The Pythagorean relationship and distance

The right angle determines the side roles

Start with two segments of lengths 3 cm and 4 cm meeting at a right angle. Join their other endpoints. How long is that joining segment? The two perpendicular sides are the legs of the right triangle, and the side opposite the right angle is its hypotenuse. We will find its length by comparing areas. We work in a Euclidean plane. Throughout the lesson, a triangle has three distinct, noncollinear vertices and positive side lengths. A right angle means 90°, wherever the triangle sits in the drawing: a tilted triangle can still be right-angled.

For triangle ABC, name the side lengths a = BC, b = CA and c = AB. If the right angle is at C, then a and b are the legs and c is the hypotenuse. If the right angle is instead at A, the hypotenuse is a and the legs are b and c. At a right angle at B, the hypotenuse is b. We often use c for a hypotenuse in a formula, but that convenient convention must not overwrite a problem's actual labels.

Choose one length unit before combining numerical lengths. A leg of 2 m and another of 150 cm have numerical lengths 2 and 1.5 in metres, or 200 and 150 in centimetres. Squaring a length produces a quantity in square units. A side of length 3 cm is not itself an area; the square built on that side has area 9 cm². A right-angle marker in an exact geometric specification supplies a hypothesis; a corner that merely looks square does not.

Before the area proof, review the areas of a square and a triangle, common length units, and positive square roots. The proof also uses the triangle angle sum of 180° and SAS congruence: two pairs of corresponding sides with equal lengths and equal included angles give congruent triangles. The converse later uses SSS congruence, where all three corresponding side lengths match. If these triangle facts are new, learn them before returning to the two proofs.

For a first reading, follow the 3, 4, 5 square-area example and the dissection, then calculate a positive missing length. Return to the separate converse when you are ready to compare the two SSS-matched triangles. The distance section is an optional next step once you can read signed Cartesian coordinates; the area proof does not require coordinates.

The square areas, rather than the side lengths, add

For a nondegenerate right triangle with positive legs a and b and hypotenuse c, the Pythagorean theorem states c² = a² + b². Equivalently, the area of a square of side c equals the combined areas of squares of sides a and b. For legs 3 and 4, those areas are 9 and 16, so the hypotenuse square has area 25 and its side length is 5. This example checks the relationship in one case; the proof below explains every positive pair of legs.

The equation immediately shows why the hypotenuse is longer than either leg. Since b² > 0, c² = a² + b² > a², and positivity then gives c > a. Similarly c > b. It does not give c = a + b. In fact, (a + b)² = a² + b² + 2ab exceeds c² because 2ab > 0, so c < a + b. That is one of the strict triangle inequalities.

The right-angle hypothesis is essential when using this theorem to calculate a side. A general triangle does not satisfy this particular equation merely because it has three sides. We will prove a separate converse that can establish a right angle from exact side data. Until that proof is available, an equality of squares should not be treated as though it were the theorem's original angle hypothesis.

A square dissection proves the relationship for every positive pair of legs

Begin with any right triangle whose legs have positive lengths a and b and whose hypotenuse has length c. Let its two acute angles be α and β. The triangle angle sum gives α + β = 90°. Draw an outer square Q₀Q₁Q₂Q₃ with side length a + b, reading its vertices consecutively around its boundary. On each directed side QᵢQᵢ₊₁, mark a point Pᵢ a distance a from Qᵢ and therefore a distance b from Qᵢ₊₁; indices wrap around after 3. Both lengths are positive, so each mark lies strictly inside its side.

Join consecutive marks P₀, P₁, P₂ and P₃. They form an inner convex quadrilateral, with one corner triangle at each outer vertex. At Q₁, for example, triangle P₀Q₁P₁ has perpendicular legs Q₁P₀ = b and Q₁P₁ = a. The other corners have the same two leg lengths and included right angle. SAS makes all four corner triangles congruent to the original right triangle. Thus every side PᵢPᵢ₊₁ of the inner quadrilateral is a hypotenuse of length c. This conclusion uses congruence, without using a coordinate distance formula or assuming the theorem we are proving.

Equal inner side lengths alone would not establish a square: a rhombus need not have right angles. At each mark Pᵢ, the two adjacent corner-triangle angles are α and β, in one order or the other. Together with the inner quadrilateral's angle they partition the straight angle along that outer side. The inner angle is therefore 180° − (α + β) = 90°. All four inner angles are right, and all four sides have length c, so the inner quadrilateral is a square with area c².

The inner square and the four corner triangles have disjoint interiors and fill the outer square: their shared edges are precisely the consecutive joining segments. Each corner triangle has area ab/2 because its legs are perpendicular. Area addition therefore gives (a + b)² = 4(ab/2) + c² = 2ab + c². Expanding the outer area gives a² + 2ab + b² = 2ab + c². Subtract 2ab to obtain c² = a² + b². Every step holds for arbitrary positive a and b, including a = b; it is not an inference from a finite set of pictures.

The converse identifies a right angle using SSS

A theorem and its converse have different hypotheses. The theorem starts with a right angle and concludes an equality of squared lengths. Its converse starts with an existing nondegenerate triangle and an equality of squared lengths, then concludes that a specified angle is right. A true implication does not automatically make its converse true; here we can prove the converse using the theorem already established.

Suppose triangle ABC has positive side lengths a = BC, b = CA and c = AB, and c² = a² + b². Construct a right triangle XYZ with its right angle at Z and legs YZ = a and ZX = b. Positive legs on perpendicular rays give a nondegenerate triangle. Let its hypotenuse XY have positive length d. The proved theorem gives d² = a² + b² = c². Since d and c are both positive, equality of their squares gives d = c.

Now AB = XY, BC = YZ and CA = ZX. SSS proves △ABC ≅ △XYZ with A ↔ X, B ↔ Y and C ↔ Z. Their corresponding angles at C and Z are equal, so ∠C = 90°. That is the required converse. The proof preserves the side opposite the claimed right angle; an equality involving a² as the sum instead gives the right angle at A. We have not derived a triangle from zero or negative side values by squaring them.

Check feasibility, then compare the actual largest side

When three proposed side lengths are given without an angle, first check that every length is positive. Then check the strict triangle inequalities a + b > c, b + c > a and c + a > b. Positive equality in one of these conditions describes a collinear boundary, not a nondegenerate triangle. If the largest length exceeds the sum of the other two, no triangle has those lengths. The right-triangle comparison is reserved for a specification that passes these checks.

For a valid triangle, let L be its largest side length and let u and v be the other two. If L² = u² + v² exactly, the converse proves that the angle opposite L is 90°. If the equality fails, the triangle is not right: were it right, the theorem would make its hypotenuse the unique largest side and require precisely this equality. This lesson only makes the right or nonright conclusion from that comparison; it does not assign acute or obtuse classifications through an unproved extra rule.

Keep the original side labels while selecting the largest length. For ordered lengths (a, b, c) = (5, 3, 4), a = BC is the largest, and a² = b² + c² reads 25 = 9 + 16. The right angle is at A, rather than at C. Sorting the numerical values can help the arithmetic, but silently sorting the names changes which vertex is opposite which side. For tied largest sides, the squared equality cannot hold with positive third side, so a valid tied-largest triangle is nonright.

A negative proposal such as (−3, 4, 5) illustrates why the first guard matters: its squares satisfy 9 + 16 = 25, but −3 is not a side length. A zero proposal also fails nondegeneracy even if a squared equality happens to hold. Reject those inputs before a rightness verdict. A failed feasibility check is a statement about whether the requested triangle exists, rather than a classification of an existing triangle.

A physical length uses the positive root and must satisfy the domain

For given positive legs a and b of a right triangle, c² = a² + b² gives c = √(a² + b²), using the positive square root. The algebraic equation z² = a² + b² has two real solutions z = ±√(a² + b²), but a side length is positive and therefore selects only the positive one. The theorem also gives c > a and c > b, useful checks on a calculated hypotenuse. Taking √(a² + b²) as a + b would discard the difference between adding lengths and adding their squares.

For a given hypotenuse c and one leg a, both supplied lengths must first be positive. Rearranging gives b² = c² − a². A nondegenerate right triangle requires c > a > 0, so its other leg is b = √(c² − a²) > 0. Those conditions are sufficient as well: construct perpendicular positive legs of lengths a and b, and the theorem makes the constructed hypotenuse √(a² + b²) = √(c²) = c. No extra approximate check is needed to establish existence.

If positive c equals positive a, the expression c² − a² is zero. The formal value b = 0 describes a collapsed boundary, not a right triangle. If 0 < c < a, the expression is negative and no real second leg can meet the specification. If either supplied length is nonpositive, reject it before squaring. For example, calling −5 a hypotenuse and 3 a leg must not produce a supposedly valid length 4 simply because (−5)² − 3² = 16.

State both the value and its geometric meaning. With c = 2 and a = 1 in one common unit, b = √3 is an exact positive length. With c = 3 and a = 4, no such real leg exists, rather than an imaginary segment in the Euclidean plane. With c = 3 and a = 3, the zero result is a degenerate limit. A value may solve a squared equation while failing the geometric constraints that gave the equation.

Keep exact roots until an approximation is needed

Some squared lengths are perfect squares: √25 = 5 and √(25/4) = 5/2. Others require an exact root expression, such as √2 or √13. A missing leg with c = 7 and a = 5 has b = √24 = 2√6. This is its exact value; a calculator gives b ≈ 4.90 to two decimal places. The equality sign records an exact relationship, while ≈ records a rounded approximation. Use the requested unit and precision and round the final answer once, rather than rounding components before squaring them.

Convert all length data to the same unit before forming the equation. If lengths are expressed in centimetres, squared terms are in cm² and the final positive square root returns centimetres. If coordinates use metres, a squared distance is in m² and a distance is in m. Do not add a numerical square in metres to one in centimetres as though their units were equal. A dimension check can reveal an error even when the arithmetic itself was performed correctly.

A Pythagorean triple is a triple of positive integers that can be the side lengths of a right triangle. For example, 3, 4, 5 and 5, 12, 13 have the required squared equality. Multiplying all lengths of a right triangle by a positive factor k preserves it, since (ka)² + (kb)² = k²(a² + b²) = (kc)². Thus 6, 8, 10 is another integer triple, while 3/2, 2, 5/2 describes a valid scaled right triangle whose lengths are not all integers. Not every right triangle has rational side lengths: positive legs 1 and 1 give hypotenuse √2.

Cartesian distance comes from two perpendicular components

Let P = (x₁, y₁) and Q = (x₂, y₂) be points in a Euclidean Cartesian plane. The axes are perpendicular, and a unit on either axis represents the same physical length. Define the signed differences Δx = x₂ − x₁ and Δy = y₂ − y₁. They describe directed changes, so either can be negative. The corresponding component lengths are |Δx| and |Δy|, which are never negative. Negative coordinates or differences are permitted; a negative proposed side length is a different matter.

When both differences are nonzero, introduce H = (x₂, y₁). Segment PH is parallel to the x-axis and segment HQ is parallel to the y-axis, so they meet at a right angle at H. Their positive lengths are |Δx| and |Δy|, and PQ is the hypotenuse. The theorem, now already proved independently, gives PQ² = |Δx|² + |Δy|² = (Δx)² + (Δy)². Taking the positive root gives PQ = √((x₂ − x₁)² + (y₂ − y₁)²). We used the right-triangle theorem to derive the distance formula, rather than using the distance formula to prove that same theorem.

If Δy = 0, the points are horizontally aligned and their distance is |Δx|. The formula gives the same result because √((Δx)²) = |Δx|. If Δx = 0, the corresponding result is |Δy|. If both are zero, P and Q coincide and the distance is 0. These cases do not form a nondegenerate auxiliary triangle, so their justification comes directly from line distance and the zero distance of a point from itself. The single distance formula includes them without pretending that a collapsed right triangle satisfies the earlier nondegeneracy hypothesis.

Reversing the endpoints replaces Δx and Δy by their negatives; their squares are unchanged, so PQ = QP. Use both coordinates from the same endpoint consistently before subtracting. A sketch can help locate the components, but its pixel lengths or display proportions do not prove the result. If the axes represent different physical length units, convert the components to one unit first; if the axes are not perpendicular, this two-square formula cannot be used without a different geometric analysis.

Model a straight distance using stated perpendicular directions

A rectangle's diagonal is the hypotenuse of a triangle formed by two adjacent sides: the rectangle's right-angle definition supplies the perpendicularity. A straight segment joining a point on a vertical wall to a point on perpendicular level ground can likewise be a hypotenuse. In either setting, identify the straight segment, the perpendicular component lengths and the common units before calculating. These are ideal geometric models; their equations alone do not establish practical conditions about a real installation.

A direct distance is not generally the length of a route made of two perpendicular segments. Between (0, 0) and (3, 4), an axis-parallel route of 3 units followed by 4 units has length 7, whereas the joining segment has length 5. In general, for positive component lengths a and b, we proved √(a² + b²) < a + b. If one component is zero, an aligned direct route has the same length as that single component. State whether the question asks for a segment, a route or a displacement magnitude.

The same idea extends to three-dimensional Euclidean coordinates with mutually perpendicular axes and a common length unit. For two points, first use the plane result for their x- and y-components: the horizontal component distance h satisfies h² = (Δx)² + (Δy)². The z-component is perpendicular to that horizontal plane, so a second right-triangle application gives d² = h² + (Δz)². Thus d = √((Δx)² + (Δy)² + (Δz)²). If one step has zero component, use its aligned or coincident distance case rather than asserting a nondegenerate triangle at that step. A rectangular box with side lengths 2, 3 and 6 has a face diagonal √13 and a space diagonal √(13 + 36) = 7.

An exact squared equality differs from numerical closeness

For exact mathematical side data, a right-angle conclusion requires exact equality after the positive-length and triangle-existence checks. Sides 3, 4 and 5 give 9 + 16 = 25 exactly. Sides 3, 4 and 5.001 form a valid triangle, but 5.001² = 25.010001 differs from 3² + 4² = 25. The triangle specified by those exact values is nonright. A small difference or an almost square drawing does not make the two values equal.

A decimal can represent an exact mathematical number or a rounded measurement, depending on the specification. When field measurements are rounded or uncertain, agreement to displayed precision supports an estimate rather than a proof of an exact 90° angle. Keep the given measurement precision and assumptions visible in the result. A model with explicitly perpendicular directions can use that right-angle hypothesis, but measurements alone must not silently introduce it.

Inspect four scenes with their conditions visible

The area scene offers seven leg pairs: (3, 4), (1, 1), (1, 2), (2, 3), (3/2, 2), (0, 4) and (−3, 4). For positive legs it shows the outer square, four congruent corner triangles and central square, together with exact outer area, combined triangle area and central area. The zero and negative proposals are rejected before constructing a right triangle or its square dissection.

The converse scene offers nine ordered side triples with a = BC, b = CA and c = AB: (3, 4, 5), (5, 3, 4), (3/2, 2, 5/2), (2, 3, 4), (4, 4, 4), (1, 2, 3), (1, 2, 4), (0, 2, 2) and (−3, 4, 5). It first checks positive lengths and all strict triangle inequalities. Only a valid triangle receives a right or nonright comparison using its actual largest side; a right result names the angle opposite that original labelled side. Degenerate, impossible and nonpositive proposals do not receive a rightness verdict.

The missing-leg scene offers eight (hypotenuse c, known leg a) pairs: (5, 3), (2, 1), (3, 4), (3, 3), (0, 3), (−5, 3), (5, 0) and (5, −3). Positive c > a gives the exact second leg √(c² − a²), while positive c = a is explicitly a zero-leg boundary. Positive c < a has no real second leg; a nonpositive supplied length is rejected before squaring. No invalid proposal receives an invented nondegenerate triangle.

The distance scene uses eight point pairs: (0, 0) to (3, 4), the reversed pair, (−2, −1) to (1, 3), (0, 0) to (1, 1), (0, 0) to (0, 4), (−2, 1) to (3, 1), the coincident pair (2, −1) to (2, −1), and (−1/2, 1/2) to (1, 5/2). Signed changes, nonnegative component lengths and exact distances are kept separate. Each scene has an equal length scale within its own graph; approximate root coordinates serve the drawing, while exact values determine the result. The next lesson uses distance and a fixed radius to begin the study of circles.

Worked examples

Try it yourself

1. Triangle ABC is right-angled at B, with AC = 10 cm and AB = 6 cm. Identify its hypotenuse and calculate BC exactly. In the notation a = BC, b = CA and c = AB, which squared-length equation applies? Why is a negative algebraic root not a second possible length?

   Show the answer and reasoning

   AC is the hypotenuse. The equation is b² = a² + c², giving BC = 8 cm. The algebraic roots ±8 of a² = 64 do not give two side lengths: only the positive value is allowed.

   The side opposite B is AC, so b = 10 is the hypotenuse and c = 6 is a leg. Rearranging gives a² = 10² − 6² = 100 − 36 = 64. The supplied lengths satisfy 10 > 6 > 0, so the missing leg is a = √64 = 8 > 0. The three resulting positive lengths satisfy 6 + 8 > 10 and the other strict inequalities. A value −8 satisfies the squared equation but fails the side-length condition.

2. In the area dissection, take positive legs a = 2 and b = 3 in one common length unit. Find the outer square's area, the combined area of its four corner triangles, and the inner square's area and side. What proves that the inner quadrilateral is a square, and why is its side not 2 + 3?

   Show the answer and reasoning

   The outer area is 25 square units, the four triangles have combined area 12 square units, and the inner square has area 13 square units and side √13 units. Congruent corner triangles give equal inner sides, while α + β = 90° gives four right inner angles. Its side is √13, not 5.

   The outer side is a + b = 5, so its area is 25. Each corner triangle has area ab/2 = 3, and four have area 12. Subtraction gives 25 − 12 = 13 for the inner region. All four joining segments are hypotenuses of congruent copies of the given right triangle, hence have the same length. At each inner vertex, the two adjacent acute corner angles total 90°, leaving an inner angle of 180° − 90° = 90°. Equal sides alone would only establish a rhombus. The positive square root of the inner area is √13; a side of 5 would instead have area 25 and leave no area for the triangles.

3. Classify these ordered proposed side triples (a, b, c), where a = BC, b = CA and c = AB: (3, 4, 5), (5, 3, 4), (2, 3, 4), (4, 4, 4), (1, 2, 3), (1, 2, 4), (0, 2, 2) and (−3, 4, 5). Give a right-angle vertex only when the data specify a valid right triangle.

   Show the answer and reasoning

   The first triple is right at C and the second is right at A. The 2, 3, 4 and 4, 4, 4 triples are valid nonright triangles. The 1, 2, 3 triple is degenerate; 1, 2, 4 is impossible. The last two proposals fail positive length and receive no rightness verdict.

   For 3, 4, 5 and 5, 3, 4, positivity and all strict inequalities hold, and the largest square is 25 = 9 + 16. The original labels put the largest side opposite C in the first case and A in the second. For 2, 3, 4, the largest length satisfies 2 + 3 > 4, but its square 16 differs from 4 + 9 = 13. For 4, 4, 4, the strict inequalities hold, but 16 differs from 16 + 16 = 32. The equality 1 + 2 = 3 is a collinear boundary; 4 > 1 + 2 prevents any triangle. Zero and negative proposed lengths are rejected before squaring, even where an algebraic equality of squares would otherwise hold.

4. A right triangle has hypotenuse 7 cm and one leg 5 cm. Find its other leg exactly and to two decimal places. Then assess the missing-leg specifications (c, a) = (3, 4), (3, 3), (−5, 3) and (5, 0). Which, if any, gives a nondegenerate right triangle?

   Show the answer and reasoning

   The first missing leg is 2√6 cm ≈ 4.90 cm. None of the four later specifications gives a nondegenerate right triangle: 3 < 4 gives no real second leg, 3 = 3 gives a zero-leg boundary, and the last two contain a nonpositive supplied length.

   Since 7 > 5 > 0, b² = 49 − 25 = 24 and b = √24 = 2√6 > 0. The displayed 4.90 is a rounded approximation, rather than an exact equality. For positive c = 3 and a = 4, the required squared leg is negative, so there is no real length. For positive c = a = 3, the formal root is zero, giving degeneracy. A negative hypotenuse is rejected before squaring; it cannot be repaired by (−5)² = 25. The supplied zero leg in (5, 0) also violates the right-triangle domain even though the formal other length would be positive.

5. Use one common coordinate length unit and perpendicular Cartesian axes. Find the distances from (−2, −1) to (1, 3), from (0, 0) to (1, 1), from (0, 0) to (0, 4), from (−2, 1) to (3, 1), and from (2, −1) to itself. Which pairs support a nondegenerate auxiliary right triangle?

   Show the answer and reasoning

   The distances are 5, √2, 4, 5 and 0 coordinate units, respectively. Only the first two pairs have both component lengths positive and therefore support the described nondegenerate auxiliary right triangle.

   The first differences are 3 and 4, giving √(9 + 16) = 5. The second differences are 1 and 1, giving √2. The third pair is vertical with distance |4 − 0| = 4; the fourth is horizontal with distance |3 − (−2)| = 5. In each aligned case one component is zero and the auxiliary configuration collapses, so line distance supplies the justification. The last endpoints coincide, so both differences and their distance are zero. Negative starting coordinates do not cause invalid lengths: absolute component differences and the final distance remain nonnegative.

6. A rectangle has adjacent sides 80 cm and 1.5 m. Find its diagonal exactly in centimetres and metres. Another rectangle is a uniform enlargement of it by the positive factor 2. Find the enlarged diagonal and explain why the squared-length relationship is preserved.

   Show the answer and reasoning

   The first diagonal is 170 cm = 1.7 m exactly. The enlarged diagonal is 340 cm = 3.4 m. Multiplying both legs by 2 multiplies their squared sum by 4 and the positive diagonal by 2.

   Convert 1.5 m to 150 cm. The diagonal square is 80² + 150² = 6400 + 22500 = 28900 cm², and √28900 = 170 cm. Alternatively, 0.8² + 1.5² = 0.64 + 2.25 = 2.89 m², whose positive root is 1.7 m. With enlarged legs 160 and 300 cm, the squared sum is 4(80² + 150²) = 4 × 170² = 340², so the positive diagonal is 340 cm. Units must agree within each sum. A positive uniform scale preserves right angles and all length ratios; the factor for squared quantities is the square of the length factor.

7. A rectangular box has side lengths 2, 3 and 6 cm. Derive its space diagonal using two successive Pythagorean calculations. Separately, do exact side lengths 3, 4 and 5.001 cm specify a right triangle because the largest side is close to 5? Would rounded field measurements alone establish an exact right angle?

   Show the answer and reasoning

   The box's space diagonal is 7 cm, from a face diagonal √13 cm followed by √(13 + 36). The exact 3, 4, 5.001 triangle is valid but nonright, since 25.010001 ≠ 25. Rounded measurements alone give an estimate rather than proof of an exact 90° angle.

   The 2 and 3 cm sides of a rectangular face are perpendicular, so its diagonal h satisfies h² = 4 + 9 = 13. The 6 cm direction is perpendicular to that face, giving the space diagonal d² = h² + 6² = 13 + 36 = 49 and d = 7 > 0. For the separate triangle, all lengths are positive and 3 + 4 > 5.001, with the other inequalities also strict. But the exact largest square is 5.001² = 25.010001, whereas the other two squares sum to 25. Numerical closeness is not equality. Rounded or uncertain measured lengths can be consistent with a right-angle model without proving that its exact angle is 90°; an explicit perpendicularity assumption would be a separate hypothesis.

What to carry forward

• A right triangle has positive legs meeting at 90° and a hypotenuse opposite that angle. Side roles follow the stated vertex labels, rather than automatically assigning c as the hypotenuse.
• For positive legs a and b and hypotenuse c, c² = a² + b² adds square areas. The general dissection proof establishes an inner square by both equal side lengths and right angles.
• The converse needs its own proof: construct a right triangle with the two given legs, use the theorem to match its positive hypotenuse, then use SSS to transfer the right angle.
• Check positive lengths and strict triangle inequalities before a rightness comparison. Use the actual largest side and identify its opposite vertex; rejected data receive no right-triangle verdict.
• A missing hypotenuse is √(a² + b²) for positive legs. A missing leg is √(c² − a²) only when c > a > 0; equality is degenerate, and nonpositive givens are rejected before squaring.
• With perpendicular Cartesian axes and common length units, distance is √((Δx)² + (Δy)²). Component lengths are absolute differences; aligned and coincident cases use line distance and zero distance without inventing a nondegenerate triangle.
• Preserve exact roots and consistent units, then round the final result when needed. An approximate drawing or near equality of measured squares does not prove an exact right angle.