Week 17

Apply the concepts of congruence and similarity, including the relationships between lengths, {area and volumes} in similar figures (2) 

This is the second of two weekly sets of tasks looking at (congruence and) similarity. The approach is quantitative this time in that we use our knowledge about similar shapes to calculate lengths or areas. [A task that involves the relation between linear scale factor and volume scale factor can be found in Week 15: Task 15C]

MONDAY: Here we use the fact that the ratio of the areas of similar shapes is the square of the linear scale factor. The challenge is to find the necessary linear scale factors.

The area of triangle A (and B) is 1 cm². If we think of the base of triangle A as the side of length 1 cm, then the corresponding bases of triangles C and D are 2 cm and √(1² + 2²) = √5 cm long. So their areas are 2² times and √5² times the area of triangle A, which gives areas of 4 cm² and 5 cm².

We can confirm that 4 copies of triangle A will exactly fill triangle C, with a diagram like the one below. It is also clear that more than 4 copies are needed to fill triangle D, though it is not immediately obvious that we need exactly 5!

TUESDAY: This task is reminiscent of Task 16C. The key is to realise that P is the point such that angle ACP is equal to angle ABC. The task allows students to confirm, and perhaps become more familiar with, the relation (linear scale factor)² = area scale factor. However, it is possible that students get too immersed in the numerical detail.

Point P has coordinates (6, 1). We can confirm that angles ACP and ABC are equal by, for example, considering their tangents or, in a more grounded way, by considering the slopes of CP and BC (see the diagram, below): we can get from C to P in steps of 3-vertical, 2-horizontal and from B to C in steps of 3-horizontal, 2-vertical.

The areas of the triangles are ½×9×6 = 27 unit squares and ½×6×4 = 12 unit squares.
Their ratio is 27 : 12 = 9 : 4 = 3² : 2².

BC = √(9² + 6²) = √117 = 3√13 units and CP = √(6² + 4²) = 2√13 units.

BC : CP = 3 : 2 (we can of course  confirm this by finding the ratio of any other pair of the triangles' corresponding sides, giving the ratios 9 : 6 or 6 : 4.)

If we square the ratio (3 : 2) of corresponding sides, we get 3² : 2², the ratio of the areas.

WEDNESDAY: Here we find the linear scale factor from the area scale factor.

The principles behind this task are simple enough but students might find it difficult to coordinate the relevant information. Thus it might help to separate out and re-orientate the similar triangles ABC and CBP, as in the diagram below.

The area of ABC is 9 times the area of CBP so its sides are √9 = 3 times as long. So AB = 3×6 = 18 cm.

We can find the length of BP in two ways. Using within triangle ratios, we can state that BP : CB = BC : AB = 6 : 18 = 1 : 3, and so BP = 2 cm. Or we can make use of areas. If we refer back to the original diagram, and think of AB and BP as the bases of our similar triangles, then they have the same height, h cm. In turn we can state that ½×18×h = 45, so h = 5, and so ½×BP×5 = 5 and so BP = 2 cm.

THURSDAY: In this task we are given the area of a trapezium and asked to find the area of another. It is tempting to think these trapezia are similar, since corresponding angles are equal. However, on closer examination it is pretty clear that they are not! Are there shapes that we can use that are similar?

The key to this task is that there are three similar triangles in the diagram, two of which contain the trapezia.
The scale factor that maps triangle ABC onto ADE is (8+2)/8 = 1¼.
So the area scale factor is (1¼)² = ²⁵⁄₁₆.
And so we can write ²⁵⁄₁₆ × Area ABC = Area ABC + 9 cm²,
so ⁹⁄₁₆ × Area ABC = 9 cm²,
and so Area ABC = 16 cm².
We can now show that Area AFG = (14/10)² × 25 cm² = 49 cm²,
and so Area DFGE = 49 – 25 cm² = 24 cm².

FRIDAY: This is a somewhat quirky task. Its intrigue (for students who are fairly fluent in manipulating algebraic symbols) lies in the fact that some of the desired formulae turn out to be quite simple.

a) Face U has dimensions a and b. The corresponding dimensions for the similar face V are b and c. So we can write a : b = b : c (note that this works whether we think of the ratios as within the shapes or between them). In turn we can write c = b²/a and Area of W = ab²/a = b².

b) If the area of face U is 20 unit squares, we can write a = 20/b.
So we can write this for the area of face V: bc = bb²/a =bb²b/20 = b⁴/20.

The cuboid looks like this (using an isometric projection) for selected values of b: