Week 3

Identify and apply circle definitions and properties, including: centre, radius, chord, diameter, circumference, tangent, arc, sector, segment.

This is a rather mundane list, though I have to admit that when it comes to 'sector' and 'segment', I still can't say with confidence which is which - though I have learnt that for me 'segment' is definitely not like the segment of an orange.

This week's tasks do not really test students' knowledge of the various terms in the list - I'm assuming they know what the terms mean. Instead, the tasks try to promote a better feel for some of the terms, but without challenging students too deeply. 

It is in Weeks 4, 5, 6, 7 and 8 that we earnestly apply circle properties This later content is based on a heading intended only for the 'more highly attaining students', though I hope it will have much wider appeal.

MONDAY: Here we are effectively being asked to compare the shapes of two segments. Are they similar?

It is quite challenging to compare curved shapes. How, for example, do we compare the 'angles' at, say, A and C? Intuitively, the 'corner' at A seems less sharp than that at C. We can look at this more analytically by thinking of the angles that the tangents at A and C make with their respective chords - do some students approach the task in this way?

In the diagram below we have enlarged the segment formed by the arc CD and chord CD, using a scale factor ×2. As we can see, the result is not congruent to the segment formed by the arc AB and chord AB.

Here's a perhaps more grounded approach (below). M is the midpoint of chord AB, so Q is the midpoint of arc AB and chord AP = chord CD. As we can see, the length of arc AP is less than the length of arc AQ.

Here are four interesting solutions posted on Twitter. The first two have similarities to the grounded approach, above. The third considers a special (and extreme) case, while the fourth uses a powerful, but far from grounded (!), argument.

TUESDAY: Here we do have an enlargement, with a scale factor ×2 and with the centre at the centre of the two circles. So the two arcs are similar, with one twice the length of the other.

WEDNESDAY: Here we are comparing an arc (minor arc, that is) with the chord joining its end-points. It is clear that the arc's length is always greater than that of the chord, but does the relationship change as the arc's length changes?

 

As the arc length gets smaller the arc and chord get closer in length: more precisely, the ratio of the lengths approaches 1. How readily do students appreciate this? As the arc length increases the ratio of arc-length to chord-length increases. Its maximum occurs when the (minor) chord is a semicircle, when of course its length is 𝜋r and the chord's length is 2r. The ratio 𝜋r/2r is about 1.5 which is less than 2.

 

THURSDAY: Here we are asked to compare the area of the non-overlapping parts of a sector and a segment of a circle.

When P is nearer to O than to C, as in the given diagram, it is fairly obvious that the area of V (the pure yellow region) is greater than that of U (the pure green region). But what happens when P is at the midpoint of OC? We can think of the two regions as two 'triangles', with the same base and with the same height. However, the yellow 'triangle' V has a convex side, so its area is greater than that of the green triangle U. As P moves closer to C, the area of the yellow region will decrease while that of the green region increases; at some instant the areas will be equal.

FRIDAY: We can use the diagram in this task to confirm some of our thinking in the previous task: here, point P is at the midpoint of OC and it is fairly clear that yellow region is larger than the green region. However, we now have to consider a third, brown, region. Is this smaller or larger than the yellow region?

It is quite difficult to compare the areas of the yellow and brown regions on a 'purely' visual basis, and it is likely that students will be divided as to which is larger. A useful, more analytical, approach is to divide the regions into triangles - moreover, triangles which are congruent, as in the diagram below. It is now fairly clear that the dark brown region is greater than the dark yellow region, so the brown region as a whole is greater than the yellow region as a whole.

If the radius of the circle is 2 units, then OP = 1 unit. Note also that angle BOP = 60˚. (Why?)

So the area (in unit squares) of triangle BOP = ½ × 1 × √3 = ½√3,
the area of triangle EOC = 4 × area of BOP = 2√3,
and the area of sector BOC = ⅙ of 𝜋2² = ⅔𝜋;

In turn, the area of the green region (G) = ½√3,
the area of the yellow region (Y) = ⅔𝜋 – ½√3,
and the area of the brown region (B) = 2√3 – ⅔𝜋.

In turn, Y – G = ⅔𝜋 – √3, which is about 2 – 1.7 which is greater than 1,
and B – Y = 2½√3 – ⁴⁄₃𝜋, which is about 4.33 – 4.19, which is (slightly) greater than 1.

 

Week 2

{Describe the changes and invariance achieved by combinations of rotations, reflections and translations}

This content, listed in curly brackets, is officially intended only for 'more highly attaining students'. However, it should be accessible, and I hope it will appeal, to a wider range of students. Our particular tasks involve pairs of transformations: two rotations, or a rotation and a translation, or two reflections.

MONDAY: This is a fairly straightforward task, which we build on in later tasks.

Students might well be able to predict that the resultant transformation is a rotation through 180˚, though it is unlikely that they will determine the position of the centre without first drawing the two images of the flag. [The centre is mid-way between P and Q and two units to the right.]

TUESDAY: This is quite a rich and subtle task. The tracing paper might help to convey the idea that when we perform a transformation, we are really transforming the whole plane, not just individual objects. It turns out that the point in the plane that returns to its starting position is the centre of rotation of the half turn that we found in Task 02A. The point that returns to its starting position is in effect an invariant point - there is only one such point in this task, which of course is the case for a rotation, where the only invariant point is the centre of rotation.

It will be interesting to see how students locate the desired point. Do they make lots of trials? Do they manage to get useful feedback from their trials? Do they use the experience gained from Task 02A?

Having found the desired point, it is easy to confirm that it is indeed 'invariant':

WEDNESDAY: Here we reverse the order of the rotations used in the previous two tasks. The resultant rotation will again be through 180˚, but will the centre be in the same place as before?

Student might have difficulty performing the rotations here, as the flag and the image being rotated are further from their centres than in the previous tasks. Also the second image goes off the page. Rather than rotating the flag and its image, do some students look for the centre by adopting the 'invariant point' approach of Task 02B?

THURSDAY: The previous task demonstrates that pairs of rotations are not (necessarily) commutative. The same applies to a rotation and translation, as becomes evident in this task.

 

It will interesting to see how students approach this task. Do they expect the two transformations to be commutative? Do they search for the desired translation by trial and improvement or do they use a more analytic approach by working backwards? If we apply the inverse of R to Flag G, we can see that we can get from Flag F to this intermediate image by a translation of 2 units to the left and 6 units up.

Students might notice that there seems to be a simple relation between the two translations. Can they express it in general terms and explain why it is true?

A nice extension to the task is to find the centre of the single rotation that would map F onto G. A further extension is to express the foot of each flag in coordinate form [(5, 3) and (11, 5), say] and to find a (general) rule for using these coordinates to find the coordinates of the centre.

FRIDAY: This task involves two reflections and is based on the fact that two reflections are equivalent to a rotation about their point of intersection,  through twice the angle between the mirrors. In this case the mirrors intersect at 45˚ and the reflections are equivalent to a 90˚ rotation. However, students don't need to know all this in order to tackle the task. Indeed, without this knowledge the task becomes a rich investigation.

There are infinitely many pairs of mirrors that will map F onto H, but probably the two most obvious ones are these. It will interesting to see whether students come up with these solutions, perhaps through a mixture of analysis, judicious trial and improvement, and a bit of luck.

Week 1

Interpret and use fractional {and negative} scale factors for enlargements

Here we consider enlargements with fractional scale factors. Some of the tasks are (or may appear) more challenging due to the fact that the centre of enlargement lies outside the confines of the diagram. The idea here is to encourage students to think about the properties of enlargement and perhaps be more analytical. Of course, they (or their teacher) can, if they wish, extend the diagram to include the centre.

MONDAY: Here the centre of enlargement lies slightly below and to the left of the given grid. However, a short pause for thought might enable students to see that the desired image can be drawn without recourse to the centre.

The segment joining points A and B is 6 units long. It's image will be ⅔ × 6 = 4 units long. It will also be parallel to AB.

TUESDAY: Here we can again find the desired image without finding/using the centre of enlargement. However, the centre is within the confines of the given diagram this time and some students might find it helpful to find it.

 

The length of A'D' is ⅔ the length of AD, which means that the scale factor is ×⅔ again. In turn, this means that C'D' = ⅔ × CD = ⅔ × 9 = 6 units. So we can now complete the image of the rectangle. However, even if we do this without first finding the centre of enlargement, it is quite nice to confirm that the lines through A and A', B and B', C and C' and D and D' all meet at a point (P, say). It is also nice to confirm that PA' = ⅔ × PA, PB' = ⅔ × PB, etc.

Note: It is highly likely that some students will complete the rectangle like this. The result looks plausible but the rectangles are not similar. One way to show this is to draw lines through the corresponding vertices A and A', B and B', etc. The lines do not all meet at a single point (the putative centre of enlargement).

WEDNESDAY: Here the 6 by 9 rectangle is 'enlarged' to become a 2 by 3 rectangle. So we can draw the two desired images without needing to find the centres. However, it is conceptually helpful to find the centres, to tie things together.

We can find the centres by simply joining selected pairs of object- and image-points. Care needs to be taken for the second centre as it is not a grid point. Students might also be challenged by the rather 'unusual' position of the first centre.

THURSDAY: Here the two desired centres are some way off the page, and it might well be useful to extend the diagram at some stage, if only to check/confirm students' answers.

Students might be able to see that for part i. the distance of A' from A (4 across, 2 up) is half as much as the distance from the centre to A, So the centre will be 8 to the left of A and 4 down from A, and so will have coordinates (2, 5). Alternatively we can make use of this relation (where C is the centre):
1.5 × CA = CA'.

For part ii. the relation (for the new C) is ⅔ × CA = CA', which leads to the finding that C now has coordinates (22, 15). We can show the centres by extending the given diagram. Notice the symmetrical nature of the centres' positions. Will that always happen for pairs of scale factors a/b and b/a?

FRIDAY: To find the desired scale factor in this task, we need to compare corresponding lengths on the object and image. The easiest to read are the heights, which look to be 2 and 5 units long. This suggests the scale factor is ×2½.

In part b), it would be quite challenging to locate the centre of enlargement, which anyway lies outside the confines of the given diagram. However, we can answer part b) without knowing where the centre is. Moving the centre 2 cm to the right means the object is a horizontal distance of 2 cm further from the centre, so the image will be a horizontal distance of 2½ × 2 cm = 5 cm further from the centre's new position. In other words, the image moves 5 cm – 2 cm = 3 cm to the left of its original position.

In the diagram below, we have assumed the grid-lines are 1 cm apart. As can be seen, moving the centre of enlargement 2 cm to the right, and keeping to a scale factor of ×2½, moves the image 3 cm to the left.

 

INTRODUCTION

 

Dietmar Küchemann

This is my second GEOMETRY blog. The first (GeometricSparks) explored the geometry curriculum for students studying the National Curriculum for England and Wales at Key Stage 3 (age 11 - 14 years). This blog explores the geometry curriculum for students at Key Stage 4 (age 14 - 16 years), ie students heading for the GCSE Mathematics Exam.

In referring to the curriculum, I have deliberately used the word explore. What I aim to do is to devise tasks for each item in the content list for Geometry and Measures that appears in the UK Government's 2014 Mathematics Programmes of Study for Key Stage 4 (see below). However, rather than attempting to devise tasks like those in the GCSE exam, my basic aim is to give students (and their teachers) a fresh look at, and perhaps a deeper insight into, the geometry content. As such, the focus is on developing understanding rather than on honing procedures.

In truth, because the KS4 geometry content is so heartbreakingly thin, I have sometimes gone well beyond the National Curriculum requirements and used the content list as a jumping off point for challenging tasks that I hope will engage students. In the process, I hope too that the tasks will help students (and their teachers) see how the meagre KS4 offering can form part of a rich and engaging area of mathematics.

The 'statements of knowledge, skills and understanding' for geometry in the National Curriculum at Key Stage 4 are listed below (from page 4 of the document ncetm_ks3_theme_6.pdf)

My first blog in this series was Algebradabra, a blog for secondary school students and their teachers. This was followed by Algeburble, which focussed on early algebra, and then by MultipliXing, which focussed on multiplicative reasoning. All these have been published in book form by ATM, the Association for Teachers of Mathematics. The GeometricSparks blog came out as an ATM book, in revised form, in October 2022.

 

Dietmar Küchemann