# Playing With Pythagoras and Trigonometry

During my first few years of teaching, I may have overused “ladder problems” for Pythagoras and trigonometry. I had other problem-solving questions involving splitting isosceles triangles in half, diagonals of rectangles, bearings, angles of elevation and other worded problems. But they were fairly predictable (especially if used during a lesson about Pythagoras or trigonometry) in that they all asked students to find a missing side or angle for a triangle that was somewhat hidden in the question. I used to find myself exhausting my collections of different setups for these sorts problems quickly. So I was keen to build up my repertoire of novel ways to disguise right-angled triangles for Pythagoras and trigonometry problems and also to find other interesting questions that might encourage students to think a little deeper about these two topics.

Since then, I’ve drawn inspiration from colleagues, textbooks, online chat and other resources to spark new ideas for getting students reasoning and problem-solving with Pythagoras and trigonometry in a variety of ways. I’d like to use this post to share a selection of my favourites. Some of them are sets of problems that focus around a particular theme (e.g. right-angled isosceles triangles), while others are ideas that just try to get students to consider the concepts in slightly different ways.

Applying The Formulae “Backwards”

In a previous post, I talked about working forwards and backwards through concepts and gave an example like the one below. In this case, rather than asking students to use Pythagoras’ Theorem to calculate the length of a missing side, the question asks them to decide whether the triangle contains any right angles.

This sort of problem could be extended further by asking students to determine whether the largest angle is acute, obtuse or a right-angle. I find this encourages students to think hard about the relationship between the lengths on the triangle and the structure of the formula in order to decide if the angle is acute or obtuse. For example, if the sum of the squares of the two shorter sides is greater than the square of the hypotenuse, an argument often breaks out about whether that means the largest angle is greater than 90 or less than 90.

Another situation where it can be interesting to “work backwards” in a similar way is with trigonometry. For example, rather than using sin to calculate a missing length or angle, a question could give the measurements for two sides and an angle then ask if the triangle is right-angled.

A Trojan Horse of Trigonometry

Craig Barton makes a great point in his book (2017, p.416) that suggests when students see problems that require Pythagoras’ Theorem in a lesson that is about Pythagoras, then they already know that there’s a good chance they’ll need to use Pythagoras’ Theorem before they even read the question. The same goes for other topics, such as trigonometry problems in a trigonometry lesson. The hardest thing about solving problems like these outside of their lessons is spotting the right-angled triangles. Therefore, when studying trigonometry, it can be useful to hold a few different types of trigonometry problems back for a later time and then sneak them into an unrelated lesson by stealth! That way, they are not already on the lookout for right-angled triangles and have to recognise the situation it for themselves.

For example, I slipped the question below into one of my starter activities with Year 10 a few weeks ago.

It had been a little while since we had done any trigonometry, so the students weren’t on the look out for right-angled triangles. Instead, this conversation happened…

Student: Can I just put “acute”?

Teacher: No, I’d like the size of the angle, please.

Student: Can I estimate?

Teacher: No, I’d like an accurate measurement given to one decimal place, please.

Student: Can I come to the board and measure it with a protractor?

Teacher: No, you don’t need to.

Student: Eh?

But once one student had spotted that it could be solved with trigonometry by visualising a vertical line like the one below, a ripple of realization made its way across the room.

This sort of idea can lead to a whole host of similar problems, like the ones below. In each case, you can make a right-angled triangle by looking for a point on the diagonal line that goes through the corner of one of the squares in the background. However, students need to then do further calculations to find the size of the red angle.

These problems can still be great to use in a lesson that is about trigonometry, as they demonstrate an interesting application of the formulae. And I’ve noticed that even when students know they need to use trigonometry, they still find the decision making around their route to the solution a little tricky. Sneaking them into an unrelated lesson, however, just adds a satisfying element of surprise.

Problems Using Right-Angled Isosceles Triangles and/or Regular Polygons

There is a lot of fun to be had with right-angled isosceles triangles. More than I think I initially appreciated. Even a basic question like the one below is interesting in its own right, because it can be solved with either Pythagoras or trigonometry based on the information given.

I often find students opt for using Pythagoras with the question above, rather than trigonometry. It’s the more obvious choice when the 45 degree angles aren’t labelled and my students have tended to find Pythagoras easier to use than trigonometry. However, when this question is turned around (like the one below), a method using Pythagoras becomes a little bit trickier than with the previous question, as students sometimes hesitate about how to get started. On the other hand, the methods with trigonometry seem just as straightforward as they did in the previous question.

Whether students use Pythagoras or trigonometry, their methods for dealing with right-angled isosceles triangles can be applied in many, many ways. For example, it allows them to solve problems involving diagonals of squares.

A problem like this can lead students to an alternative formula for calculating the area of a square: “area of square = 0.5d2“. This formula could also be discovered before students learn Pythagoras by drawing two diagonals to split the square into four congruent right-angled triangles.  However, I find the Pythagoras route neater!

Dealing with the diagonal of a square can also give some interesting opportunities for mathematical reasoning with Pythagoras’ Theorem and irrational numbers, such as with the question below.

Students often start this problem with trial and error by choosing a value for x and seeing if they get a whole number for y. But then when they start to work solely with the algebra that is given to them, they can deduce that y is equal to root2x, meaning that at least one of the two lengths must be irrational. This problem can also provide a nice segue into some history of mathematics and the story of Hippasus (an example of this story can be found on nrich).

The hypotenuse of a right-angled isosceles triangle could also disguise itself as a chord in a circle, like with the problem below. This question asks for the area of the circle, but alternatives could be to ask for the circumference, the length of the minor arc AB, area of the segment, and so on. Regardless of the end product, the start of the problem is still essentially the same as the second example in this section of the post.

The two radii and right-angle marker provide visual clues that the question involves a right-angled triangle. The entry into this problem could be made trickier by presenting the information without these things drawn, like in the example below. Students have to wrestle with the terminology a little first and deduce that the arc AB is one quarter of the circumference before they can be sure that the angle at the centre is 90 degrees.

A question like this could catch students out when the fraction given in one quarter rather than one third, like with the question below. Now, the angle at the centre is 72 degrees rather than 90 degrees, which prompts a change in strategy. However, seeing the word “quarter” in the question makes it very tempting to think that the arc is a quarter of the circumference, like in the problem above.

The hypotenuse of a right-angled isosceles triangle could also hide itself on the side of a regular octagon. Once again, solving this problem involves dealing with the same triangle as in the second example from this section, but visualising it can be a little tricky until students start drawing extra lines on the diagram.

Exploring regular polygons leads to a whole other exiting avenue for creating interesting Pythagoras and trigonometry problems (and not just ones with right-angled isosceles). For example, one tweak to the question above can make the problem require a different strategy.

The route to the solution is different but still has many options. The most direct way would be recall the interior angle size of a regular octagon and use the cosine rule.  However, if students haven’t met the cosine rule and have only learned right-angled trigonometry then they could split the shape into two congruent right-angled triangles and use half the interior angle. If students have only learned Pythagoras, then they could solve it by using the formula twice: once with GH as the hypotenuse and then again with HB as the hypotenuse.

There really is a whole Aladdin’s Cave to be of wonders to explore in problems that hide right-angled triangles in regular polygons.

The Two Small Areas Add Up to the Big Area

Another interesting avenue of Pythagoras problems can be started with the example below.

This problem is already a little bit quirky because it gives the lengths of the squares rather than the triangle itself. But it doesn’t take much for students to move those labels to the short sides of the triangle, calculate the length of the hypotenuse and then calculate the area of the biggest square. At least that is what I find most students want to do when they see this question (and then kick themselves afterwards for bothering to square root the 100). Alternatively, students who know their Pythagorean triples well will spot that the hypotenuse is 10 cm straight away and then square it to get the blue area.

However, this problem can become more thought provoking if we made the hypotenuse an irrational number and didn’t allow students to use a calculator.

What I like about this problem is that it requires students to stop thinking of Pythagoras’ Theorem as just a formula that you plug numbers into and get a number out; it encourages them to think more deeply about what the theorem tells us. It doesn’t just give us a relationship between three lengths, but also the relationship between the areas of three similar shapes. By looking at the problem through this lens, it becomes easier than students might initially think because they can just add up the two green areas to get the blue area.

Now there is a whole host of new questions that a class can explore with this message at the heart of the problem. For example, the last question could be made slightly more complex by making students work a little harder to get the areas of the smaller squares first, like below.

It can also be interesting to explore using shapes other than squares around a right-angled triangle.

Students who see this for the first time often solve it by substituting the lengths 16 cm and 12 cm into Pythagoras’ Theorem to calculate the length of the hypotenuse, half it and use that to calculate the area of the blue semicircle. They then have the nice surprise that they can also get the same answer by calculating the areas of the two green semicircles and adding them together.

Related to this are problems based on The Lunes of Alhazen, such as the one below. This comes with the delightful surprise at the end that the total green area is equal to the area of the triangle. The problem could be extended further by considering whether that result is just the case for this particular problem or if it can be proven to be true for all possible measurements on a right-angle triangle.

There are other problems that can be solved simply when students use Pythagoras’ Theorem to focus on relationships between areas rather lengths. For example, calculating the length of the hypotenuse in the problem below would be a nightmare at high school level!

The final fun thing I like to do with this idea is to break the shapes away from a right-angled triangle altogether. For instance, if the lengths for the three squares below satisfy the formula a2 + b2 = c2 while they are connected around a right-angled triangle…

Then those same lengths would continue to satisfy the formula if the three shapes were pulled apart…

So the areas of the two green squares still add up to the area of the blue square because the lengths satisfy the Pythagorean formula.

This means that if students are good at spotting Pythagorean Triples, then they could use this idea to solve problems that don’t even have a triangle in sight!

This next problem could be solved by calculating scale factors between two of the trapezia. But if it’s posed as a non-calculator question, then the calculations could get very fiddly! However, because they are similar shapes and the three lengths make a Pythagorean Triple, the smallest area is the difference between the other two areas.

We could also start to think 3D…

If students need a little more convincing about this last problem, then it could be explored by looking at nets for cubes around a right-angled triangle.

The limits to creating ir finding interesting problems using Pythagorean Triples is endless and there are so many ways to surprise students with their applications!

If you would like to use or share any of the images from this blog, feel free to use this Power Point to help you: Playing With Pythagoras and Trigonometry

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