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Posts Tagged ‘Area’

Area 48

July 10, 2019 Leave a comment

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.

Nice area puzzle

April 23, 2019 2 comments

Yesterday evening I came across this lovely area puzzle on twitter:

The puzzle is from Gerry McNally (@mcnally_gerry) he says its his first, and I hope that’s “first of many”.

I reached for the nearest pen and paper and had a quick go:

As you can see, I misread the puzzle originally and thought the lower quadrilateral was a square. The large triangle is isosceles as given in the question. This allowed me to use the properties of similar triangles and the base lengths given to work out the areas of the square, both right angled triangles and the whole triangle. This then allowed me to calculate the area of the shaded quadrilateral and hence that area as a fraction of the whole.

Then I went to tweet my solution to Gerry and realised that nowhere does it say that the bottom quadrilateral is a square. I had added an assumption. This made me ponder the question some more. Instincts told me that it didn’t have to be a square, but that the solution would be the sane whether it was a square or not. But I didn’t want to leave it at that, I wanted to be sure, so I had another go.

I sketched out the triangle again:

This time I called the height of the rectangle x.

This made it trivial to find the area’s of the rectangle and the triangle GCD. Triangle HAB was easy enough to find using similar triangle properties.

and then I found the area of the whole shape again using similarity to discover the height.

This allowed me to find the shaded area:

Then when I put it as a fraction the xs cancelled and it of course reduced to the same answer.

I really like this puzzle, and would be interested to see how you approached it, please let me know in the comments or on social media.

An interesting area puzzle 

July 5, 2017 12 comments

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed.

It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations:

I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject:

Then I needed dw:

Then I needed cx:


I thought “now I need ay”, then realised I had it:

This meant I could.sub back in through the equations:

So the area:

I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker. 

If you did it, I’d love to hear your approach, especially if you spotted something I missed!

An interesting area problem

April 27, 2016 7 comments

Here’s an interesting little question for you:

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Have you worked it out? How long did it take you to see it?

It took me a few seconds at least, I had screenshotted the picture and was reaching for the pencil when the penny dropped, and that’s why I thought it was an interesting question.

The answer is, of course,  100pi. This follows easily from the information you have as the diagonal of the rectangle is clearly a radius – the top left is on the circumference and the bottom right is on the centre.

So why didn’t I spot it immediately?

I think the reason for me not spotting it instantly might be the misdirection in the question, the needless info that the height of the rectangle had me thinking about 6, 8, 10 triangles before I had even discovered what the question was.

I see this in students quite often at exam time, they can get confused about what they’re doing and it links to this piece I wrote earlier about analogy mistakes. The difference is I wasn’t constrained by my first instinct but all too often students can be and they can worry that it must be solved in the manner they first thought of.

Earlier today a student was working on an FP1 paper and he was struggling with a parabola question, he had done exactly this, he had assumed one thing which wasn’t the right way and got hung up in it. When he showed me the problem my instinct was the same as his, but when I hit the same dead end he had I stepped back and said “what else do we know”, then saw the right answer. I’m hoping that by seeing me do this he will realise that first instincts aren’t always correct.

I’m going to try this puzzle on all my classes tomorrow and Friday and see if they can manage it!

How quickly did you see the answer? Do you experience this sort of thinking from your students? I’d love to hear any similar experiences.

Cross-posted to Betterqs here.

A surprising find

April 18, 2015 3 comments

The other day I my timehop showed me this lovely little post from last year. It includes “Heron’s Formula” for calculating the area of a triangle, as I read it I remembered thinking it was a little strange that not many people had heard of it before.

Today I was looking through a number of textbooks trying to find a decent set of questions on area, perimeter and volume for my year nines as I wanted to consolidate their learning at the start then move onto surface area. I’m not a fan of textbook misuse- ie “copy the example and try the questions” but I do sometimes use them for exercises as we have a very limited printing budget and some of them have superb exercises. For a fuller picture on.my view of textbooks, read this.

I was looking in one of my favourite textbooks:

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And I happened across this:

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There it is! Plain as day! Heron’s Formula! In a KS3 textbook!

I was disappointed that its function was described and its name wasn’t and there was no mention of why this worked. It basically reduces the question down from a geometry one to a purely algebraic substitution task and I would question the appropriateness of including it in an exercise on area, but still, I was incredibly exciting to find it there!

Are you a fan of Heron’s Formula? Had you even heard of it? Do you have a favourite textbook? I’d love to hear your views.

Area Puzzle – Squares and Circles

January 31, 2015 1 comment

This morning I came across this puzzle from Ed Southall (@solvemymaths):

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And it seemed interesting. Looking at it is seems that A is the midpoint of the arc, so equidistant from b and the lower left corner of the square. This would mean that an isosceles triangle exists in the semi-circle, which in turn implies the square has side length 4rt2 (from Pythagoras’s Theorem, as the side of the square is a diameter).

This means the square has area 32cm^2. The semi-circles have radius 2rt2 and as such have area 4pi. That leaves just the white circle. The diagonal of the square is a diameter of this, and as the side lengths are 4rt2 the diagonal must be 8 (again via Pythagoras’s Theorem). This means the circle must have a radius of 4 and hence an area of 16pi.

So the sum of the orange areas must equal the square add the 4 Semi-Circles subtract the circle add the square.

Or:

2(32) + 4(4pi) – 16pi

Which simplifies to 64cm^2.

A lovely solution, and one which shows us that the areas of the orange crescents equal the area of the square.

Towards the end of the working I realised I could have used another property of Pythagoras’s Theorem, namely that the sum of the areas of semi-circles on the two shorter sides equals the area of the semi-circle on the hypotenuse. By splitting the square into two right angled triangles I could have reasoned that the 4 smaller semi-circles provide the same area as the large circle. Which means we would get the total area to be the area of the square, add the area of the circle (from the semi-circles) subtract the area of the circle (from the circle) add the area of the square. Which again simplifies to twice the area if the square. A much more elegant solution.

A Puzzling Heptagon

January 4, 2015 Leave a comment

At somepoint Ed Southall (@solvemymaths) posted this heptagon puzzle:

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I saved it in my phone to solve later, and forgot about it until the other day.

At first I looked at it and wasn’t sure where to start. I attacked the problem by sketching what I knew:

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The heptagon is split along the base of the red shape into an isosceles trapezium and a pentagon. Because I know the interior angle of a heptagon (900/7) I know enough about this trapezium to work out the other angles (51 3/7).

This means I can deduce the angle BAE (77 1/7):

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Because the heptagon is regular I know that AE = BE, and as such ABE = BAE. This means I can use the angle sum of a triangle and the sine rule to calculate each angle and each side of the triangle.

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From here I could calculate the area of the triangle ABE either using absin(c)/2 or Heron’s Formula. I chose Heron’s Formula:

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I know that triangles ABE and ADE are congruent so the area I’m looking for is double this area subtract the area of the overlap.

I considered the Pentagon above the line AE and how the triangles split it up into 4 triangles and a rhombus. I briefly considered the rhombus:

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And quickly realised that using the rhombus properties and opposite angles I had enough information to calculate the area of the overlap:

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First calculating the height, then the area:

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Then I could find the area I was looking for:

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I thought this was a great puzzle, and it got me thinking about which of my students would be able to attack it. The skills needed are all skills that higher level GCSE students should have, and I think that some of my year ten class would give it a good go, but I also worry that some of my sixth formers may struggle with it. I think by exposing students to these problems early, and by sharing our own thought processes, we can start to build the resilience and mathematical thinking needed to succeed.

When I first attempted this problem it was late, and I made some daft errors transfering working from one line to another, and got a very wrong answer, this showed me that it’s easy to do, and we need to make sure we reiterate often the importance of checking our work to our students.

There was also the problem of rounding, as we were dealing with angles that didn’t give us nice exact trig ratios I had to round, through my working I rounded where it seemed sensible, but all these errors would have built up so I decided 1dp would be a good limit to round my final answer to, although now I feel nearest whole number would have been better.

I’m fairly sure that there is a much better and more concise way of solving this problem, but I currently can’t see it. If you do spot one, or solve it an even more long winded way, please let me know.

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