Pythagorean Theorem Calculator: Hypotenuse, Legs and Area

About this calculator

The Pythagorean theorem calculator finds the third side of a right triangle from any two: given the two legs $a$ and $b$ it returns the hypotenuse $c = \sqrt{a^2 + b^2}$ ; given the hypotenuse and one leg it solves for the other leg. It also reports the area of the triangle, $ab/2$ .

Whenever a problem contains a right angle—a wall and the floor, the width and height of a screen, two displacements along coordinate axes, a ladder against a wall—you can use it: measure two sides and the third needs no measuring. It is also the starting point for the distance between two points, the law of cosines and the length of a vector; understanding it means understanding how “length” is defined in Cartesian coordinates.

What it does not do: it does not handle triangles without a right angle (those need the law of cosines or the law of sines), it does not judge whether three given sides can form a triangle, and it does not convert units—all three sides must use the same length unit.

Formula and variables

Let the two legs of a right triangle be $a$ and $b$ and the hypotenuse (the side opposite the right angle, and the longest side) be $c$ :

a^2 + b^2 = c^2

This gives three solving directions and the area:

c = \sqrt{a^2 + b^2},\qquad a = \sqrt{c^2 - b^2},\qquad b = \sqrt{c^2 - a^2},\qquad S = \tfrac{1}{2}\,ab

Symbol	Meaning	Calculator field	Unit	Range and notes
$a$	One leg	Leg a	length	$a > 0$ ; default 3
$b$	The other leg	Leg b	length	$b > 0$ ; default 4
$c$	Hypotenuse	Hypotenuse c	length	$c > a$ and $c > b$ ; when solving, no solution if $c \le$ the known leg
$S$	Area	Area	length²	Output, $S = ab/2$ ; meaningful only when both legs are known

All three sides must use the same length unit (all metres, all centimetres, all pixels, …); the area is in the square of that unit.

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Pythagorean theorem” (/#/mathtools/formula?calculator=calculator.pythagorean). It opens with the first two fields filled with the defaults $a = 3$ , $b = 4$ , and the hypotenuse and area computed.
Find the hypotenusechange “Leg a” and “Leg b”; fields marked “input” are yours, fields marked “computed” update live. The hypotenuse $c$ and the area are given together.
Solve for a legclear “Leg a” (or b), enter the hypotenuse in “Hypotenuse c” and keep the other leg as an input; the cleared leg switches to “computed” and shows the result.
Read the resultsvalues are shown to at most 8 decimal places with trailing zeros removed, e.g. $\sqrt{2}$ appears as 1.41421356. If, when solving, the hypotenuse is not longer than the known leg, the result field is left blank—the inputs do not form a right triangle.
Copy the formulathe copy button to the right of the panel title copies the formula text a² + b² = c² to the clipboard for pasting into homework or documents.

Worked examples

All four examples were recomputed by the engine’s pythagorean compute routine; the display convention is at most 8 decimal places with trailing zeros removed.

Example 1: the classic 3-4-5

Inputs: $a = 3$ , $b = 4$ .

$a^2 + b^2 = 9 + 16 = 25$ .
$c = \sqrt{25} = 5$ .
Area $S = 3 \times 4 / 2 = 6$ .

Interface shows: hypotenuse 5, area 6. This is the smallest integer Pythagorean triple; the ancient Chinese text Zhoubi Suanjing refers to it as “gou 3, gu 4, xian 5”.

Figure 1: the right triangle of Example 1. The hypotenuse is the side opposite the right angle and is always the longest

Example 2: solving for a leg from the hypotenuse

Inputs: $c = 13$ , $b = 5$ (a cleared).

$c^2 - b^2 = 169 - 25 = 144$ .
$a = \sqrt{144} = 12$ .

Interface shows: leg a 12. Note that the “Area” field shows 0 here: the engine computes the area only when both legs are inputs, and the solved $a$ does not take part. To see the area, enter 12 back into “Leg a” and clear the hypotenuse; the area is then 30.

Example 3: the ladder problem

A ladder 6.5 m long leans against a wall with its foot 2.5 m from the wall. How high does it reach? Here the hypotenuse is the ladder and the legs are the distance from the wall and the height.

Inputs: $c = 6.5$ , $a = 2.5$ (b cleared).

$c^2 - a^2 = 42.25 - 6.25 = 36$ .
$b = \sqrt{36} = 6$ .

Interface shows: leg b 6. Conversely, knowing 2.5 m from the wall and 6 m high, entering $a = 2.5$ , $b = 6$ gives hypotenuse 6.5 and area 7.5.

Example 4: irrational results and screen diagonals

Inputs: $a = 1$ , $b = 1$ : $c = \sqrt{2}$ , shown as 1.41421356 (8 decimal places), area 0.5. This was the first length in history proved not to be a fraction.

Inputs: $a = 1920$ , $b = 1080$ (the pixel width and height of a 1080p screen): diagonal 2202.90717008 pixels. If the screen is sold as 27 inches (a 27 in diagonal), solving with $c = 27$ and $b = 13.24$ (height) gives a width of 23.53088184 in—“27 inches” is the diagonal, not the width.

The inputs of Example 1 can be entered directly into the panel to reproduce it; set both a and b to 1 and the hypotenuse becomes 1.41421356.

Principles and derivation

Why the squares of the legs add up to the square of the hypotenuse

The most visual proof is a rearrangement. Take a large square of side $a + b$ ; there are two ways to cut it up:

First cut: mark a point on each of the four sides so that the large square splits into four congruent right triangles (legs $a$ , $b$ ) and a tilted square in the middle. The sides of the tilted square are exactly the hypotenuse $c$ of the triangles, so the central area is $c^2$ .
Second cut: pair the same four triangles into two $a \times b$ rectangles and place them in two corners of the large square; what remains is exactly one $a \times a$ square and one $b \times b$ square, area $a^2 + b^2$ .

Both cuts use the same four triangles and the same large square, so the remaining areas must be equal: $c^2 = a^2 + b^2$ .

Figure 2: two ways of cutting the same (a+b)². The four triangles are identical, so c² = a² + b²

Euclid’s Elements, Book I, Proposition 47 gives a different proof—squares are constructed outward on the three sides and the areas of the two smaller squares are shown to equal the area of the large one; Proposition 48 proves the converse: if a triangle satisfies $a^2 + b^2 = c^2$ , it must be right-angled. The converse is the basis of the builder’s “3-4-5” method for laying out a right angle: a triangle formed from lengths of 3, 4 and 5 has an exact 90° corner.

It is really the definition of distance

In Cartesian coordinates, the two displacements $\Delta x$ and $\Delta y$ from $(x_1, y_1)$ to $(x_2, y_2)$ are perpendicular, so the straight-line distance between the points is the hypotenuse: $d = \sqrt{\Delta x^2 + \Delta y^2}$ . This is the formula of the distance calculator; extending it to three dimensions is just one more application of Pythagoras (first the diagonal of the base, then a new right triangle with the height). The magnitude of a vector, the modulus of a complex number and the standard deviation in statistics (the square root of a sum of squared deviations) are all the same expression.

Relationship to the law of cosines

For any triangle, $c^2 = a^2 + b^2 - 2ab\cos C$ . When the included angle $C = 90°$ , $\cos C = 0$ , the last term vanishes and the law reduces to the Pythagorean theorem. The theorem is thus the special case of the law of cosines for a right angle; conversely, comparing $c^2$ with $a^2 + b^2$ tells you whether the angle is acute ( $c^2 < a^2+b^2$ ) or obtuse ( $c^2 > a^2+b^2$ ).

Pythagorean triples

Right triangles whose three sides are all integers are called Pythagorean triples. The smallest few:

$(a, b, c)$	Check
(3, 4, 5)	$9 + 16 = 25$
(5, 12, 13)	$25 + 144 = 169$
(8, 15, 17)	$64 + 225 = 289$
(7, 24, 25)	$49 + 576 = 625$
(20, 21, 29)	$400 + 441 = 841$

Euclid’s formula generates every primitive triple: take coprime positive integers $m > n$ of opposite parity and set $a = m^2 - n^2$ , $b = 2mn$ , $c = m^2 + n^2$ . $m = 2, n = 1$ gives (3, 4, 5); $m = 3, n = 2$ gives (5, 12, 13). Any triple scaled by a constant is still a triple—(6, 8, 10) and (0.3, 0.4, 0.5) give equally tidy results in the calculator.

History

The Babylonian tablet Plimpton 322 (c. 1800 BCE) already lists many Pythagorean triples. In China, the Zhoubi Suanjing records “gou 3, gu 4, xian 5”, the “Gougu” chapter of the Nine Chapters on the Mathematical Art systematically solves problems with the theorem, and Zhao Shuang’s commentary contains the “hypotenuse diagram”, which is precisely the rearrangement proof above. The Pythagorean school (c. 6th century BCE) is credited with the earliest general proof, and with the discovery, through $\sqrt{2}$ , that some lengths cannot be written as fractions. Gou and gu are the classical Chinese names for the shorter and longer legs of a right triangle; xian is the hypotenuse.

It holds only in the plane

The Pythagorean theorem is a theorem of Euclidean plane geometry. On a sphere, two perpendicular great-circle arcs $a$ and $b$ and the third side $c$ satisfy $\cos c = \cos a \cos b$ (unit sphere), which approaches $a^2 + b^2 = c^2$ only when all three sides are much smaller than the sphere’s radius. Over a few tens of kilometres on a map, Pythagoras is safe; for intercontinental distances, use spherical formulas.

Assumptions

It really is a right triangle: the angle between $a$ and $b$ is exactly 90°. Even a few degrees off and the hypotenuse departs from $\sqrt{a^2+b^2}$ (the law of cosines estimates the deviation).
Plane geometry: the three sides lie in one plane, and that plane is Euclidean; the theorem fails on curved surfaces (large regions of the Earth’s surface).
One unit: the three sides are entered in the same length unit; the area unit is automatically its square.
Positive finite real numbers: side lengths are positive; the engine does not block negative or zero values, but they have no geometric meaning.
Area convention: the area is computed only from the two input legs; in solve mode the area field shows 0, which does not mean the area is zero.

Scope and limitations

Can calculate: the third side and area of any right triangle with two known sides, including solving for a leg; results to 8 decimal places (trailing zeros removed).
Approximate only: when the measurements themselves carry error, the error of the result is about $\Delta c \approx (a\,\Delta a + b\,\Delta b)/c$ ; the 8 displayed decimals do not reflect measurement precision.
Cannot calculate: non-right triangles (use the law of cosines), one side plus one angle (use trigonometric functions), or “triangles” on spheres or other curved surfaces.
Does not: check the triangle inequality, convert units, or output angles.
Numerics: extremely large values (such as $10^{200}$ ) overflow to infinity when squared; the engine uses $a^2 + b^2$ directly rather than the numerically safer hypot, which makes no difference at everyday magnitudes.

Common mistakes

Entering the hypotenuse in a leg field: the hypotenuse is always the longest side and is opposite the right angle. Putting 5 in a and 3 in b gives a “hypotenuse” of 5.83, which is not a side of the original triangle at all.
Entering a hypotenuse shorter than a leg when solving: $c^2 - b^2$ is negative and the result is blank. Check which side is the hypotenuse first.
Mixing units between sides: one side in metres and another in centimetres gives a meaningless result. Convert to one unit before entering.
Forgetting the square root: $3^2 + 4^2 = 25$ is $c^2$ , not $c$ ; the calculator takes the root for you, but by hand this is the most commonly skipped step.
Applying it to a non-right triangle: the third side of an isosceles or arbitrary triangle cannot be found this way unless you have first shown there is a right angle.
Area 0 in solve mode: see Assumptions—this is the engine’s convention, not a zero-area triangle.
Taking “27 inches” as the width: the quoted size of a screen or TV is the diagonal; width and height follow from the Pythagorean theorem and the aspect ratio.

Typical use cases

Teaching: from numbers to shapes

Enter a few triples first (3-4-5, 5-12-13) to see integer results, then 1-1 to see $\sqrt{2}$ , then hold $a$ fixed and increase $b$ step by step to watch the hypotenuse approach $b$ —an intuitive feel for sums of squares, roots and limits.

Building and fitting out: squaring corners, checking diagonals

Mark 3 m and 4 m on the floor; if the diagonal is exactly 5 m the corner is square. When tiling or building cabinets, measure the diagonals to check for squareness. Conversely, given a room’s length and width, find the diagonal to judge whether a whole sheet of material will fit through the door at an angle.

Screens and frames

From pixel width and height find the diagonal in pixels (and with the quoted size, the pixel density in PPI), or from the diagonal in inches and the aspect ratio recover the actual width and height.

Coordinates and navigation

The straight-line distance between two points in the plane, the resultant displacement of a two-axis robot, and the impedance triangle in circuits (resistance, reactance, impedance) are all direct applications of the theorem.

FAQ

Can I solve it with one side and one angle?

Not with this calculator—it needs two sides. One side plus an acute angle calls for trigonometry: opposite = hypotenuse × sin(angle), adjacent = hypotenuse × cos(angle).

I know all three sides; how do I tell whether the triangle is right-angled?

Take the longest side as $c$ and test whether $a^2 + b^2$ equals $c^2$ (allowing a small discrepancy for measurement error). Equal means a right angle (the converse theorem); smaller means obtuse, larger means acute.

Why do results have so many decimal places?

Most right triangles have an irrational hypotenuse with infinitely many non-repeating decimals; the interface keeps at most 8 and drops trailing zeros. In practice, round to your measurement precision.

Why does the area show 0 when I solve for a leg?

The engine computes the area only from the two input legs and does not feed the solved side back. Enter the solved value into the leg field by hand and clear the hypotenuse, and the area shows correctly.

What happens if I enter a negative number?

Squaring removes the sign, so the hypotenuse is the same as for the positive value, but the area becomes negative. Negative lengths have no geometric meaning; enter positive numbers.

How do I find a diagonal in three dimensions?

First find the diagonal of the base from its two sides, then treat that diagonal and the height as the two legs of a new right triangle; or enter three-dimensional coordinates directly into the distance calculator.

References and further reading

The pythagorean definition in the CalcX engine source src/data/formulas.ts (compute(known, solveFor) supports solving for a / b / c and the area); every figure on this page was recomputed by that engine.
Euclid, Elements, Book I, Proposition 47 and Proposition 48, edited and annotated by D. E. Joyce, Clark University （访问日期：2026-09-08）—the classical proofs of the theorem and its converse.
Wikipedia, Pythagorean theorem （访问日期：2026-09-08）—many proofs, generalisations and history.
Wikipedia, Pythagorean triple （访问日期：2026-09-08）—Euclid’s generating formula and primitive triples.
Wikipedia, Zhoubi Suanjing （访问日期：2026-09-08）—the Zhoubi Suanjing and “gou 3, gu 4, xian 5”.
Nine Chapters on the Mathematical Art, Chapter 9 “Gougu”, with Liu Hui’s commentary, and Zhao Shuang’s commentary on the Zhoubi Suanjing (“diagram of the hypotenuse, circle and square”)—the original Chinese sources for the hypotenuse-diagram proof.

Privacy

All inputs and calculations run inside your browser and are never uploaded to a server. The text on this page is static content; it neither contains nor records any user input.

Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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