Apply the concepts of congruence and similarity, including the relationships between lengths, {area and volumes} in similar figures (1)
This
is the first of two weekly sets of tasks looking at congruence and similarity. The approach here is qualitative rather than quantitative in that the focus is on whether or not figures are congruent or similar, rather than on using such knowledge to calculate lengths, areas or volumes.
MONDAY: This task is quite challenging because one of the edges is curved in each shape.
If we flipped the smaller shapes so they overlapped their larger companion as in the diagram below, then if the pairs of shapes are similar the larger shape would be an enlargement of its companion. Is that the case? Are the curved edges 'parallel'? In a sense they are as they are arcs of concentric circles, but in the case of the yellow shapes it is not clear how points on the two arcs 'correspond'. Moreover, if we consider the ratio of corresponding straight edges, then it is clear that XA/XA' is not equal to XB/XB'. We can also argue that the corresponding 'angles' in the yellow 'triangles' are not equal. For example, the tangent to the smaller circle at A makes a different angle with XA' than the tangent to the larger circle at A'.
TUESDAY: This is quite a complex task, made more challenging by the fact that we don't know enough about the lengths of the shapes' sides to be able to say with certainty that a pair of shapes are congruent or similar. On the other hand, the marked angles allow us to say with certainty that some pairs of shapes are not congruent or similar.
The only two shapes that look as though they might well be congruent are C and D.
Shapes A and B look as though they might be similar, and E might well be similar to the congruent pair C and D.
WEDNESDAY: A necessary condition for two shapes to be similar is that corresponding angles are equal. For triangles this is a sufficient condition; moreover, if two pairs of corresponding angles are equal, then all three will be equal.
Triangles ACE and ECB are also similar. Some students might choose the pair of triangles ECB and ECD as well, but this would only be true if the angles at C were right angles.
THURSDAY: This is a very subtle task and quite complex....
A powerful way of solving this task is to use the notion of enlargement. As the pairs of L-shapes have the same orientation (corresponding sides are parallel), one shape can be mapped onto the other by an enlargement if the shapes are similar. Moreover, the centre of enlargement will have to be at the top right vertex of the rectangle that envelops the two Ls (since the top edges of the two Ls lie on the top edge of the rectangle and their right-hand edges lie on the right-hand edge of the rectangle).
In the first diagram, both shapes look like 'proper' Ls, in that for each L the horizontal and vertical 'limbs' are the same thickness. However, the Ls are not similar because the limbs of the green L are the same thickness as those of the yellow L, and so would become thinker if we enlarged it to achieve the same height, say, as the yellow L. Note also that if we joined corresponding inner corners of the Ls, the line (shown in red, below) does not go through the top-right corner of the enveloping rectangle.
For the second pair of Ls, the equivalent red line does go through the top-right corner, but the Ls are still not similar. The scale factor required to map the inner edges of the green L onto the yellow L does not map its outer edges onto the outer edges of the yellow L (see the shape with the green outline, below).
It looks as though the third pairs of L-shapes might well be similar: the red line (below) goes through the top-right corner of the enveloping rectangle and the limbs of the green L are thinner than the corresponding limbs of the yellow L.
The last pair of L-shapes are not similar, even though the limbs of the green L are thinner than the corresponding limbs of the yellow L. If we enlarge the green L so its limbs are the same thickness as those of the yellow L, the shapes don't fully overlap (below).
FRIDAY: This task again involves L-shapes but the shapes are much simpler and easier to compare. In Thursday's task, a necessary (but not sufficient) condition for two L-shapes to be similar was that for each individual L, the shape formed by the inner two edges had to be similar to the shape formed by the corresponding two outer edges. Here we have reduced the task to just comparing such pairs of rectilinear shapes.
So, for example, the first pair of L-shapes below are formed from the two inner and the corresponding two outer edges of the first yellow L-shape in Thursday's task.
The task can help students achieve greater clarity if they struggled with aspects of Thursday's task.
It turns out that all the pairs consist of similar L-shapes, except the first, where a line through the corner of each L-shape (shown in red, below) does not go through the top-right corner of the enveloping rectangle.
The last two pairs are easy to assess, as the L-shapes together with the slanting line form triangles with equal corresponding angles. If one thinks in terms of enlargement, the centre is where the line through the corners meets the slanting line.
