 × 54,24 CA = 𝑅3 796,80 ✓answer/ antwoord CA (3) 11.3 Volume = area of base × height/opp van basis × ℎ𝑜𝑜𝑔𝑡𝑒 ✓ SF A L2 =8×4 = 32𝑚3 ✓ 32𝑚3 CA = 32 000 litres/𝑙𝑖𝑡𝑒𝑟𝑠 ✓ 32 000 𝑙 CA (3) 11.4 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 2 ℎ ✓F A L4 22 ✓ 1003 CA 32 × 1003 = (152 )ℎ 7 448000 ✓answer/ ℎ= 99 antwoord CA = 45 252,53 cm (3) OR/OF OR/OF 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋𝑟 2 ℎ ✓F A 22 32 = (152 )ℎ 7 112 ✓ 1003 CA ℎ= × 1003 2475 448000 ℎ= ✓answer/ 99 = 45 252,53 cm antwoord CA (3) [14] 22

11092/25 QUESTION/VRAAG 12 CL/ DV 12.1 𝑜1 + 𝑜2 ✓F A L2 𝐴𝑇 = 𝑎 ( + 𝑜2 + 𝑜3 + ⋯ + 𝑜𝑛−1 ) 2 ✓ SF A 12 + 9 = 8( + 6,5 + 7,2 + 8,1 + 7,9 + 6,7) 2 1 ✓answer/ = 375 5 antwoord CA = 375,2 𝑐𝑚2 (3) OR/OF OR/OF A T = a ( m1 + m2 + m3 + ... + mn ) ✓F A  12 + 6,5 6,5 + 7, 2 7, 2 + 8,1 8,1 + 7,9 7,9 + 6, 7 6, 7 + 9  ✓ SF A A = 8 + + + + +   2 2 2 2 2 2  A = 375, 2 units 2 ✓answer/ antwoord CA (3) 12.2 8 Area/Opp = 𝐿 × 𝐵 ✓100 A L2 8 = (2,2) (6 × 100) ✓ SF CA = 1,056 𝑚2 ✓ 1,056 𝑚2 CA (3) [6] TOTAL/TOTAAL : 150 23