Introduction

Teachers notes KS1 KS2 KS3 KS4 Higher Shape These units are designed to make it easy for a pupil to find their way to some mathematics - but not so easy for them to find this page! The teacher may be sitting one-to-one with the pupil, or allowing groups of pupils to work alone. If the teacher has the appropriate equipment - a large projected computer screen - they could use these materials with the whole class. This page contains suggested tasks and further questions you could ask the pupils on each topic. Try to introduce and use words from the listed vocabulary - which are highlighted here in blue the first time they are mentioned. The emphasis is on properties of lines and angles, particularly in relation to circles. Interactive exercise will use interactive buttons. See the page on buttons. These notes - and the units themselves - are still in an early stage of development. Any comments would be appreciated - particularly from KS4 teachers. Overall there are 17 topics arranged in four sections: KS1 1. Recognising 2. Describing 3. Creating KS2 4. Lines 5. Angles 6. Measures 7. Classifying KS3 & KS4 Foundation 8. Angles & lines 9. Polygons 10. Circles 11. Constructing KS4 Higher 12. Angles and lines 13. Circle theorems 14. Circles and angles 15. Congruence 16. Locus 17. Think!	12. Angles & lines >> a1 This activity is entended to ensure that students understand the terms and what they mean. a2 The angles are all either 30° or 150°. a3 In the blue triangle the angles are 70°, and in the green they are 75° a4 The green triangle is equilateral so its interior angles are all 60°. The other green interior angles are 30° and 60°. In the blue diagram all angles are either 75° or 30°. b1 All angles are either 50° or 80°. b2 Angle BDC=38°, ADB=52°. b3 45° and 135°. b4 ACB=64°, ACD=52°, DCF=64°, ABE=116° c1 42°, 48° and 90°. c2 The interior and exterior angles are 66° and 114°. c3 36° and 108°. c4 The interior angles of the parallelogram iteself are 74° and 106°.
	13. Circle theorems >> Six theorems are presented. First the student has to realise what the property is, then they are given 3 steps which should help them to understand (or prove) why this property holds. a1 The angle in a semicircle is a right angle. a2 Triangle ABD is isosceles therefore the two marked angles are equal. a3 Triangle BDC is isosceles therefore the two marked angles are equal. a4 In triangle ADC the four marked angles add up to 180°. Therefore the two at D must add up to 90° b1 The angle subtended at the centre of a circle by a chord is twice the angle subtended at the circumference by the same chord. b2 The red angle equals the sum of the other two yellow angles (which are identical). b3 The blue angle equals the sum of the other two green angles (which are identical). b4 Red and blue together must equal twice the sum of the yellow and green. c1 Angles in the same segment are equal. c2 Red is half the blue c3 Yellow is half the blue c4 Red and yellow are both half the blue and so equal. d1 The opposite angles of a cyclic quadrilateral add up to 180° d2 Because of Circle theorem 3! d3 Red + blue + yellow + green=180° d4 Look at opposite angles: red + blue + yellow + green=180° e1 The tangents from a point outside a circle to a circle are equal in length. e2 They are both 90° e3 They are equal (radii). e4 Triangles ADB and ACB are identical, so DB=CB. f1 The angle between a chord and a tangent is equal to the angle in the alternate segment. f2 It is 90° - see Circle theorem 1 f3 Green + red=90° f4 Green + blue=90°, so blue=red which equals yellow.
	14. Circles and angles >> These problems assume familiarity with the 6 Circle theorems above. a1 Green triangle angles: 90° and 65°; blue triangle angles: 90° and 72°. a2 Blue triangle angles: 88°, 46° and 46°; purple triangle angles: 20°, 20° and 140°. a3 Blue angle=30°, green angle=43°. a4 Yellow angle=20°, red angle=50°. b1 Yellow angle=63°. b2 They are 45°. b3 B=120°, C=75°, F=140°. b4 40°. c1 They are all 70°. c2 27.5° and 125°. c3 Angle ABD=angle BDC (alternate angles). Angle BDC=angle BAC (angles in same segment). Therefore angle ABC=angle BAC and the orange triangle is isosceles. c4 Angle BAD=30°, therefore angle BCD=30°. Therefore angle ACD=30+60=90° and so the line AD is a diameter.
	15. Congruence >> The concept of congruence (or identical) is suggested through activities involving constructing triangles. In every case, the student should be encouraged to understand whether the construction can be made in more than one way. Where appropriate the conventional descriptions SSS, SAS, SSA or RHS are referenced. All these tasks are interactive exercises a1 SSS a2 SAS a3 ASA a4 RHS b1 ASS b2 SSA. The diagram given uses the "angle in a segment" circle theorem. There are two triangles possible for every set of values for AB, BC and angle C. b3 SSS. b4 SSS. c1 The interactive geometry will not respond other than to measure lengths and move points. The congruent triangles are the blue and pink ones (SSS). c2 The brown and orange triangles are congruent (ASA). c3 Use the parallel line button (ASA). c4 Use the compass to measure the sides of the green triangle and transfer them to the new triangle (SSS).
	16. Locus >> Four loci are presented. Each one is introduced in three stages. The final task asked the student to construct one point on this locus. a1 The locus of points a fixed distance from a given fixed point. a2 is... a3 ... a circle, with radius equal to that fixed distance. a4 Interactive exercise. Use the compass to copy the length AB to O and thus construct a circle. b1 The locus of points equidistant from two given fixed points. b2 is... b3 ...a straight line, the perpendicular bisector of the line joining the two points. b4 Interactive exercise. Use the compass to construct the perpendicular bisector, then put a point on this line. c1 The locus of points equidistant from two given fixed lines. c2 is... c3 a straight line that bisects the angle between the two fixed lines. c4 Interactive exercise. Follow the instructions given. Use the compass to construct point Y. d1 The locus of points equidistant from a given fixed point and a given fixed line. d2 is... d3 ... a parabola! (a curve whose equation is quadratic) d4 Interactive exercise. Use the compass to construct the perpendicular bisector of OX.
	17. Think! >> Problems about area which are "easily" solved if approached in the right way. a1 The area is maximised when M and N are positioned at top and bottom left. The area is then 0.5 x 8 x 12 = 48 cm² a2 As long as x is between 0 and 6, the green area is constant and equal to 0.5 x 6 x 12 = 36 cm². This is most easily seen by making x zero. a3 This problem appeared as a GCSE question some years ago. It was assumed that students would find the areas of the two circles and subtract them, and then use the fact that r²+10²=R². A simpler approach is to make r zero and then it can be seen that the area = 3.14159 x 10² a4 This problem appears in a recent GCSE Higher text book, where it is assumed that the four areas will be calculated in terms of x and y, and then added together, whereupon the x's and y's cancel, leaving the area = 0.5 x 5.6 x 9.4 = 26.32. A simpler appreach is to make both x and y zero. b1 The diagram shows clearly that the yellow square is one fifth of the total area of the cross of five squares. b2 The diagram shows that the longest side of the triangle, c, can be expressed as a-r + b-r, hence c = a+b-2r. b3 The area of the largest square is the same as the total area of the other two, as can be seen by comparing the individual parts of each square. b4 The area of the two triangles together is ab; the area of the rectangle below is r x (a + b + c). c1 Rotate each triangle 90 degrees anti-clockwise and observe that it must have the same area as the central triangle. c2 The grid shows the tesselation pattern is based on a square grid. c3 Each vertex of the cube is at the centre of a circle. c4 Each marked angle is equal, which implies that arc C'C = arc D'D and arc C'D' = arc CD. Therefore C'D' = CD.