Pythagorean Theorem Animation: A Visual Area Proof

What this animation proves

For a right triangle with perpendicular sides a and b and hypotenuse c, a² + b² = c². The animation compares two arrangements of four identical triangles. The outside boundary stays a square of side a + b. Because neither that boundary nor the total triangle area changes, the uncovered areas in the two completed arrangements must be equal.

In this example, a = 3, b = 4, and c = 5. Each triangle has area 3 × 4 ÷ 2 = 6. Four triangles occupy 24 square units inside an outer square of area 49. The remaining area is 25: first one tilted square, then two squares with areas 9 and 16.

The movement is a way to compare the endpoints. The triangles overlap during the transition, so the uncovered area is not constant in every intermediate frame. Pause at the start and end to inspect the actual tilings.

Why the middle shape is a square

Each edge of the tilted middle shape is a triangle's hypotenuse, so all four edges have length c. The two acute angles of a right triangle sum to 90°. At each corner of the middle shape, those two angles and the middle angle form a straight angle. The middle angle is therefore 180° − 90° = 90°.

This matters: four equal sides alone would establish a rhombus, not necessarily a square. Equal sides plus right angles let us name its area c².

From the 3–4–5 example to any right triangle

The numbers make the diagram easy to check, but the area argument does not depend on those particular lengths. For positive perpendicular side lengths a and b, the outer square has area (a + b)² and the four triangles total 4 × ab/2 = 2ab.

Subtracting the triangles leaves (a + b)² − 2ab = a² + b². In the first arrangement the same remaining region is the square of side c, so its area is c². The diagram plus this argument establishes the relationship for right triangles, rather than merely checking one numerical example.

Use the animation in a lesson

1. Select Start arrangement. Ask students to identify the four congruent triangles and the tilted square before introducing the formula.
2. Ask for a prediction: if the triangles move without changing size, what can stay equal at the two endpoints?
3. Play the animation once, then select End arrangement. Find the two uncovered squares and connect their side lengths to 3 and 4.
4. Check 9 + 16 = 25. Then replace the numbers with a, b, and c and discuss why the argument needs a right angle.

A useful check question is: “Why do we add the areas of the squares rather than the side lengths?” A student who answers 3 + 4 = 5 has confused length with area. A second check is whether the learner can explain why the outside square and all four triangles must stay the same size.

A short storyboard for your own explanation

Use four beats: identify the right triangle; assemble the outer square; move the four colored triangles; compare the uncovered areas. Keep each triangle's color consistent and hold both endpoints long enough to read the labels.

Suggested narration: “The same four triangles fit inside the same square in two ways. Removing them leaves equal areas. One arrangement leaves c squared; the other leaves a squared plus b squared.” This is a proposed lesson script, not a recording or an automatically generated video.

For the broader planning and review process, follow how to make math animations. Browse animation examples for other teaching patterns, or explore the unit circle animation to see a different kind of visual explanation.

Further exploration

The University of Waterloo hosts an animated Pythagorean proof, and the GeoGebra Content Team's activity offers another visual investigation. The diagram and controls on this page are our own implementation of the rearrangement argument; third-party artwork has not been copied.