Paper 23- Shape- Geometric Reasoning- (26 marks) Question Paper Q1. Diagram NOT accurately drawn

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1 Paper 23- Shape- Geometric Reasoning- (26 marks) Question Paper Q1. ABC and DEF are parallel lines. BEG is a straight line. Angle GEF = 47. Work out the size of the angle marked x. Give reasons for your answer. Diagram NOT accurately drawn (Total for Question is 3 marks) Q2. The diagram shows 3 sides of a regular polygon. Each interior angle of the regular polygon is 140. Work out the number of sides of the regular polygon. Diagram NOT accurately drawn...

2 (Total for Question is 3 marks) Q3. * ABC is parallel to DEF. EBP is a straight line. AB = EB. Angle PBC = 40. Angle AED = x. Work out the value of x. Give a reason for each stage of your working. Q4. (Total for Question is 5 marks) *

3 CDEF is a straight line. AB is parallel to CF. DE = AE. Work out the size of the angle marked x. You must give reasons for your answer. (Total for Question is 4 marks) Q5.

4 The diagram shows a square and 4 regular pentagons. Work out the size of the angle marked x.... (Total for Question is 3 marks) Q6. ABC, PQR and AQD are straight lines. ABC is parallel to PQR.

5 Angle BAQ = 35 Angle BQA = 90 Work out the size of the angle marked x. Give reasons for each stage of your working. x = (Total for Question is 4 marks)

6 Q7. ABCDE and AFGCH are regular pentagons. The two pentagons are the same size. Work out the size of angle EAH. You must show how you got your answer.... (Total for Question is 4 marks)

7 Paper 23- Shape- Geometric Reasoning- (26 marks) Examiners Report Q1. Many candidates gained two marks for finding x = 133 but disappointingly few candidates achieved the third mark for giving correct reasons. Many either wrote no reason at all or, most commonly, just wrote one reason e.g. 'angles on a straight line add up to 180'. Most candidates were not able to give full clear statements with the correct naming of the types of angles. The minimal phrases that were often used were insufficient to gain any credit. Some candidates used the terms 'F angles', 'Z angles' and 'C angles' in their reasons rather than 'corresponding angles', 'alternate angles' and 'co-interior angles'. These terms are not acceptable. A small number of candidates thought that x was 47 despite the fact that it was clearly an obtuse angle. Q2. Where candidates calculated the correct exterior angle, the correct answer usually followed although = 8 was quite common. Some candidates added that the shape was a nonagon. Many candidates chose the less efficient and more error prone strategy of listing multiples of 140 to compare with a list of the multiples of 180. Some did not appreciate that only part of a regular polygon was shown and instead drew horizontal and/or vertical lines to close the shape and form a trapezium or hexagon. Q3. Many candidates were able to score the three marks for finding x = 70. The two marks for giving correct reasons proved more elusive as many candidates simply described the process they had used to reach 70 but failed to give any correct geometrical reasons. Most candidates were not able to give full clear statements with the correct naming of the type of angles used. Some gained one mark for giving at least one correct reason (quite often this was 'opposite angles'). The minimal phrases that were often used, e.g. 'straight line' rather than 'angles on a straight line add up to 180 ' and 'isosceles triangle' instead of 'base angles of an isosceles triangle are equal', were insufficient to gain any credit. Some candidates referred to 'F angles' and 'Z angles' instead of 'corresponding angles' and 'alternate angles' and this is not acceptable. Q4. Candidates' performances on this starred question gave a good differentiation of marks. Only a quarter of candidates gained 1 mark for stating either that angle AED was 38 or that angle AEF was 142 ; a further mark was gained for a correct method to find one of the base angles of isosceles triangle ADE. A large number of candidates realised the triangle was isosceles but then failed to identify the correct pair of equal angles possibly because they thought 'base angles' are those at the bottom of the diagram. Problems arose when candidates had to give their reasons. The most successful candidates were those who wrote their reasons next to the working the reason applied to. It was a pity that only a very small percentage of candidates gained both marks for a correct answer with a full set of reasons, but some gained 1 mark for one correct reason. Many candidates knew the correct

8 reasons but failed to write enough, eg 'angles in a triangle' without stating they add up to 180. Candidates are realising that Z angles will not gain them the mark but often confuse corresponding angles with alternate angles, some resorting to talking about parallel lines. Many candidates failed to score as they did not identify the correct angles in the working, by using correct angle notation or by showing them on the diagram. Q5. Although few candidates gave a fully correct answer to this question, there was much misunderstanding of the relevance of dividing 360 by 5. A small number of candidates found 108 as the interior angle in a regular pentagon but could make no further progress and those who understood the question but showed inaccurate calculations scored 2 marks. It was also clear that many candidates did not use the diagram, as they did not appreciate that the interior angle of a regular pentagon was obtuse and could not be 72. Q6. Nearly all candidates worked within the right angled triangle to find angle ABQ, and most then went on to give angle x as 55 The mark for giving an appropriate reason within the context of the question was not always earned since a geometrical reference had to be precise such as "alternative" or "corresponding". Hence merely stating "parallel lines" or "Z angles" was insufficient. It is always useful to show the angles on the diagram as well as in working.

9 Q7. This question was well attempted by most candidates but many only achieved M2 for correcting calculating the interior angle of a pentagon as 108. More candidates chose use the quadrilateral ABCH to work out the size of angle EAH and they were the most successful. Others used the pentagon AHCDE and provided they calculated the reflex angle at H correctly usually achieved at least M3. Candidates who chose to use the hexagon AFGCDE were the least successful, often failing to realise that the hexagon was not regular. Several weaker candidates assumed the internal angles to be 120 or split the diagram into triangles and labelled their angles 60 assuming them to be equilateral. A common error was to divide 108 by 3 or 72 by 2 which led to the correct answer but was incorrect method so did not achieve full marks.

10 Paper 23- Shape- Geometric Reasoning- (26 marks) Mark Scheme Q1. Working Answer Mark Notes Q M1 for A1 for 133 C1(dep on M1) for full reasons e.g. angles on a straight line add up to 180 and alternate angles are equal OR correspondingangles are equal and angles on a straight line add up to 180 OR vertically opposite angles (or verticallyopposite angles) are equal and allied angles (or cointeriorangles) add up to 180 Working Answer Mark Notes (= 40) 360 "40" 9 3 M1 for (= 40) M1 (dep) for 360 "40" A1 cao Q3. Working Answer Mark Notes Angle ABE = 40 (vertically opposite angles are equal) Angle BAE = angle BEA = (180 40)/2 = 70 (base angles of an isos triangle are equal) x = 70 (alternate angles on parallel lines are equal) OR Angle ABE = 40 (vertically opposite angles are equal) Angle BAE = angle BEA = (180 40)/2 = 70 (base angles of an isosceles triangle are equal) Angle BEF = 40 (corresponding angles are equal) x = = 70 (angles as a straight line add up to 180 ) Q B1 for angle ABE = 40, could be marked on the diagram M1 for (180 '40')/2 (= 70) A1 for 70 identified as the angle x C2 for fully correct reasons: 'vertically opposite angles are equal' or 'vertically opposite angles are equal' 'base angles of an isosceles triangle are equal' 'alternate angles on parallel lines are equal' (C1 for just one correct reason quoted) OR B1 for angle ABE = 40 or angle BEF = 40, could be marked on the diagram M1 for (180 '40')/2 (= 70) A1 for 70 identified as the angle x C2 for fully correct reasons: 'vertically opposite angles are equal' or 'vertically opposite angles are equal' 'base angles of an isosceles triangle are equal' 'corresponding angles on parallel lines are equal' 'angles on a straight line add up to 180' (C1 for just one correct reason quoted)

11 Working Answer Mark Notes Angle AED = 38 alternate angles are equal Angle ADE = (180 38) 2 = 71 x = base angles of an isosceles triangle are equalangles in a triangle add up to 180angles on a straight line sum to 180 OR angle AEF = 142 allied angles/cointeriorangles add up to 180 ADE = = 71 base angles of an isosceles triangle are equalexterior angle of a triangle is equal to the sum of the interior opposite angles, x = angles in a straight line add to 180 OR Angle AED = 38 alternate angles are equal for angles BAE and AED and BAD and ADC (x) Angle DAE= (180 38) 2 = 71 base angles of an isosceles triangle are equalangles in a triangle add up to 180 Or Angle AED = 38 alternate angles are equal Angle ADE = (180 38) 2 = 71 base angles of an isosceles triangle are equal and angles in a triangle sum to 180x = alternate anglesbad and ADC(x) are equal x = B1 for angle AED = 38 or AEF = 142 M1 for a complete method to find one of the base angles of the isosceles triangle C2 (dep M1) for x = 109 with complete reasons (C1 (dep M1) for one reason correctly used and stated) Q5.

12 Answer Mark Notes Q M1 for or 108 seen as the interior angle of a pentagon M1 (dep on previous M1) for '108' 90 A1 for 54 cao OR M1 for 180 (5 2) (= 540) 5 or 108 given as the interior angle of a pentagon M1 (dep on previous M1) for '108' 90 A1 for 54 cao Answer Mark Notes 55 4 M1 for a correct method to find a different angle using 35 M1 for setting up a complete process to calculate angle x A1 cao B1 states one of the following reasons relating to their chosen method: Alternate angles are equal; Corresponding angles are equal; Allied angles / Co-interior angles add up to 180; the exterior angle of a triangle is equal to the sum of the interioropposite angles. Q7 Answer Mark Notes 36 4 M1 for (=72) or (2 5 4) 90 (=540) or (5 2) 180 (=540) M1(dep) for 180 "72" (=108) or (=108) (could be marked on the diagram) M1 for complete method to find angle HAB (360 2 "108") 2 oe or angle EAH + angle HCD 540 ("108" + "108" + (360 "108")) oe or angle EAF 720 ("108" 4) 2 oe A1 cao

Higher Unit 5a topic test

Name: Higher Unit 5a topic test Date: Time: 50 minutes Total marks available: 49 Total marks achieved: Questions Q1. (i) Find the size of the angle marked x.... (ii) Give a reason for your answer.... (Total