Posts Tagged ‘Triangle’

4 squares and a rectangle

May 14, 2020 7 comments

Today’s puzzle comes from Catriona Shearer (@cshearer41):

It’s one that gave me a little trouble early on but finished up with a lovely solution. Have a go before you read on.

Here’s one of a few aborted efforts:

So my solution started, as usual, with a sketch. I worked on my sketch and came up with this:

Which proved to be the key. But I realise this is not a very easy to follow solution so I’ll try to explain a bit. Here is a second diagram, this time I have included some vertex labels:

Starting with triangle ABF I labelled side FB as a and FA as b, I know the side AB is 1 aa it is a side of a square of area 1. Angle BFA is a right angle so if angle FAB is x then angle ABF IS 90 – x. As angle ABC is 90 and FG is a straight line that means angle DBG is also x (180 – 90 – (90 – x)).

This means triangles ABF and BDG are similar (both have one angle 90, one x and the other 90-x). As the hypotenuse of BDG is 2 and the hypotenuse of ABF is 1 (from the side length of the squares being 1) then we know that GB is 2b, and GD is 2a. At this point I considered reflecting DBG to get BED but decided against that root.

I worked out that angle JDI (where J is directly above I such that JI and GK are perpendicular) is x. (Again 180 – 90 – (90- x)). IJD is 90 by design and the hypotenuse DI is 2 hence triangle IJD is congruent to BGD which means IJ is 2a.

However, angle JIK is also x (HIK is 90, HIJ is 90- x so JIK is 90 – (90-x)). IJK is 90 and the hypotenuse IK is 1 so triangle IJK is congruent to triangle ABF. So IJ is b.

This means as IJ is 2a, but IJ is also b we can say that b = 2a.

Back our original diagram now. Using these similar and congruent triangle I can see that the vertical side of the rectangle is a + 2b, and the horizontal one is 3a + 2b.

Subbing b = 2a in we get the horizontal side to be 7a and the vertical side to be 5a. So the area is 35a.

Almost there. We also have a number of right angled triangles we could use for the next be I chose the top left one BGD in the other diagram.

By Pythagoras’s Theorem we can see that a^2 is 1/5.

So the area is 7.

I think this is a lovely solution to a problem I very much enjoyed wrestling with. I imagine there are plenty of other approaches that you could use to solve it. Please let me know your solution either in the comments, by email or via social media.

Rectangles and RATS.

April 24, 2020 Leave a comment

Today’s post comes from Ed Southall (@edsouthall) and is one I found fun to think about. Here it is:

Have a go at it if you haven’t already.

Here’s how I approached it. Firstly I had a brief lapse of sense and thought I needed to check if the pick rectangles were congruent, but realised they have to be for the diagram to make sense. Then I drew it out:

I realised that the whole thing was symmetrical and as such I could reduce the problem to looking at this:

It becomes a square split into a rectangle and two pairs of congruent triangles, each of which are similar to the other. All triangles are isoceles and right angled (this is known by the fact the angle between the 2 pink rectangles is a right angle as Ed’s second tweet states).

I set the side length of the smaller triangles as x and used Pythagoras’s theorem and rules of similar shapes to fill in the gaps:

This made finding the area simple:

And also finding the fraction simple:

I think this is a lovely problem and a nice concise solution. Obviously given the xs will cancel I could have used an arbitrary value rather than x, but I sometimes like to work algebraically and show that it works whatever side length you assign.

Afterwards I considered whether there is a better more concise solution, but I’ve yet to come up with one. If you have one please let me know. I would love to hear how you approached it anyway. Please let me know in the comments or on social media.

More triangle puzzles

April 23, 2020 Leave a comment

Today’s puzzle is another brilliant puzzle from the brain and pen of Catriona Shearer (@cshearer41):

If you haven’t done it, have a crack.

I tried to do it in my head first, but I came up with the answer 1/9, and decided that that really couldn’t be right, so dug out the pen and paper. I started, got half way through, was working from my own sketch not the initial problem and decided I couldn’t assume it was an equilateral triangle, only an isoceles so started again. Then I looked at the question again and realised it wouldn’t be an assumption as its given information!

Here’s my sketch, as you can see I changed the orientation, I’m not really sure why I died that. Perhaps I just must remembered the diagram and this is a more familiar orientation to me.

I worked with radius r, again I’m not sure why, they are all going to cancel anyway so I could just have easily used any arbitrary value, but in this instance I went with r.

I decided that the best approach would probably be to find a length scale factor and square it. And I decided that as I had the height of the trapezium (2r) that the perpendicular height of the triangles would be the best things to find to get my scale factor. Using some basic trig and properties of equilateral triangles it was easy enough to work out a side length of the larger triangle, (2rt3 + 2)r.

From there it was a simple piece of trig to find the perpendicular height of the large triangle, and to subtract 2r from it to find the perpendicular height of the small triangle. (Ed. I appear to have dropped the r early in my scribbles, apologies, it should appear in both the perpendicular heights and also in the fraction, where the r in the numerator and the denominator would cancel).

I made that a fraction and simplified it, using some surds knowledge. Then squared my answer:

At this point I realised that while doing it in my head I was probably losing a root somewhere in the surd simplification and that’s why I kept getting 1/9.

I really like this puzzle. It uses a number of GCSE topics (trig, surds, scale factors, circle theorems etc) and combines them. It’s the sort of question I like to use with the top end in year 10 and 11 to see how they get on. I also use them as starters with the sixth formers I teach to get their minds working. I’m going to set it for some students as part of remote working to see if they can manage it.

How did you approach it? I’m sure that there are other solutions and I may think about them further, but I’d love to see and hear about your methods. Please let me know in the comments or via social media.

A nice little triangle puzzle

December 5, 2016 8 comments

Here is a nice little puzzle I saw from on Facebook.

Have you worked it out yet?

Here’s what I did:

First I drew a diagram (obviously).

And worked out the area of the triangle.

Then the area of each sector.

Leaving a subtraction to finish.

Categories: #MTBoS, Maths, Starters Tags: , ,

Exact Trigonometric Ratios

November 8, 2015 4 comments

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).


Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.


Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.

UKMT Maths Challenge Q16

November 7, 2014 1 comment


Above is a photo of question 16 from yesterday’s UKMT Senior Maths Challenge. And what I fantastic question it is!

It shows (incase you can’t make it out on the picture) a rectangle and a circle which have the same centre. The two shorter sides of the rectangle are tangents to the circle, and the rectangle is 6×12. The task is to find the area that is inside both.

After the challenge had been completed one if my Y13s came to ask about it, and it took me a little while to work it out. I played around with some similar triangles, thinking that I would need to calculate the bit of the rectangle outside the circle and subtract them from 72. Then I had the realisation that I could easily split it up into triangles and sectors that I could calculate the area of with great ease.

First I sketched:


Showing the rectangle split into areas and sectors. I then considered the triangles, if I split then into right angled triangles I had a hypotenuse of 6 and a height of 3, so the cosine of the top angle must be 0.5, which means the angle must be pi/3.


I then used the sine of this angle ((rt3)/2) to calculate the opposite side as 3rt3. Then thought “why didn’t I use Pythagoras’s Theorem?!” A quick check using Pythagoras’s Theorem shows us the side is rt27, which simplifies to 3rt3. The area of the triangle then must be 9rt3.


Knowing that the angles at the centre in the two triangles are 2pi/3, that angles round a point add up to 2pi and that the sectors are congruent means we can deduce the angle if each sector is pi/3. We also know the area of a sector is (r^2(theta))/2, which gives the area of each sector as 6pi.


Finally we sum the areas and get 18rt3 + 12pi.

The student had looked at the answers and estimated them using the knowledge that rt3 is around 1.7 and pick is a but more than three. This meant he could eliminate c,d and e, but he didn’t get to the final answer.

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