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## Area 48

Today I was looking at some of Ed Southall’s (@solvemymaths) puzzles on his website. I saw this one that I had not seen before:

I thought I’d give it a crack. You should too….. go on…. Did you get an answer? Well here is how I approached it:

First I did a little sketch, as I always tell my students to do:

I labelled the points with letters as this is normally quite a good way of keeping track of things.

I then decided to let AB = 1 (I chose that bit to be 1 as I knew a unit square would lead to lots of fractions, in hindsight this also let to fractions and AB = 2 would have been better.)

This gave me a few lengths straight off the bat, and I could find BD by Pythagoras’s Theorem and hence had the area of the larger square – which I need to answer the question.

I also noticed I had a RAT (ABD) and I knew the perpendicular sides, and therefore could work out the area.

I then looked at the triangle BCH. This looked like it would be similar to ABD but I took a couple of moments to justify it to myself before moving on, just in case….

If angle ABD is x then as DAB and BCH are both 90 and the angle sums of a triangle and on a straight line are both 180 then CBH and BDA must both equal 90 – x and CHB must equal x, hence they are similar.

They are similar and the scale factor is 2 (as BC is half of AD and they are corresponding sides).

Hence the Area scale factor is 4 and the area pf BCH is a quarter of the area ABD. As Area ABD = 1 then Area BCH = ¼.

From here I took the area of the two triangles away from the area of the square ACED to get the shaded area and put it over the area of the larger square. (Well, after momentarily putting it over the area of the smaller square like a fool!).

So here I had an answer, 11/20. I clicked on the comments on Ed’s website and saw some answers that were not what I had. This had me second guessing myself, so I thought about a different approach.

I went for a coordinate geometry approach (coordinate geometry seems to have taken over from trig as my brains go to method).

I chose the origin as the common corner of the two squares and called the point where the vertex meets the horizontal point B. This mean B’s coordinates were (1,2). I called this line l1 and could spot its equation was 2x. Part of the shaded area is the area under this curve between x = 0 and x = 1 so I calculated that area to be 1.

The perpendicular through B is the other line that bounds the top of our shaded region. I know the perpendicular gradiens multiply to -1 and I know it goes through point (1,2) so I could work out the equation of this line easy enough:

Then calculate the area below it between the values x = 1 and x = 2. This gave an area of 7/4.

So I had a total shaded area of 11/4 and could divide this by the area of the large square to get 11/20 again.

I felt happier now that I had the same answer though two different methods, and I stress to my students that this is what they should be doing with any extra time in exams. Doing different methods and seeing if they get the same answer!

I hope you tried Ed’s puzzle, and if you did, please let me know how you approached it.

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