Analyze and justify an area-rearrangement proof of the Pythagorean Theorem and its converse

Learning goal: Analyze an informal proof of the Pythagorean Theorem built from four congruent right triangles placed inside one square, justify every step of it in writing, and analyze and justify an informal proof of the converse, which tells whether a triangle with three known side lengths has a right angle.

Vocabulary and the statement. A right triangle is a triangle with one right angle, an angle of exactly 90 degrees. The two sides that form the right angle are the legs. The side opposite the right angle is the hypotenuse, and it is always the longest side, because the largest angle of a triangle faces the longest side. Call the legs a and b and the hypotenuse c. The Pythagorean Theorem states: in a right triangle, the square of one leg plus the square of the other leg equals the square of the hypotenuse, a² + b² = c². The condition matters. The relationship is between the squares of the lengths, not the lengths themselves, and it is only promised when the angle between a and b is a right angle. A theorem is a statement that has been proved; a proof is a chain of reasons, each one justified, that starts from things already known and ends at the statement.

Why "informal" proof. The proof below uses area and rearrangement instead of formal two-column reasoning. It is still a proof, because each step gives a reason. Your job in this lesson is to analyze the proof, meaning to identify what each step claims and what it depends on, and to justify it, meaning to say why the claim must be true.

The proof, step by step.

Step 1. Take one right triangle with legs a and b and hypotenuse c, and make four exact copies. The four triangles are congruent because they were traced from the same original: every side and angle matches.

Step 2. Draw a square whose side is a + b. Place one triangle in each corner so that going around the square, a leg of length a is followed by a leg of length b from the next triangle. Each side of the big square is covered exactly, since a + b is its length. Justification: the four right angles of the triangles sit in the four corners of the square, and the legs lie along the edges.

Step 3. The uncovered region in the middle is bounded by the four hypotenuses, so it has four sides of length c. That alone makes it a rhombus; to call it a square we need right angles. Look at any point where two triangles touch on an edge of the big square. Three angles meet there on a straight line: one acute angle from each of the two triangles, and one angle of the inner figure. The two acute angles of a right triangle add to 90 degrees, because the three angles of any triangle add to 180 degrees and the right angle already uses 90. So the inner angle is 180 − 90 = 90 degrees. All four inner angles are right angles, and the inner figure is a square of area c².

Step 4. Count the area two ways. The big square has area (a + b)². It is also covered, without overlap, by four triangles of area ½ab each and the inner square of area c². Therefore (a + b)² = 4 × ½ab + c². The left side expands to a² + 2ab + b², and 4 × ½ab is 2ab, so a² + 2ab + b² = 2ab + c². Subtract 2ab from both sides: a² + b² = c². Justification for the counting: the area of a region equals the sum of the areas of non-overlapping pieces that cover it exactly.

A second arrangement with no algebra. Put the same four triangles back into the same a + b square, but now pair them into two rectangles, each a by b, placed in opposite corners. The uncovered space is now two squares, one a by a and one b by b, with area a² + b². The four triangles have the same total area as before, and the big square is the same, so the uncovered area must be the same as before: a² + b² = c². Comparing the two arrangements is the whole idea; the diagram shows both.

A model with numbers. Take legs 3 and 4 and hypotenuse 5. The big square has side 3 + 4 = 7 and area 49 square units. Four triangles each have area ½ × 3 × 4 = 6, total 24. The leftover in the first arrangement is 49 − 24 = 25, which is 5², the hypotenuse squared. In the second arrangement the leftover is 3² + 4² = 9 + 16 = 25. Both agree. The numbers illustrate the proof; they do not replace it, because the letters a, b and c stand for the legs and hypotenuse of any right triangle at all.

The misconception, and why it is wrong. Students often write a + b = c, adding the legs to get the hypotenuse. In the 3-4-5 triangle that gives 7, but the hypotenuse is 5. The hypotenuse is shorter than the two legs together, because a straight path between two points is shorter than a bent path. A second error is to apply a² + b² = c² to a triangle with no right angle. A triangle with sides 4, 5 and 6 is real, but 4² + 5² = 41 while 6² = 36, so the equation fails. The equation is a property of right triangles only.

The converse, and its proof. The converse swaps the "if" and the "then": if a triangle has sides a, b and c, with c the longest, and a² + b² = c², then the angle opposite c is a right angle. Informal proof: build a second triangle with legs a and b meeting at a right angle. By the theorem, its hypotenuse squared is a² + b², which equals c² by assumption, so its hypotenuse is exactly c. The two triangles now have the same three side lengths. In Grade 7 you found that three side lengths that can form a triangle determine one and only one triangle, so the two triangles are congruent, and the original triangle has the same angles as the built one, including the right angle between a and b.

Using the converse. Test 8, 15, 17: 64 + 225 = 289 and 17² = 289, so this is a right triangle. Test 5, 6, 8: 25 + 36 = 61 but 8² = 64, so it is not. Always square the longest side alone; testing 15² + 17² = 8² is meaningless because 8 cannot be the hypotenuse.

Theorem versus converse. The theorem starts from a known right angle and tells you about lengths. The converse starts from known lengths and tells you about the angle. A justification that says which direction you are using is part of a complete answer.

Worked area check with 5, 12, 13: The big square has side 5 + 12 = 17 and area 17² = 289. Each triangle has area ½ × 5 × 12 = 30, and four of them cover 120. First arrangement: leftover 289 − 120 = 169 = 13², the hypotenuse squared. Second arrangement: leftover 5² + 12² = 25 + 144 = 169. Both leftovers agree, as the proof says they must, and 169 = 13² confirms a² + b² = c² for this triangle.

Worked justification of the inner right angle: Suppose the right triangle has acute angles x and 90 − x degrees; they add to 90 because 180 − 90 = 90. Where two triangles meet on the edge of the big square, the angles along the straight edge are x, the inner angle, and 90 − x. Since angles on a straight line add to 180, the inner angle is 180 − x − (90 − x) = 90 degrees. The reasoning did not use a value for x, so it holds for every right triangle, and the inner quadrilateral with four equal sides and four right angles is a square.

Worked converse decision: Is a triangle with sides 7, 24 and 25 a right triangle? The longest side is 25. Compute 7² + 24² = 49 + 576 = 625 and 25² = 625. They are equal, so by the converse the angle opposite the 25 side is a right angle. Now sides 6, 7 and 9: 36 + 49 = 85 and 9² = 81. They differ, so the triangle is not a right triangle; the converse gives a yes only when the equation holds exactly.

Cut-and-rearrange record: On centimeter grid paper cut four congruent right triangles with legs 6 cm and 8 cm; measure the hypotenuse and confirm it is 10 cm. Draw a 14 cm square. Record its area (196 cm²) and the total area of the four triangles (4 × 24 = 96 cm²). Arrange the triangles the first way and record the leftover square: side 10 cm, area 100 cm², equal to 196 − 96. Arrange them the second way and record the two leftover squares: 36 cm² and 64 cm², total 100 cm². Write one sentence stating what the two equal leftovers show.

Independent performance: Write a proof paragraph for the Pythagorean Theorem that another student could follow, using letters a, b and c rather than specific numbers. Success criteria: it states the theorem with its condition (a right triangle); it justifies why the four triangles fit inside the a + b square; it justifies why the inner figure is a square using the angle sum of a triangle; it shows the area equation and the algebra step that removes 2ab; and it ends with a converse test of the sides 9, 40 and 41 (81 + 1600 = 1681 = 41²), stating which direction, theorem or converse, the test uses.