Grade 12 Assessment Exemplars

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2 Grade Assessment Eemplars Learning Outcomes and. Assignment : Functions - Memo. Investigation: Sequences and Series Memo/Rubric 5. Control Test: Number Patterns, Finance and Functions - Memo 7. Project: Finance - Guideline 9.5 Eam A: Paper - Memo.6 Eam B: Paper - Memo 8 Learning Outcomes and. Assignment: Recap of Grade Data Handling - Memo 6. Investigation: Polygons - Guideline 8. Control Test: Geometry and Trigonometry - Memo. Project: Transformation Geometry - Guidelines.5 Eam A: Paper - Memo 5.6 Eam B: Paper - Memo Grade - - Educators Guide 8

3 Grade Assignment: Functions Marks: In eac case for eac value of tere is only one associated output value a b c Tose points were te function values are zero wic means tey are on te - ais/ Te points were te function cuts te -ais Te possible number of -intercepts are determined by te degree of in te function st degree: one -intercept nd degree: two possible -intercepts rd degree: tree possible -intercepts In te case of nd and rd degree functions two of te roots may be te same or non-real. 5 y is of te first degree in eac case ence only one y-intercept 6 a min. value b no min. or ma. value c local min. and local ma. values. Te points were a function as a min or ma value are also called stationary points. Tese are determined were te gradient of a tangent to te function (derivative equals zero. To determine weter te point is a min. or ma. one as investigate te derivative/gradient on eiter side of te stationary points. Min or ma. can also be determined by te value of te nd derivative and and f ( not a function one-to-many mapping 7. domain needs to be restricted to eiter or 7. g ( is a function one-to-one mapping Grade - - Educators Guide 8

4 7. and f ( not a function one-to-many mapping 7. domain needs to be restricted to or or 8 ( and g( gives te function value wen. ( and g ( gives value of te gradient (derivative wen. 9 / // If f ( ten f ( and f ( tus no points of inflection. / If g ( 5 ten g ( 5 tus constant gradient so no points of inflection. / // // If ( 6 ten ( and ( 6. ( wen tus ( as a point of inflection wen.. k( is te reflection of k( in te -ais. k( is te reflection of k( in te y-ais. k(y is te reflection of k( in te line y Grade - - Educators Guide 8

5 Grade Investigation: Te Koc Snowflake Marks: Section A t stage snowflake on dotty paper ( Section B. marks for first 5 rows (in any form for eac error or omission ( X 5 Snowflake A:Lengt of side B: Number of sides C:Perimeter n - n. Last row of table above ( 5 X. Column A: geometric sequence:, Column B: geometric sequence: Column C: geometric sequence any valid observations (. Te perimeter at eac stage is a term of a diverging geometric sequence since r. Hence te perimeter increases witout limit. Correct answer and valid eplanation (5 Grade Educators Guide 8

6 Section. marks for first 5 rows ( Stage A: Area of eac B: number of C: Increase in area D: Total area added triangle triangles added 5 n. ie. were and (5. As, and te total area (. As te number of stages gets very large, te perimeter of te snowflake increases witout limit, but te area approaces a fied limit. ( Grade Educators Guide 8

7 Grade Test: Number Patterns, Finance and Functions Time: our Marks: 5... substitute (; ( f ( a 6 a a a 6 (.. ( 6 ; (; (... p ; q ( [7].. Te area of te rectangle depends on te value of. For eac value of tere is only one value for te area ( 5( 5 ( invalid (.. -intercepts: (-5; and (7; y-intercept: (;5 turning point: (;6 ais of symmetry: Note: full marks if Quad I only. (-5; (7; y (;5 (;6 ( < < 7 (... > (... 6 square units ( Grade Educators Guide 8

8 ... a 6; r ( [7].. 6.( n n n 58 5 n 6 ( [6]. Stage Stage Stage Stage Stage n Number of patterned tiles n Number of black tiles 9 6 n Number of wite tiles 6 n n Total number of tiles n n Patterned tiles : Black tiles : Total number of tiles : T T T T n n n n ( n n n n n n ( by inspection n n n (5.. n n n t stage will require patterned tiles. ( 5... Area.5(n n m ( [] 5.. A P n ( i (.7 R59 5 ( ,95,95,95 ( ( (,95 6 ( (,95 R77.8 [ ],95... to 6 terms (6 Grade Educators Guide 8

9 Grade Project: Finance Marks: 5 Notes to Educator. Te data seet for tis project sould be updated to reflect current veicle prices, interest rates and rate of inflation.. Current and istorical data on te fied and linked lending rates of banks can be obtained from te Sout African Reserve Bank internet site: ttp:// Te required document codes are: KBP8M AND KBP8M. Before learners commence te project, te following concepts and terminology sould be discussed in class: loan, interest, compound interest, annuity, present value, future value, outstanding balance, depreciation, sinking fund, inflation and rate of inflation. Assessment Guidelines. repayment calculation for fied- and linked rate options: substitution simplification payment total (8. balance calculation: formula substitution simplification answer repayment calculation: substitutionsimplificationpayment total comment (9. calculation of balloon payment calculation of loan amount repayment calculation: substitution simplification payment total (6. calculation of difference between payments annuity calculation: substitution simplification value of n ( substitution answer ( 5.. substitution answer ( 5.. Fv of sinking fund substitution simplification answer ( 6. Metod used: (5 Deducts Fv of T, T, T6, T8 simplification answers Any oter metod tat as matematical merit and correct calculations, but does not consider te Fv of te oliday payments tat are missed 7. Valid comment substantiation using calculated figures from above ( X (8 in respect of eac of te following: fied rate versus linked rate balloon payment payment oliday creation of a sinking fund Grade Educators Guide 8

10 Grade Matematics Eam Paper Time: ours Marks: 5 No. Solutions Comments.. ( ( multiplying 7 7 ( ( 5 standard form factors.. or 5 6 ± (( 6 ± 5 6,87 or, 5 values standard form formula subst simplification values ( (5... ( 9 9 > > 5 > ( ( > < or > y 5 and y 5 y simplification standard form factors (grap or any oter metod used critical values/inequality (6 making y te subject subst simplification standard form factors ( 6 6 y 8 value y value (7 [] Grade - - Educators Guide 8

11 ..... A P( i r 7 5 r,7%,7 i i,9 % n Loan amount R 8 7 Let montly instalment be,, 8 6, R6, 5 OR 6 Equating te current value of te loan wit te current value of repayments we get:, 8 R 6,5(to nearest cent 6, After years balance:, 8, 6.5, R 775, 97 OR Calculating te current value of te sum of te outstanding payments:, 6.5, 6,,... 6,5 6 for calculating nominal interest rate for calculating effective interest rate [5] for numerator for denominato r answer [8] for balance Grade - - Educators Guide 8

12 6.5,,,, R 775,97 OR n n 6 R77 5,8 n, 5,, 76,9,, n, n n,,679998,,5766 log,5766 n, log n 5,5 monts ,8,, log n n n, ,8, 5 Ten n 5,6, log As above tere are 5 payments of R 5 and a final lesser payment. [8].... Te number of HIV positive people increase by, mill people per year. H, t,7 H, (6,7,9 million Grade - - Educators Guide 8

13 .. For a series to converge < r < ; r tus series converges. ( S S a r 5( 9 5( 67,5 9 67,7 67,5 67,7, 6 ; 6 Second common difference T n an bn c Second common difference a T n n bn c n : T b c 6.. (i n : T b c.. (ii (ii (i b b c n n T n formula subst ( answer ( formula subst answer ( answer ( ( n n 6 n n 6 ( n 5( n 6 n 5 (5 equation standard form factors value of n Grade - - Educators Guide 8

14 .. ( Grade - - Educators Guide 8

15 . q value of q 5 k k value of k. f ( a( f ( a( a 9 subst value of a. g is a one-to-many relation. [ ; ( ; ].5 f : k y k log ( k y y.6 ( ( 9 [] 5.. y y / c : y / c : o f ( sin or f ( cos( o ; 6 o [] Grade Educators Guide 8

16 Grade Educators Guide f f f ] 5 [ lim 5 lim lim 5 ( 5( lim ( ( lim ( / 8 d d or Tangent: y ( ; is on f : a ( b ( a b b a b a f b a f ( ( / / a b a a b 5 (5 power form derivatives ( (7

17 ( 7. A ( 7. For min A: A / cm (to nearest cm A / ( 8. f ( f ( f ( f ( - is a factor of f ( ( ( ( ( ( f / ( 6 6 ( ( or ( ; 5 7 y quadratic factor linear factors derivative factors -values y-values Grap: y-intercept turning points sape 8.5 f / /( 6 f // f // answer [6] Grade Educators Guide 8

18 9. 6 y y 5 5y 5 y 5 y (one mark for eac constraint 9. 5 y 9. (one for mark for eac grap (for searc line 9. (for feasible region 5 9. P y 9.5 Plasma s and LCD s [5] Grade Educators Guide 8

19 Grade Matematics Eam Paper Time: ours Marks: 5 QUESTION... ( or 9 ± 7 bot sides ± Solution Multiply & simplify Factorise Solution (.. ( ( ± 6,8 or 6,6 (5... or 5 ( Sape Intercepts (.. 5 (. sub y in 6 y 6 6 ( ( or and y or y (5 Grade Educators Guide 8

20 . Values.. and y And (.. or y or ( [9] QUESTION... T 6 5 5(,5 R65,9 Formula Subst Answer (.. S 6 6 5(,5,5 R9, Earnings for year : R9, R8688,65 (5... T T n 6 6(, n 6 6(,78 R68. (.. 7% of book value R557,6 Less tan book value Will not be written of f (.. ( F 8 (,8...( R77,6,75 8,75 Formula Subst Calculation ( [9] Grade - - Educators Guide 8

21 QUESTION... 7; 9; (.. Numbers from to 5 not divisible by (.. Subtract number of terms in following two sequences: ; ; ; 5 and ; 6; 9; 5 5 ( n n 5 n 5 Number of terms in sequence 5 5 (5. k k k (... T T T T T n Side 8 8( n Perimeter 6 8( n (.. Sum of perimeter of inner triangles 6... cm Tus te limit of te sum of te inner triangles is cm and te sum will not eceed cm, wic is te perimeter of te outer triangle. ( [] Grade - - Educators Guide 8

22 QUESTION. Because te given function is a many-to-one function, its inverse is one-to-many and ence is not a function.. y f (. ( ;]. At points of intersection: Hence or 8 ( ^y O > (,5;,5.5 [ f ( ] f ( f [ ] f And f ( f [] Grade - - Educators Guide 8

23 QUESTION 5 5. m and c 5 Terefore equation is : ( 5 ( 5. Cable 5 ( (5( touces 996 ( 99 ground or 99 again at 99m (5 [6] QUESTION 6 6. ( -8 o -5 o -9 o -5 o 5 o 9 o 5 o 8 o - 6. Te grap repeats itself every 8 degrees. ( 6. Eac -intercept would be degrees more tan it currently is. ( 6. ( [5] Grade - - Educators Guide 8

24 Grade - - Educators Guide 8 QUESTION 7 7. ( ( 7.. [] QUESTION 8 8. ( 8. ( ( ( ( ( 5 ( 8. -intercepts: or - y-intercept: y ( ( lim ( lim ( lim lim ( ( lim ( f f f ( 6 7 ( f f ( ( ( f f f ( 5 ( 5 ( f of factor a is f f

25 8. Stationery points at : f ( 5 ( 5( 5 or ( Turning points: ( ;6; 9,5 or (; f( 5 f ( Local ma Local min 8.5 Point of inflection at 6 C: (-.6, 9.5 ( Sketc ( <,6 or > ( 8.8 Sift grap more tan units to rigt. ( Or: Sift grap down more tan,6 units and rigt more tan one unit. D: (.,. 8.9 g( g( 5 5 B: (-.,. A: (.,. ( - 8. Avg rate of cange f ( ( 8. f '( ( ( 5 9 f ( 9 Eqn of line : y 9 ( ( y 5 ( ( ( ( 6 ( ( ( [] Tangent cuts curve at : or Grade Educators Guide 8

26 QUESTION 9 ( b 9. b 7 b 7 b b 6 9. lengt - breadt 6 ( 9. A( ( ( ( 6 8 cm 9. A ( 6 Ma at A' ( 6 9 <9 9 >9 A ( - Local ma ( [9] QUESTION. A : y B : C : y 55 D : y 5 A (. on diagram ( C D. A : y (. Must be integer solutions, terefore feasible solutions are (; and B (; (.5 More ball skills coaces. Terefore coose (; (.6 Cost 5 R7 ( [] Grade Educators Guide 8

27 Grade Assignment: Recap of Grade Data Handling Marks: 5 Question.. 76, 9... σ 89, Cumulative Frequency Class Interval Tally Frequency less tan 999 University Fees Ogive Cumulative Frequency - 99 l l - 99 ll l 5-99 lllll llll llll 8-99 llll lll llll lll l l ll ll l l l l Fees in Rand Grade Educators Guide 8

28 .5 University Fees Skewed to te rigt, more clustered to left of median tan to te rigt, large range, smaller interquartile range, a few ig figures skewing te data (any reasonable eplanation Question.. σ, Values ( ( 8 Var,8 Question Question. How far from te mean most figures are, te average deviation from te norm. large range, most learners in ecess of % away from te mean, couple of ig marks skewing data. Boys: 9 6, 5. m 6 Girls: 6 6, 86 m 6 5 Learners attending University. Boys ave large range, some ig and some low, very little bundling in te middle. Girls marks don t start as low but ave fewer girls scoring as ig as te boys. Mean and median very similar but marks very different. Low scores of boys are offset by ig scores. (any reasonable eplanation No of Learners Year. Eponential Grade Educators Guide 8

29 Grade Investigation: Polygons wit Matces Marks: Some answers and Marking Rubric Triangles Isosceles triangle wit sides, 5, 5 Area,5 base t,5 6 Rigt angled triangle wit sides,, 5 Area,5 6 Equilateral triangle wit sides,, Area,5 sin > 6 6 > 6 So te equilateral triangle as te greatest area. Tere are no oter triangles because te longest side must be sorter tan te sum of te oter two sides, oterwise tere is no triangle! So longest side < 6 If longest side 5, ten te sum of te oter sides is 7. Tey can be 5 and or and. If te longest side is, ten te sum of te oter sides is 8 and te only possibility is te equilateral triangle,,. Quadrilaterals wit equal angles If all angles are equal, eac is 9 and te quadrilaterals will be rectangles Te possible rectangles are,, 5, 5,,,,,, Te by square as te greatest area. 9 > 8 > 5 Grade Educators Guide 8

30 Pentagons Witout breaking te matces, it is not possible to create a regular pentagon (or a regular polygon wit n sides were n is not a factor of. It is also not possible (as can be proved using te sine and cosine rules to create a pentagon in wic all angles are 8 and te perimeter matces. A matces matces B 9 9 E matces F matces C matces D Tis pentagon as an area of 6 wic can be calculated as follows: Let AF be te altitude from te verte A to te opposite side, CD. In ABC : AC (T. of Pytagoras And in ACF AF ( (T. of Pytagoras Hence altitude AF Area of pentagon ABC ACD ADE 9,6... (bigger tan te square. Heagons A regular eagon as sides eac matces long and angles of. Tis can be broken down into 6 equilateral triangles wit all sides matces long. Area 6,5 sin 6 6,9... getting bigger! Oter eagons ave areas less tan tis. Tere are endless possibilities. Twelve sides 8 Eac angle in te regular sided polygon is 5 Tis polygon can be broken into twelve isosceles triangles, eac aving equal sides of r (calculated using te cosine rule wit te included angle being. Hence te area is,5 r sin,96... Again oter sided polygons can be constructed but none as an area as large as tis Te circle wit a perimeter of as a radius r and te area is π,59... π π π π It sould be clear tat te biggest possible area tat can be enclosed wit n matces is a regular n sided polygon: as close to a circle as possible. Grade - - Educators Guide 8

31 Rubric Communication Special cases Generalisation and justification Presentation Total -: Clear, coerent, logical eplanations, supported by appropriate sketces and tables of results.all main ideas covered: te more sides, te greater te possible area; regular polygons larger tan oters wit te same number of sides; te limiting area is tat of te circle wit a perimeter of sides. Some logical etension. 8 to 5: Interesting, compreensive set of special cases. All calculations accurate. 6 to : Quality arguments to support all claims. 5: Neat, striking visual impact 8 to : Clear, logical eplanations, supported by appropriate sketces and tables. One main idea migt be ecluded or no etension 5 to 7: Areas and sketces of te tree triangles, te square, te regular eagon and te regular sided polygon correctly sown and some oter eamples to support te regular is biggest teory. to5: Quality arguments to support almost all claims. : Neat pleasing impact. to 7: Clear logical eplanations, supported by appropriate sketces and tables. One main idea missing and no etension. to : Areas and sketces of te tree triangles, te square, te regular eagon and te regular sided polygon correctly sown. to : Quality arguments to support most claims. : Neat to : Satisfactory eplanations wit appropriate sketces and tables. More tan one main idea missing. 8 to : Areas and sketces provided of te tree triangles, te square, te regular eagon and te regular sided polygon. Accurate calculations wit at most one minor error. to Some valid attempts at eplanations and justifications. : Neat 6 to 9: Adequate eplanations wit some sketces and tables. At least one main idea satisfactorily covered. to 7: Areas and sketces provided of te tree triangles, te square, te regular eagon and te regular sided polygon. Accurate calculations wit at most two minor errors. 8 to 9: Attempts at justifications but flawed or inadequate. : Rater untidy. to 5 Some logical discussion wit at least some sketces and tables. to : Areas and sketces provided of te tree triangles, te square, te regular eagon and te regular sided polygon. Accurate calculations wit at most tree minor errors or one major error or omission. 6 to 7: no valid eplanations or justifications for claims. : Very untidy. to : No main ideas satisfactorily covered. Sketces, tables omitted, flawed or inadequate. to : Major errors and/or omissions. to 5 : A mess: clearly no effort made. Grade - - Educators Guide 8

32 Grade Test: Geometry and Trigonometry Time: our Marks: 5. y ( 5 ( 69. ( ( 69. y 5 5 y 6 y 6. Line troug te origin, wic is perpendicular to AB will cut te circle at te points of contact of te required tangents. Tis line is y Points of intersection are at points were 69 ie 69 ± One tangent is: y te oter is y ± y y y. Reflection in te y-ais. Glide ( to te rigt and reflection about te -ais. Rotation of about te origin. Anti-clockwise rotation of 6 about te origin.5 Rotation troug 8 (because k < Enlargement by a factor of k ( or reduction by a factor of k Reflection about te line y. sin 5 cos( 9 cos( 5 9 ± ( 5 k.6 ; k Z 5 k.6 or 75 k.6 5 k. or 75 k.6 k Z. Yes, Mvuyo is correct: 55 5 (or substitute Grade - - Educators Guide 8

33 tan tan 5 tan tan 5 tan tan tan. tan tan tan tan tan tan tan tan tan tan tan. Identity not valid for 5 9 k.8 ; k Z. tan( 5 ie for is not defined for tese values of θ Also not defined for 9 k.9 since tan is not defined for tese values of θ 5. In ABC,5 k.9 as tan ( 5 BC..cos BC 5. Area ABC Area ACD Area ADB,5 sin Area BCD,5.sin 6,5 ( square units Total surface area ( Grade - - Educators Guide 8

34 Grade Project: Escer and Transformation Geometry Marks: (All M.C. Escer works (c 7 Te M.C. Escer Company - te Neterlands. All rigts reserved. Used by permission. Some Guidelines Te object of tis project is to sow tat transformation geometry is not just a lot of rules and formulae, but tat te concepts are used in te fascinating work of te Dutc artist, M.C. Escer. It is suggested tat a rubric is drawn up, wit agreement from te learners about te need to sow a clear understanding of te transformations learnt in te course of teir years at scool. It is oped tat te learners will enjoy investigating te various patterns. Te solutions below sould be elpful to te marker: Task Tis is Escer s version of filling te plane wit tis fis. Learners may come up wit oter options. Measure eac on its merit.. Mark te point of te tail (not te etreme point, te one about 5mm from te etreme point, A.. Place A at te origin so tat te fis faces upwards. Trace te fis and call it.. Rotate te fis anti-clockwise, about A, troug 9. Trace te fis again and call it.. Rotate again troug 9 around A and call tis fis. 5. Rotate for a tird time and call te resulting fis. 6. Place your template fis on fis and ten slide (translate it down and sligtly to te rigt to fit between fis and fis. Trace te fis again and call it Repeat steps to 5 and call te resulting fis 6, 7 and Place te template on fis. Translate it to te rigt and sligtly up until it lies between fises and. Call tis fis 9. And so on Grade - - Educators Guide 8

35 Task Tree sapes are used: a fis, a bird and a lizard/gecko. Te fis meet at te mout and eac is a rotation of anoter troug about te point were te mouts meet. Te birds fit between te fis and are also rotations of eac oter. Te geckos fit into te tails of te fis and te birds. Te geckos are reflections of eac oter about a vertical line troug te centre of te fis. Tere is also an ais of reflection at an angle of 6 to te vertical. Te complete motifs of all tree creatures can be translated orizontally and vertically so tat te geckos fit on to eac oter. Task Te birds and fis are repeated by a series of glide reflections. Te ais of reflection is a orizontal line troug te middle of te frieze. Te glide is a translation about cm to te rigt. Task Start wit a 6cm by 6cm square. Place te first large gecko in te bottom rigt and corner as sown in te given art work. Rotate tis motif tree times, eiter clockwise or anti-clockwise troug te central point (8cm from te bottom and 8cm from te left and trace copies in te oter tree corners. Te oter eigt boundary geckos are a reduction of te biggest geckos by a factor of tree, but tey first need to be flipped over. Place a small gecko orizontally at te nose of te bottom left big gecko and ten translate it orizontally to form te oter. Repeat tis process (eiter vertically or orizontally to complete te outer ring. 5cm from te bottom, to te rigt of te central vertical line, is a reduction of te largest gecko by a 5 factor of. Tis gecko is ten rotated seven times, about te central point, toug an angle of 5 7 eac time. 5cm from te bottom is anoter ring of geckos, tis time reductions of te largest gecko by a factor of 5. Tis gecko is also rotated seven times troug 5 to form te ring. Te net ring of black geckos is formed in te same way. In between te rings of black geckos, te space is filled by a collection of reductions (and in some cases also reflections of te original. Eac one is rotated troug 9 tree times to create its identical partners. Grade Educators Guide 8

36 Grade Matematics Eam Paper Time: ours Marks: Equation is. At point of intersection of AC and BD: and But te mid-point of BD is And gradient of AC gradient of BD Hence AC is te perpendicular bisector of BD. Area of kite ABCD area ( (,5 5 square units.5 Te inclination of AB.6 Te inclination of AD (correct to decimal place ˆ B AD 7,9 5 7 (correct to nearest degree. Te equation of te circle is or. Te mid-point of PQ is Te gradient of PQ Hence te perpendicular bisector as te equation Substituting te co-ordinates of te centre into tis equation: Hence te perpendicular bisector of PQ passes troug te centre of te circle.. Te radius of te given circle is or Te distance between R and is Since, te point R lies outside te circle. Grade Educators Guide 8

37 . Te co-ordinates of are and te co-ordinates of are.5 Area of original circle Area of enlarged circle centre is Area original circle : area of enlarged circle. D is as been reflected about te line. F is te point as been rotated anticlockwise about te origin troug an angle of. H is as been reflected about te y ais ie. as been rotated anticlockwise about te origin troug an angle of and ten reflected about te y ais.. J is te point is reflected first about and ten about te y ais.... Grade Educators Guide 8

38 ..... or or (correct to decimal place 5. Bearing metres Race distance metres 6. Grade Educators Guide 8

39 6. 6. In In 7. or or ; For 7. Function values correct to decimal place were applicable. marks for eac grap 7. Grade Educators Guide 8

40 8. marks for Boys and marks for Girls (need not split te s into - and 5-9 Boys Girls Bo and wisker plot of Grade Girls Masses 8. Standard deviation, (correct to decimal place 8. Modal mass5 kg 8.5 Interquartile range 5 5kg 8.6 9% of 5 is,5. On te ogive, te weigt tat corresponds to,5 is approimately 7,5 kg 9. True: 5% of te data items are witin te inter-quartile range compared to 68% witin one standard deviation. 9. True: te data is spread more to te left of te median tan to te rigt. 9. False: te greater te time, te lower te water level, so te correlation is negative. 9. C t False: te scatter plot will be a straigt line: is te eigt. C is te original πr capacity of te dam, t te number of ours and r te radius of te dam. All values ecept and t are constants. is a linear function of t wit a negative gradient. Grade - - Educators Guide 8

41 Grade Matematics Eam Paper Time: ours Marks: 5. ( 5 ( AC ( ( AB ( 5 ( 68 BC Hence ABAC and te triangle is isosceles And BC AB AC and ence te triangle is rigt angled at A Rigt angle can also be proved by sowing tat gradient AB gradient AC. Area ABC 7 square units 5. M, te mid-point of BC is ; ( ;. Circle is ( ( ( ( 7.5 Gradient BC 5 gradient of tangent y equation of tangent is 5 i.e. y Gradient MA Gradient of tangent gradient MA ance tese lines are parallel. Radius. y Equation of AB is y 6 i.e. y 5. At points if intersection: ( 5 ± and y 8 y ± So te points of intersection are ( ; and ( ; Grade - - Educators Guide 8

42 . ACB ˆ inclination AC inclination BC tan tan 7, ,78... Hence.. ( 5;.. ( ; 5 5 AOB ˆ ACˆ B P P.. Te co-ordinates of te vertices of te images are reduced by a factor of : i.e. tey are alf te distance of te original vertices from te origin. Te area of te transformed triangle is te area of te original triangle... Te vertices are reflected about te -ais and translated units to te rigt.. Tis transformation is a glide reflection wic is rigid and ence te area of te transformed triangle is equal to te area of te original triangle. P' ^y P(a;b O 6 α > a OP cosα and b OP sinα and ence O P cos α 6 te co-ordinate of P ( OP [cosα cos6 sinα sin 6 a b OP cosα. OP sinα. Grade - - Educators Guide 8

43 . and te y co-ordinate of OP [sin ( α 6 sin cos ( α cos( 9 α α cos( 8 α P O P sinα. cosα. b a ( sinα ( sinα cosα ( cosα sin α cos α tan α.. cos 7 t sin tan5 t t cos cos 6 sin 7 t cos 5 cos 7 tan 7.. (. tan( 75 sin 7 t 75 5 k.8 ; k Z k.8 k.6 ; k Z. sin5 sin5 cos( 8 5 cos95 ( cos5.sin5 sin sin5 cos cos5 cos 5 ( Grade - - Educators Guide 8

44 5. In ADC : r r k cosθ r. r r k r 5. In ABD : r ( ( r ( k cos 8 θ. r.r 5r k cosθ r k 5r cosθ r 5. r k k 5r Hence r r r k k 5r 9r 6k and k r r r So cosθ r 6. A B θ C α metres β D 6. In ACD AC sin β sin[ 8 ( α β ] sin β AC sin α β ( AB In ABC : sinθ AC sin β AB.sinθ sin α β ( 6..sin 7.sin5 Heigt sin, metres (correct to decimal places or metres (correct to te nearest metre Grade - - Educators Guide 8

45 6. Area ACD AC. CD.sinα sin β..sinα sin α β or 665 ( 66,97 m (correct to decimal places m (correct to te nearest m 7. At A sin k.9 ; k Z A ( 9 ; B is ( ; so cos( 6 6 At C y C is ( ; 7. cos( 6 sin( 7. At G: DE cos 6 sin 6 At H: sin( sin 6 y y cos 6 Hence te equation of te new curve is ( g 8 ( i 85 σ 5,98 g (correct to decimal places i 8 or 6 g (correct to te nearest g 8. Standard deviation is a measure of dispersion: te larger te standard deviation, te wider te spread of data items. Grade Educators Guide 8

46 9. B A C 9. Median 59% (read at A Lower quartile 5 (read at B Upper quartile 7% (read at C Mean 5 6,8% (correct to decimal places 9. Mean 6%, median 59% and mode 55 %. Because all tese values are approimate, and te differences are not significant, we can say tat te distribution is fairly symmetric (te data is not skewed Grade Educators Guide 8