Combining Two Different Squares

Suppose you have two square pieces of paper, one with side 3 cm and the other with side 4 cm. You want to turn all the paper into a single square, using every piece exactly once. The smaller square contributes 9 cm² and the larger contributes 16 cm², so the new square must have area 25 cm². In this particular case you can guess its side. But how could you construct the right square when the numbers are less convenient?

Your earlier method worked for two identical squares. Now the squares may have different sizes, so the equal-triangle construction needs to be extended. The useful connecting shape is still a right triangle, but its perpendicular sides no longer have to be equal.

We will follow the pieces carefully. An area equation becomes convincing when you can identify which regions remain in place, which regions move, and why their new boundary is a square.

Making the Connecting Right Triangle

Use the side of one square for one perpendicular side of a right triangle. Use the side of the other square for its second perpendicular side. If the original sides are a and b, the triangle’s perpendicular sides are a and b.

Call the hypotenuse c. The construction we want is a square of side c. The claim is that this new square has exactly the combined areas of the two original squares. At this stage it is a claim to explain, not a formula to assume.

If a and b happen to be equal, the triangle becomes isosceles and its hypotenuse is the diagonal of either original square. The proposed construction therefore agrees with the doubling construction you already understand.

Following Baudhāyana’s Construction

Place the smaller square to the left of the larger square with their bottom edges on the same straight line. Let their side lengths be a and b, with b greater than a. Together their bottom edges have length a + b, and the top boundary has a step.

Label the upper-left corner of the small square A and the upper-right corner of the large square D. On the bottom edge of the large square mark a point P that is a units from the far-right corner. Consequently P is b − a units to the right of the join between the squares.

The horizontal distance from the far-left bottom corner to P is a + (b − a) = b. Draw AP and PD. The two outer triangles cut off by these segments each have perpendicular sides a and b. They are congruent, even though one is turned relative to the other. Each cut segment is a hypotenuse of length c.

The remaining middle piece is an irregular region; call it V. Call the two cut-off triangles W and X. The original two squares have been divided into W, V, and X. Their areas have not changed, and nothing has been discarded.

Why the New Figure Is a Square

Extend the vertical join between the two original squares upwards. Move W to the left of that extended line, above A, and move X to its right, above the larger square. In the new positions call them T and U. Their long vertical and horizontal edges fit against the step of V; the outside boundary consists of four hypotenuses.

All four outside sides have the same length c because they belong to congruent right triangles. Equal sides alone would establish a rhombus, but a rhombus is not necessarily a square. We must also establish the angles.

Let the two acute angles of each right triangle be x° and (90 − x)°. At a new outer corner, these complementary angles combine to make 90°. At a corner where the corresponding two angles lie along a straight line beside the interior, the interior angle is 180° − x° − (90 − x)° = 90°. Thus every outside corner is a right angle.

The new figure has four equal sides and four right angles. It is therefore a square of side c. The proof depends on the congruence and angle relationships, not on whether a hand-cut paper model looks perfectly square.

Comparing Areas by Rearrangement

Before rearranging, the total area is area(W) + area(V) + area(X). After rearranging, it is area(T) + area(V) + area(U). T is the moved copy of W, and U is the moved copy of X. Their respective areas are equal, while V is unchanged.

The first arrangement consists of squares with areas a² and b². The second is one square with area c². Since the same pieces make both arrangements without gaps or overlaps, a² + b² = c².

This is a proof for any positive side lengths a and b for which the construction is made, not merely a check using the numbers 3, 4, and 5. When the two original squares are equal, use the equal-square construction from the earlier lesson as the special case.

Combining Two Squares Using Paper

For this model, the two starting areas are 16 cm² and 36 cm², so the new area is 52 cm². You do not need a whole-number side length for the construction to work. The exact new side is √52 cm.

As an additional check, draw a right triangle with perpendicular sides 4 cm and 6 cm and construct a square on its hypotenuse. Your three pieces should cover that square. Small gaps caused by inaccurate cutting are measurement errors; the geometric argument describes the ideal construction.

Stating the Baudhāyana–Pythagoras Theorem

Read the equation in terms of areas: a² is one square’s area, b² is another square’s area, and c² is the combined area. The letters themselves represent lengths. Confusing a length with its square is one reason students incorrectly write a + b = c.

The right-angle condition matters. We constructed the connecting triangle using perpendicular sides. A triangle that merely has two labelled sides a and b does not automatically satisfy this relationship. Identify the right angle before applying the theorem.

Connecting the General and Special Cases

If the perpendicular sides are both a, the theorem gives a² + a² = c². Combining the equal terms gives 2a² = c², exactly the relationship for an isosceles right triangle.

The general theorem has therefore not replaced the earlier reasoning. It includes it. Doubling a square is the equal-area case of combining two different square areas, and the diagonal of a square is the equal-side case of a right triangle’s hypotenuse.