Historically, the Pythagorean theorem has been taught as a memorized formula. It has become so emblematic that Gilbert and Sullivan lampooned it as an example of arcane but impractical knowledge. Fortunately, things are improving a bit. Students in modern mathematics classes are likely to at least be shown a proof of the theorem rather than expected to memorize it.

I want to do better. Here’s a narrative I have refined and used with some success for leading 7th or 8th grade students along the whole journey of discovery, discovering and understanding the key ideas behind the Pythagorean theorem. The goal is not just to know or use the formula, but to leave them feeling that they could have come up with it on their own.

You will likely recognize this as a well-known and familiar proof. What I want to focus on here, though, is not the argument as a proof of the formula after it’s been stated, but as a plausible path of discovery.

Step 1: Framing the problem

The first step to a good journey of discovery is to have an interesting and important question. This is the first place where I think we often get off on the wrong foot. I’ve seen students bewildered by the sudden concern for right triangles and especially their hypotenuses. These terms are abstractions, and abstraction should be justified before it is used.

Most students by this point, though, have internalized the notion of the coordinate plane, and points as x and y coordinates on the plane. A question that feels much more authentic and important to them is finding the distance between two points. So I start there. Here are two points on the coordinate plane. How far apart are they?

Left to their own devices, students will often attempt to measure the line, either precisely or by just eyeballing it. That’s fine, but they should be aware that their answer is an estimate, not an exact solution. Even careful measurement with a ruler can only promise that the result is close. If asked to reason about an exact answer, most students will eventually compute the distance separately in the x and y dimensions. They will often see that the two-dimensional distance is greater than the distance in either single dimension, but less than their sum because the diagonal is a shortcut. These are great things to notice, but students will then get stuck.

Step 2: Introduce area as an intermediate step

Since the question is happening in two dimensions, it turns out to be useful to think about areas instead of distance. This is the single greatest insight in this whole exercise! All the rest is pretty straight-forward reasoning, but the recasting of a question about distance into a question about area is the revolutionary concept here. So it’s worth taking slowly.

This is a good time to quickly review the relationship between the area of a square and the length of its side.

Students will already know this, of course, so it’s not necessary to spend too much time here, but you want students to think about squaring and square roots as inverse conversions between side-length and area of a square, so that if they know one, they can easily compute the other.

With this in mind, your next task is to draw a square with side length equal to the distance between those points. With enough foresight to leave room, you can do this on the same coordinate plane you started with.

If you have the time, you might pause to ask students for ways to determine the area of this square. As soon as they aren’t making progress, you can lift the veil by drawing a larger square…

Step 3: Add more structure

Here’s the picture that will get many of your students squirming to answer the question:

For most students, the strategy will now be in reach. They will calculate the area of the larger square — in this case, 17², or 289 — and then subtract the areas of the four colored triangles — in this case, 30 each, for a total of 120. The difference — in this case, 169 — is the area of the white square in the center. They should then be able to perform a square root to get the original distance that they wanted — in this case, 13.

This doesn’t look much like the Pythagorean theorem. That’s okay! This is an answer that your students really understand, and you can work out the next steps once their understanding is reinforced. At this point, I would ask them to work through an example together with me, and then on their own or with a partner. While most students can follow the reasoning and do the area calculation once the diagram is drawn, drawing the diagram itself for a pair of (x, y) coordinates takes them a few attempts.

Step 4: Simplify

To work toward the more conventional Pythagorean theorem, one can start with the diagram above, and shift the triangles around.

With luck (or shrewd planning), your students will have seen this picture before. It is just the visual approach to squaring a sum: (a + b)² = a² + 2ab + b². Since they only want the white portion, they can calculate it as a² + b². Now you’re done.

I love how the simplification can be done in two ways here, with algebraic expressions or by sliding around triangles in the diagram. The original strategy of computing the larger area and then subtracting off the triangles was computing (a + b)² - 2ab. Sliding the triangles corresponded to applying the distributive property, so that the 2ab terms cancel, leaving a² and b² intact.

I have seen some students who, after going through this progression, still prefer to solve distance questions for a while by drawing the first picture and calculating the answer by squaring the sum and then subtracting off the triangles. That’s okay with me for a while, because it makes sense to them. Eventually, of course, I hope they will become comfortable with the conventional form, so that it will be familiar to them for more advanced mathematics.