# AQA Further Maths question

This isn’t new mathematics; instead it is a question taken from the AQA Level 2 Further Mathematics qualification (like a GCSE), which I will be sitting next month here in England:

How they want you to do it (gets the marks):
Setting y = 0 in 2x + y = 8 yields x = 4, so A is at (4, 0)
Setting x = 0 in 2x + y = 8 yields y = 8, so B is at (0, 8)
AB : BC = 2 : 1, so C is at (-2, 12)
EC : CD = 1 : 2 so E is at (-3, 0) and D is at (0, 36)
So area of triangle AEC = 1/2 × (4 + 3) × 12 = 42
Area of triangle BCD = 1/2 × (36 – 8) × 2 = 28
So area of triangle AEC : area of triangle BCD = 42 : 28 = 3 : 2
OK, it’s not that bad in terms of the algebraic fortitude required, but it doesn’t excite me (funnily enough).

However, the question actually tells you more than you need to know.
I solved it in a different way because I saw it faster than the method above.
It actually took me quite a long time to work through the coordinate method … maybe that’s just me …

The way I did it (probably wouldn’t get any marks):
I don’t really want to work with the line equation.
I’ll work with just AB : BC = 2 : 1 and EC : CD = 1 : 2
Now observe the following fact:
If two triangles have the same perpendicular height, then the ratio of their areas is the same as the ratio of their bases.

So, let’s apply this:

Let [XYZ] denote the area of a triangle XYZ.
[AEC] : [ACD] = EC : CD = 1 : 2
[ACD] : [BCD] = AC : BC = 3 : 1
Write [AEC]  = 3n; then [ACD] = 6n so [BCD] = 2n
So [AEC] : [BCD] = 3 : 2
Now that is more to my taste.
It’s just a shame that the examiner probably wouldn’t be impressed …