Chapter Summary and Practice

Imagine opening this chapter again after several weeks. You may remember a² + b² = c², but the important question is whether you remember what the letters mean, why the relationship works, and when to use addition or subtraction. A formula is most useful when it reconnects to its diagram and its reasoning. This lesson rebuilds those connections so that you can return to any of the chapter’s problem types with confidence.

Start with the square constructions, then follow the ideas through square roots, the theorem, triples, applications, and dot grids. You do not need to memorise every example. Instead, notice the decision that makes each example work: identifying a hypotenuse, using half a diagonal, preserving a stem’s length, or expressing an area as two square areas.

Try explaining each main idea aloud before reading its worked example. If an explanation feels uncertain, the surrounding paragraphs supply the missing reasoning. The mixed questions at the end require you to choose a method, not just repeat the last calculation you saw.

From Square Areas to Triangle Lengths

A square of side s has area s². If its side doubles, its area becomes (2s)² = 4s². If its side halves, its area becomes (s/2)² = s²/4. Changing a side and changing an area are different operations.

A square built on an original square’s diagonal has twice the original area. The reason is that the original square can be divided into two congruent right triangles, while the square on its diagonal consists of four triangles of that same size. Congruent means equal in size and shape, so corresponding pieces have equal area.

Joining neighbouring side midpoints gives a square with half the original area. Four equal corner triangles are removed. Their combined area is half the original square’s area. The inner figure really is a square: its sides are equal by triangle congruence, and each inner angle is 180° − 45° − 45° = 90°.

Paper constructions support these arguments when pieces are moved without stretching, gaps, or overlaps. Counting pieces only compares areas correctly when the counted pieces have equal areas.

Exact Lengths and Approximate Values

The hypotenuse is the side opposite a right triangle’s 90° angle. Locate that angle before assigning letters; rotation of the drawing does not change the hypotenuse. A square’s diagonal is the hypotenuse of either of the two triangles it creates.

The square root √N is the positive number whose square is N, for N > 0. A unit square’s diagonal has square 2, so its exact length is √2. Because √2 is not a ratio of integers, it is called irrational. The radical is an exact name, not an instruction that must always be replaced by a decimal.

To find bounds, compare squares of positive candidates. We have 1.4² < 2 < 1.5², then 1.41² < 2 < 1.42², then 1.414² < 2 < 1.415². Thus 1.414 < √2 < 1.415. Write an approximation with ≈, such as √2 ≈ 1.414, instead of asserting an exact equality.

Why can no terminating decimal equal √2? After removing trailing zeros, a terminating decimal has a nonzero last digit; its square also has a nonzero last decimal digit, so it cannot be the integer 2. But non-termination alone does not show irrationality, because a fraction such as 1/3 also has a non-terminating decimal.

The fraction argument is stronger. Assuming √2 = m/n leads to m² = 2n². In a perfect square each prime factor occurs an even number of times. The extra factor 2 on the right makes its number of factors 2 odd, while the left has an even number. That impossible equality rules out the fraction.

The Isosceles Right-Triangle Relationship

An isosceles right triangle has equal perpendicular sides a and a. Its hypotenuse c satisfies c² = 2a², or c = a√2. To work backwards, divide c² by 2 to find a², then take the positive square root.

For example, if a = 4, then c² = 32 and c = 4√2. If c = √50, then a² = 50/2 = 25, so a = 5. The relationship requires both equality of the perpendicular sides and a right angle; it does not apply to an arbitrary isosceles triangle.

The Theorem and Its Area Proof

The Baudhāyana–Pythagoras theorem states that the square on a right triangle’s hypotenuse has area equal to the sum of the squares on its perpendicular sides. If those sides are a and b and the hypotenuse is c, the relationship is a² + b² = c².

In the three-piece proof, two squares are joined and cut into two congruent right triangles and a middle region. Moving the two triangles around the unchanged middle region creates a new square. The new boundary consists of equal hypotenuses, and the triangle angles establish right angles at the corners.

The original pieces cover area a² + b²; the rearranged pieces cover area c². They are the same pieces without overlaps or gaps, so the areas are equal. This explanation tells you both why the theorem works and why a + b = c is the wrong relationship.

Choosing the Right Relationship

If both perpendicular sides are known, add their squares to obtain the hypotenuse square. If the hypotenuse and one perpendicular side are known, subtract that side’s square from the hypotenuse square. In either case the last step is a positive square root.

Use the same length unit throughout. Keep exact square roots when requested, and use bounds or a final approximation when a numerical estimate is useful. The hypotenuse must exceed either perpendicular side because its square includes both positive side squares.

Triples, Scaling, and Primitive Forms

A Baudhāyana–Pythagoras triple consists of positive integers a, b, and c with a² + b² = c². Common examples are (3,4,5), (5,12,13), (8,15,17), (7,24,25), and (12,35,37). A right triangle may have non-integer lengths, but those lengths do not form an integer triple under this definition.

Multiplying every entry by the same positive integer k gives another triple. Algebraically, (ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)². Lengths multiply by k, while the square areas multiply by k². There is no corresponding rule that adding a fixed number to every entry preserves a triple.

A primitive triple has greatest common factor 1 across its three entries. Primitive does not mean that each entry is prime. For a non-primitive triple, divide all three entries by their greatest common factor to obtain its primitive form.

For example, (15,36,39) has greatest common factor 3 and reduces to (5,12,13). Dividing the equation by 3² preserves equality. Every triple is either already primitive or a positive-integer scaled version of one. Unlimited scale factors give infinitely many triples.

Generating Triples and Exploring Higher Powers

The first n odd numbers sum to n². The final added odd number is 2n − 1, so (n − 1)² + (2n − 1) = n². When that odd number is itself a square, the identity becomes a right-triangle triple.

Choose 81 = 9². Solving 2n − 1 = 81 gives n = 41, so 40² + 9² = 41². In general, for odd m > 1, the method gives (m, (m² − 1)/2, (m² + 1)/2). The last two entries are consecutive and therefore have no common factor greater than 1; consequently the triple is primitive.

The construction generates infinitely many primitive triples, but not every primitive triple. For instance, (8,15,17) has no side one less than its hypotenuse and cannot arise from this method. A method can be valid and useful without being complete.

Fermat asked whether the square-power equation could work with larger integer powers. Fermat’s Last Theorem states that xⁿ + yⁿ = zⁿ has no positive-integer solutions for integer n > 2. It is a solved theorem, proved by Andrew Wiles in 1994, not a problem that remains open. Its advanced proof is outside our work here; checking a few failed examples is not a substitute for that proof.

Finding Hidden Right Triangles

A rectangle’s diagonal forms a right triangle with adjacent sides. Thus a square with side s has diagonal s√2, while a rectangle with length l and width w has diagonal √(l² + w²). Integer triples give rectangles with integer diagonals.

A rhombus’s diagonals bisect each other at right angles. Use their halves as perpendicular sides; the rhombus side is the hypotenuse. For diagonals d₁ and d₂, a side has length √((d₁/2)² + (d₂/2)²). The full diagonals belong to a triangle twice the required size.

In an equilateral triangle, the altitude is perpendicular to the base and bisects it. The two small right triangles are congruent because their hypotenuses are equal and their altitude is shared. With original side s, use half-base s/2 and hypotenuse s to find the height. Then area = ½ × base × height.

The lotus problem uses a model: fixed root, unchanged straight stem length, initially vertical stem, and horizontal water surface. If the depth is x and the tip is 1 unit above water, the stem length is x + 1. A sideways reach of 3 units gives x² + 9 = (x + 1)². Expanding and cancelling x² gives x = 4 units. The stem itself is 5 units long.

Area Constructions and Dot Grids

To construct a square with three times a starting area A, first obtain a side for area 2A using a diagonal, then combine the side lengths for areas A and 2A as perpendicular sides. The hypotenuse square has area 3A.

For five times the starting area, use perpendicular lengths s and 2s, where s is the original side. Their square areas are A and 4A. To construct a difference L² − s², make a right triangle with hypotenuse L and one perpendicular side s, then build the square on the remaining perpendicular side.

On a unit dot grid, a side connecting two grid points has whole-number horizontal and vertical displacements p and q. Its squared length, and therefore the square’s area, is p² + q². Rotate that movement through right angles to complete the other sides.

Allow p or q to be zero for grid-aligned squares, but not both. Areas 2, 4, and 5 are possible through step pairs (1,1), (2,0), and (2,1). Area 3 is impossible because any displacement 2 already contributes 4, while displacements 0 or 1 cannot give a squared sum of 3. This grid restriction does not forbid an ordinary square of area 3.

Explaining and Correcting Mistakes

Find the Colours!

For a final reasoning puzzle, imagine three closed boxes. One contains only red balls, one only blue balls, and one only green balls. The boxes are labelled RED, BLUE, and GREEN, but every label is wrong. You may open only one box. How can you determine the correct contents of all three?

Open the box labelled RED. It cannot contain red balls. If it contains blue balls, the box labelled GREEN cannot contain green by the condition and cannot contain blue because blue has already been located. It must contain red, leaving green for the box labelled BLUE.

If the opened box contains green instead, the box labelled BLUE cannot contain blue and cannot contain green, so it contains red. The remaining box labelled GREEN contains blue. One observation resolves the whole arrangement because the condition that every label is wrong rules out the alternatives.