Mathematics · Geometry

Pythagorean theorem

Q: How do you calculate a leg when the hypotenuse and the other leg are known?

Rearrange the theorem for the required leg: $a=\sqrt{c^2-b^2}$. So subtract the square of the known leg from the square of the hypotenuse and take the root of the result. The order matters, because here there is a minus, not a plus as for the hypotenuse. Example: for $c=13$ and $b=5$ it follows that $a=\sqrt{169-25}=\sqrt{144}=12$. Make sure to put the square of the hypotenuse first; if the expression under the root is negative, the side identification, data or right-angle assumption is inconsistent. The hypotenuse is always the longest side, so its square is the largest.

The Pythagorean theorem holds for right triangles: the square of the hypotenuse equals the sum of the squares of the two legs.

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Formula

Calculate

a^{2} + b^{2} = c^{2}

a, b, c in the same length unit; squares in the corresponding area unit

Variables & units

Variables & units
Symbol	Meaning	Unit
a, b	Legs (the two sides adjacent to the right angle)	m, cm, etc.
$c$	Hypotenuse (the longest side, opposite the 90° angle)	same unit

LaTeX

a^2 + b^2 = c^2

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Diagram: a right triangle with a square on each side; the two leg squares a² and b² together equal the hypotenuse square c². — For a right triangle: the sum of the squares on the legs a² and b² equals the square on the hypotenuse c².

Pythagorean theorem calculator

Find the hypotenuse or a missing leg of a right triangle, plus distances from coordinates, pyramid heights and a check for a right angle.

Calculation model

Find

Leg a

One side next to the right angle, e.g. 3 cm.

Leg b

The other side next to the right angle, e.g. 4 cm.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for c, substitute, evaluate

1. Rearrange for c
$c=\sqrt{a^2+b^2}$
2. Substitute values (in SI units)
- a = 3 cm = 0.03 m
- b = 4 cm = 0.04 m
$\sqrt{0.03^2+0.04^2}$
3. Result
c = 5 cm

Result: Hypotenuse opposite the right angle c

c = 5 cm

With a = 3 cm and b = 4 cm, c = 5 cm.

Quick tips

The hypotenuse is opposite the right angle and is always the longest side. For a leg you subtract squares: a = √(c² − b²), not c − b.
Pythagorean triples as a check: 3, 4, 5 and 5, 12, 13 and any multiple such as 6, 8, 10 work out exactly.

Assumptions and limits

Established right angle in Euclidean geometry; a and b meet at the right angle, c is opposite. Positive nondegenerate lengths in a common spatial unit.

Derivation & background

The theorem bears the name Pythagoras. A classical area proof appears in Euclid’s Elements, Book I, Proposition 47; Proposition 48 states the converse. The cosine rule $c^2=a^2+b^2-2ab\cos\gamma$ generalises to arbitrary triangles. At $\gamma=90^\circ$ the additional term vanishes.

Exam blueprint

Validity range

Applies only to right triangles in Euclidean geometry.

Derivation steps

A square of side a + b partitions into four right triangles and a central square of side c.

1Choose four congruent right triangles with legs a,b and hypotenuse c.
2Arrange them inside an outer square of side a + b with their hypotenuses facing inward.
3The inner shape has four sides c and right angles because the acute triangle angles sum to 90°.
4Compare areas: $(a+b)^2=4\cdot\frac{ab}{2}+c^2$ .
5Expand: $a^2+2ab+b^2=2ab+c^2$ .
6Subtract 2ab on both sides: $a^2+b^2=c^2$ . A right angle remains essential.

Rearrangements

Calculate hypotenuse

c = \sqrt{a^2+b^2}

c is always opposite the right angle.

Task variant

A triangle has legs 5 and 12. Find c.

$c=\sqrt{25+144}=\sqrt{169}=13$ .

Common mistakes

Mistaking a leg for the hypotenuse.

Find the right angle first; c is opposite it.

Exam context

Often used in geometry, vector lengths, distances and coordinate tasks.

Formula cluster

Right angles and distance

Connects side squares, perpendicular coordinates and the cosine rule as a generalisation.

|v⃗| = √(v₁² + v₂² + v₃²)

Magnitude of a Vector (Length)

c² = a² + b² − 2ab·cos γ

Law of Cosines

a⃗·b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Dot Product

sin α = GK/Hyp, cos α = AK/Hyp, tan α = GK/AK

Trigonometry in the Right Triangle

Worked example

A carpenter needs the diagonal of $a 3 m \times{} 4 m$ room: $c=\sqrt{(3\,\mathrm m)^2+(4\,\mathrm m)^2}=5\,\mathrm m$ . The "3-4-5 triple" is a classic Pythagorean triple.

Applications

Architecture, distance in planar coordinates, computer graphics, vector calculations

How do you calculate with Pythagorean Theorem?

Here is how to work through a typical Pythagorean Theorem (a² + b² = c²) task step by step:

1
Task
A triangle has legs 5 and 12. Find c.
Solution path
$c=\sqrt{25+144}=\sqrt{169}=13$ .

Understand and apply the equation

The Pythagorean theorem relates the three side lengths of a right triangle. It is also an area statement and the basis of Euclidean distance. Correct use requires identifying the hypotenuse, establishing a right angle and using consistent length units. This guide connects a proof, rearrangements and original problems.

Identify the side you need

A ladder, a screen diagonal or the straight connection between two points: a right angle makes Pythagoras the appropriate tool. The longest side lies opposite that angle. You work with squared lengths; taking the square root returns a length.

Prerequisites: identify right triangles, convert length units, square numbers and take the positive square root.
A sketch helps identify sides. An angle that merely looks right does not replace a stated or established right angle.

Legs, hypotenuse and squares

The sides adjacent to the right angle are legs a and b. The opposite side c is the hypotenuse and longest side. Letters can change: a problem may call the hypotenuse h or d. Identify it by its position relative to the right angle, not by its letter.

a^2+b^2=c^2,\qquad a,b,c>0

This equation applies to a nondegenerate right triangle in Euclidean geometry.[1]

Squares can be interpreted as areas: a square with side a has area a². The two leg-square areas add to the hypotenuse-square area. Side lengths themselves do not satisfy $a+b=c$ . The original diagram above shows these three squares.

Trigonometry in a right triangle adds angle relationships to the side relation. Opposite and adjacent legs depend on the acute angle being considered. The hypotenuse remains opposite the right angle.[5]

A proof by rearranging areas

Take four congruent right triangles with legs a and b. Place them in the corners of a large square of side a + b, with their hypotenuses bounding a central quadrilateral. Each inner side has length c. The two acute angles of a right triangle sum to 90°, making the central quadrilateral a square.

(a+b)^2=4\cdot\frac{ab}{2}+c^2

The outer area consists of four triangle areas $\frac{ab}{2}$ plus the central square c².

a^2+2ab+b^2=2ab+c^2\quad\Longrightarrow\quad a^2+b^2=c^2

Expand and subtract 2ab from both sides.

The proof relies on the right angle. An oblique triangle cannot be substituted under the same assumptions. Euclid’s classical proof in Book I, Proposition 47 uses another area construction. A proof establishes a general relation; one numerical example alone does not.[1]

Calculating a hypotenuse or a leg

\begin{aligned}c&=\sqrt{a^2+b^2}\\a&=\sqrt{c^2-b^2}\\b&=\sqrt{c^2-a^2}\end{aligned}

For geometric lengths, use the positive square root.

To find c, add the leg squares. To find a leg, subtract the other leg square from the hypotenuse square. Subtracting lengths directly is incorrect. A negative $c^2-b^2$ indicates inconsistent side identification, data or right-angle assumptions. A zero radicand gives a degenerate triangle.

Convert every length to the same unit first. Squared lengths have the corresponding area unit; the square root returns a length unit. For $a=1.20\,\mathrm m$ and $b=50\,\mathrm{cm}$ , use $b=0.50\,\mathrm m$ , not the number 50 alongside metres.

Three examples with different approaches

Diagonal of a rectangular panel

Given: Sides 1.20 m and 50 cm; adjacent sides are perpendicular.
Find: Calculate the diagonal.

Approach: The right-angle corner makes the diagonal the hypotenuse. $d=\sqrt{a^2+b^2}$ . Convert the second length to metres.
$b=0.50\,\mathrm m$
Add squares; the diagonal is the hypotenuse.
$c^2=(1.20)^2+(0.50)^2=1.69\,\mathrm{m^2}$
Take the positive square root.
$c=\sqrt{1.69}\,\mathrm m=1.30\,\mathrm m$

Result: The diagonal is 1.30 m long.

Check: It exceeds either side but is less than $1.20\,\mathrm m+0.50\,\mathrm m=1.70\,\mathrm m$ . Squaring returns 1.69 m².

Finding a leg from a diagonal

Given: Hypotenuse 3.9 m; one leg 1.5 m.
Find: Find the other leg.

Approach: The hypotenuse is the longest side; select the positive length. $a=\sqrt{c^2-b^2}$ . Subtract the known leg square from the hypotenuse square.
$a^2=(3.9)^2-(1.5)^2=15.21-2.25=12.96\,\mathrm{m^2}$
Take the positive square root.
$a=\sqrt{12.96}\,\mathrm m=3.6\,\mathrm m$
Check using the original equation.
$(3.6)^2+(1.5)^2=12.96+2.25=15.21=(3.9)^2$

Result: The missing leg is 3.6 m long.

Check: Direct subtraction $3.9\,\mathrm m-1.5\,\mathrm m=2.4\,\mathrm m$ would be incorrect. The calculated leg remains shorter than the hypotenuse.

Two coordinate changes, one distance

Given: On a planar map with axes in metres, $A=(-1,2)$ and $B=(7,-4)$ .
Find: Find the straight-line distance and distinguish it from a route along the axes.

Approach: horizontal and vertical coordinate differences are perpendicular components. Initially retain their signs.
$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$
Subtract the starting coordinate from the ending coordinate.
$\Delta x=7-(-1)=8,\qquad\Delta y=-4-2=-6$
Add the squares and choose the positive square root as a length.
$d=\sqrt{8^2+(-6)^2}\,\mathrm m=10\,\mathrm m$

Result: The straight-line distance is 10 m. Travelling horizontally and then vertically gives a 14 m route.

Check: The distance exceeds either component magnitude but is shorter than their sum. The negative vertical change describes direction only.

View diagram: coordinate differences form a right triangle

Straight-line distance and a route along the axes

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The perpendicular segments are 8 m and 6 m long; the direct connection is 10 m. Both axes use the same scale. The change

\Delta y=-6\,\mathrm m

is signed, while its geometric length is positive. This diagram accompanies the third worked example.[3]

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Checking a right angle from side lengths

The converse also holds: if triangle sides satisfy $a^2+b^2=c^2$ with c the longest side, the angle opposite c is right. First check that positive lengths and the triangle inequality permit a nondegenerate triangle.[2]

Original check: 12, 35 and 37 satisfy $12+35>37$ and $12^2+35^2=144+1225=1369=37^2$ . The triangle is right. For 12, 35 and 36, the longest-side square is 1296 rather than 1369, so the relation does not hold.

Approximate equality from measured lengths does not prove an exactly right angle. Measurement uncertainty and rounding affect the squares. Use a tolerance consistent with measurement accuracy and state the result accordingly.

From a triangle to distance

Two points in Cartesian coordinates differ horizontally by Δx and vertically by Δy. These changes form perpendicular legs, so Euclidean distance is the square root of $(\Delta x)^2+(\Delta y)^2$ . Coordinate differences may be negative; their squares are nonnegative. Both axes must use the same spatial unit.[3]

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For three perpendicular spatial axes, add $(z_2-z_1)^2$ under the square root.

Original points $A=(-2,1)$ and $B=(4,9)$ give $\Delta x=6$ and $\Delta y=8$ , hence d = 10 length units. A cuboid with edges 2, 3 and 6 has space diagonal $\sqrt{2^2+3^2+6^2}=7$ . Directly applying this formula to latitude and longitude on the curved Earth is not an exact distance calculation.

In a right regular square pyramid, the apex is vertically above the base centre. Let a denote the base side and k the edge from apex to a base vertex. Half the base diagonal is $\frac{a}{\sqrt2}$ ; together with vertical height h and edge k, it forms a right triangle. The slant height of a triangular face is a different segment.[1]

h=\sqrt{k^2-\frac{a^2}{2}},\qquad a>0,\;k>\frac a{\sqrt2}

The positive root gives vertical height; equality would make the pyramid degenerate.[1]

Original calculation: for $a=6\,\mathrm m$ and $k=5\,\mathrm m$ , $h=\sqrt{25-18}\,\mathrm m=\sqrt7\,\mathrm m\approx2.646\,\mathrm m$ . Check: $h^2+\frac{a^2}{2}=7+18=25$ . A pyramid-volume calculation needs this perpendicular height.[1]

Not every triangle is right

c^2=a^2+b^2-2ab\cos\gamma

The cosine rule includes angle γ between sides a and b.[4]

For $\gamma=90^\circ$ , $\cos\gamma=0$ gives Pythagoras as a special case. At other angles, the added term is generally nonzero. An equilateral triangle of side 2 has $c^2=4$ whereas $a^2+b^2=8$ . Merely knowing two sides does not establish a right angle.

Do not simplify $\sqrt{a^2+b^2}$ to a + b. For $a=b=1$ the result is √2, not 2. Also round at the end; insufficient precision in intermediate values can distort the check.

The cosine rule extends the side calculation to a specified non-right angle. Use that relation when a right angle is not established; the appearance of a drawing is insufficient evidence.[4]

Five problems with solutions

Calculate · Foundation

Legs are 8 cm and 15 cm. Find the hypotenuse.

Show hint

Add squares, then take the square root.

Learning goal: Find hypotenuse from legs.

Show solution and calculation

Approach: The hypotenuse is opposite the right angle, so $c=\sqrt{a^2+b^2}$ .
Substitution gives $c=\sqrt{(8\,\mathrm{cm})^2+(15\,\mathrm{cm})^2}=17\,\mathrm{cm}$ .

Check: $(17\,\mathrm{cm})^2=64\,\mathrm{cm^2}+225\,\mathrm{cm^2}=289\,\mathrm{cm^2}$ . Taking a square root converts an area unit back to a length unit.

c=\sqrt{(8\,\mathrm{cm})^2+(15\,\mathrm{cm})^2}=17\,\mathrm{cm}

Calculate · Application

Hypotenuse 25 mm and one leg 7 mm. Find the other leg.

Show hint

Subtract the leg square from the hypotenuse square.

Learning goal: Find a missing leg.

Show solution and calculation

Approach: Rearrange $a^2+b^2=c^2$ for the missing positive length: $a=\sqrt{c^2-b^2}$ .
Substitution gives $a=\sqrt{(25\,\mathrm{mm})^2-(7\,\mathrm{mm})^2}=24\,\mathrm{mm}$ .

Check: $(24\,\mathrm{mm})^2+(7\,\mathrm{mm})^2=(25\,\mathrm{mm})^2$ .

a=\sqrt{(25\,\mathrm{mm})^2-(7\,\mathrm{mm})^2}=24\,\mathrm{mm}

Calculate · Application

A rectangle has sides 90 cm and 1.20 m. Find the diagonal in metres.

Show hint

Convert 90 cm to 0.90 m first.

Learning goal: Calculate rectangle diagonal with unit conversion.

Show solution and calculation

Approach: The rectangle diagonal is the hypotenuse, so $d=\sqrt{a^2+b^2}$ .
First convert $90\,\mathrm{cm}=0.90\,\mathrm m$ .
Then $d=\sqrt{(0.90\,\mathrm m)^2+(1.20\,\mathrm m)^2}=1.50\,\mathrm m$ .

Check: $(1.50\,\mathrm m)^2=2.25\,\mathrm{m^2}$ ; the diagonal is longer than either side.

d=\sqrt{(0.90\,\mathrm m)^2+(1.20\,\mathrm m)^2}=1.50\,\mathrm m

Interpret · Application

A nondegenerate triangle has sides 5, 6 and 8 cm, with 8 cm largest. A sketch makes its opposite angle look right. Determine from lengths whether this is possible and whether that angle is acute or obtuse.

Show hint

Compare the longest-side square with the sum of the other squares.

Learning goal: Test right-angle geometry from side lengths.

Show solution and calculation

Approach: The converse theorem requires equality of squares for a right angle.
Here 64 exceeds 61, so the angle is obtuse rather than right.
The cosine rule confirms a negative cosine.
The lengths also satisfy the triangle inequality.
An unscaled sketch cannot replace the length test.

\begin{aligned}8^2&=64>5^2+6^2=61\\{}\cos\gamma&=\frac{5^2+6^2-8^2}{2\cdot5\cdot6}=-\frac1{20}<0\end{aligned}

[2][4]

Error analysis · Application

A right-angled route runs 9 m east then 12 m north. Someone calls 21 m the direct distance by adding both segments. State path length and direct distance separately and explain the mistake.

Show hint

The hypotenuse joins start and finish rather than following the route.

Learning goal: Reject adding lengths in place of squares.

Show solution and calculation

Approach: Add lengths for the actual route, but add squared perpendicular components for the direct connection.
Path length is 21 m; distance is 15 m.

Check: the direct connection is shorter than the detour and longer than either component. The error confuses two distinct quantities.

\begin{aligned}s_{\mathrm{path}}&=9+12=21\,\mathrm m\\{}d&=\sqrt{9^2+12^2}\,\mathrm m=15\,\mathrm m\end{aligned}

[3]

What to remember for your solution

First establish the right angle and identify the opposite hypotenuse. Convert all lengths to one unit. Add squared legs to find the hypotenuse; subtract the known squared leg from the squared hypotenuse to find the other leg. The positive square root gives its length. Finally check by substitution and the triangle inequality.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added square-pyramid height with setup and check, connected relevant side, angle and volume relations and added a source on right-triangle trigonometry; preserved all prior exercises and figures.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified the introduction and prerequisites; added a third worked example, an original figure and a summary; made supplemental material selectively expandable.

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Quanta. Pythagorean Theorem. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/pythagorean-theorem

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Frequently asked questions about Pythagorean Theorem

How do you calculate the hypotenuse with the Pythagorean theorem?+

Square the two legs, add the squares and take the square root: $c=\sqrt{a^2+b^2}$ . The hypotenuse c is always the longest side and lies opposite the right angle. Example: for legs of 5 and 12, $c=\sqrt{25+144}=\sqrt{169}=13$ . All lengths must be in the same unit. Make sure to square and add first and only then take the root; the root of a sum is not the sum of the roots. The theorem holds only in right triangles, so you first identify the right angle and the side opposite it.

How do you calculate a leg when the hypotenuse and the other leg are known?+

Rearrange the theorem for the required leg: $a=\sqrt{c^2-b^2}$ . So subtract the square of the known leg from the square of the hypotenuse and take the root of the result. The order matters, because here there is a minus, not a plus as for the hypotenuse. Example: for $c=13$ and $b=5$ it follows that $a=\sqrt{169-25}=\sqrt{144}=12$ . Make sure to put the square of the hypotenuse first; if the expression under the root is negative, the side identification, data or right-angle assumption is inconsistent. The hypotenuse is always the longest side, so its square is the largest.

When does the Pythagorean theorem not apply?+

The theorem applies only in right triangles in Euclidean geometry; the triangle itself lies in a plane. In a triangle without a right angle, $a^2+b^2$ does not equal $c^{2}$ ; there you need the law of cosines instead, which as a generalization contains an extra term with the included angle. On curved surfaces, for example on a sphere, the theorem also does not hold, because there the sum of angles in a triangle differs from 180°. A common mistake is to apply the theorem to arbitrary triangles. Therefore always check first whether there really is a right angle before using $a^2+b^2=c^2$ .

How do you use Pythagoras to calculate the distance between two points?+

The distance between two points in the plane is the hypotenuse of a right triangle whose legs are the differences of the coordinates. For the points $(x_1,y_1)$ and $(x_2,y_2)$ , $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . So you form the differences in the x and y directions, square them, add them and take the root. In space the term $(z_2-z_1)^2$ is simply added under the root. This distance formula is a direct application of the Pythagorean theorem and the basis for vector lengths, the magnitude of a displacement and many problems in analytic geometry. This assumes perpendicular Cartesian axes using the same spatial unit, rather than latitude and longitude.

What is a Pythagorean triple?+

A Pythagorean triple consists of three positive integers a, b and c that satisfy the equation $a^2+b^2=c^2$ exactly. The best known is 3, 4, 5, since $9+16=25$ . Others are 5, 12, 13 and 8, 15, 17. Multiples of a triple, such as 6, 8, 10, are also triples. Such numbers are handy because they give clean, whole-number results in problems and let craftsmen set out a right angle without a protractor, for example with a twelve-knot rope. If you recognize a known triple in a problem, you can read off the third side immediately without having to compute the square root.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Prealgebra, §9.3: Use Properties of Angles, Triangles, and the Pythagorean Theorem

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Pythagorean theorem

The Pythagorean theorem holds for right triangles: the square of the hypotenuse equals the sum of the squares of the two legs.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

a, b

Legs (the two sides adjacent to the right angle)

m, cm, etc.

c

Hypotenuse (the longest side, opposite the 90° angle)

same unit

Derivation & background

Exam blueprint

Validity range

Applies only to right triangles in Euclidean geometry.

Derivation steps

A square of side a + b partitions into four right triangles and a central square of side c.

1Choose four congruent right triangles with legs a,b and hypotenuse c.
2Arrange them inside an outer square of side a + b with their hypotenuses facing inward.
3The inner shape has four sides c and right angles because the acute triangle angles sum to 90°.
4Compare areas: $(a+b)^2=4\cdot\frac{ab}{2}+c^2$ .
5Expand: $a^2+2ab+b^2=2ab+c^2$ .
6Subtract 2ab on both sides: $a^2+b^2=c^2$ . A right angle remains essential.

Understand and apply the equation

Identify the side you need

Prerequisites: identify right triangles, convert length units, square numbers and take the positive square root.
A sketch helps identify sides. An angle that merely looks right does not replace a stated or established right angle.

Legs, hypotenuse and squares

a^2+b^2=c^2,\qquad a,b,c>0

This equation applies to a nondegenerate right triangle in Euclidean geometry.[1]

A proof by rearranging areas

(a+b)^2=4\cdot\frac{ab}{2}+c^2

The outer area consists of four triangle areas $\frac{ab}{2}$ plus the central square c².

a^2+2ab+b^2=2ab+c^2\quad\Longrightarrow\quad a^2+b^2=c^2

Expand and subtract 2ab from both sides.

Calculating a hypotenuse or a leg

\begin{aligned}c&=\sqrt{a^2+b^2}\\a&=\sqrt{c^2-b^2}\\b&=\sqrt{c^2-a^2}\end{aligned}

For geometric lengths, use the positive square root.

Three examples with different approaches

Diagonal of a rectangular panel

Given: Sides 1.20 m and 50 cm; adjacent sides are perpendicular.
Find: Calculate the diagonal.

Approach: The right-angle corner makes the diagonal the hypotenuse. $d=\sqrt{a^2+b^2}$ . Convert the second length to metres.
$b=0.50\,\mathrm m$
Add squares; the diagonal is the hypotenuse.
$c^2=(1.20)^2+(0.50)^2=1.69\,\mathrm{m^2}$
Take the positive square root.
$c=\sqrt{1.69}\,\mathrm m=1.30\,\mathrm m$

Result: The diagonal is 1.30 m long.

Check: It exceeds either side but is less than $1.20\,\mathrm m+0.50\,\mathrm m=1.70\,\mathrm m$ . Squaring returns 1.69 m².

Finding a leg from a diagonal

Given: Hypotenuse 3.9 m; one leg 1.5 m.
Find: Find the other leg.

Approach: The hypotenuse is the longest side; select the positive length. $a=\sqrt{c^2-b^2}$ . Subtract the known leg square from the hypotenuse square.
$a^2=(3.9)^2-(1.5)^2=15.21-2.25=12.96\,\mathrm{m^2}$
Take the positive square root.
$a=\sqrt{12.96}\,\mathrm m=3.6\,\mathrm m$
Check using the original equation.
$(3.6)^2+(1.5)^2=12.96+2.25=15.21=(3.9)^2$

Result: The missing leg is 3.6 m long.

Check: Direct subtraction $3.9\,\mathrm m-1.5\,\mathrm m=2.4\,\mathrm m$ would be incorrect. The calculated leg remains shorter than the hypotenuse.

Two coordinate changes, one distance

Given: On a planar map with axes in metres, $A=(-1,2)$ and $B=(7,-4)$ .
Find: Find the straight-line distance and distinguish it from a route along the axes.

Approach: horizontal and vertical coordinate differences are perpendicular components. Initially retain their signs.
$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$
Subtract the starting coordinate from the ending coordinate.
$\Delta x=7-(-1)=8,\qquad\Delta y=-4-2=-6$
Add the squares and choose the positive square root as a length.
$d=\sqrt{8^2+(-6)^2}\,\mathrm m=10\,\mathrm m$

Result: The straight-line distance is 10 m. Travelling horizontally and then vertically gives a 14 m route.

Check: The distance exceeds either component magnitude but is shorter than their sum. The negative vertical change describes direction only.

View diagram: coordinate differences form a right triangle

Straight-line distance and a route along the axes

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The perpendicular segments are 8 m and 6 m long; the direct connection is 10 m. Both axes use the same scale. The change

\Delta y=-6\,\mathrm m

is signed, while its geometric length is positive. This diagram accompanies the third worked example.[3]

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Checking a right angle from side lengths

From a triangle to distance

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For three perpendicular spatial axes, add $(z_2-z_1)^2$ under the square root.

h=\sqrt{k^2-\frac{a^2}{2}},\qquad a>0,\;k>\frac a{\sqrt2}

The positive root gives vertical height; equality would make the pyramid degenerate.[1]

Not every triangle is right

c^2=a^2+b^2-2ab\cos\gamma

The cosine rule includes angle γ between sides a and b.[4]

Do not simplify $\sqrt{a^2+b^2}$ to a + b. For $a=b=1$ the result is √2, not 2. Also round at the end; insufficient precision in intermediate values can distort the check.

The cosine rule extends the side calculation to a specified non-right angle. Use that relation when a right angle is not established; the appearance of a drawing is insufficient evidence.[4]

Five problems with solutions

Calculate · Foundation

Legs are 8 cm and 15 cm. Find the hypotenuse.

Show hint

Add squares, then take the square root.

Learning goal: Find hypotenuse from legs.

Show solution and calculation

Approach: The hypotenuse is opposite the right angle, so $c=\sqrt{a^2+b^2}$ .
Substitution gives $c=\sqrt{(8\,\mathrm{cm})^2+(15\,\mathrm{cm})^2}=17\,\mathrm{cm}$ .

Check: $(17\,\mathrm{cm})^2=64\,\mathrm{cm^2}+225\,\mathrm{cm^2}=289\,\mathrm{cm^2}$ . Taking a square root converts an area unit back to a length unit.

c=\sqrt{(8\,\mathrm{cm})^2+(15\,\mathrm{cm})^2}=17\,\mathrm{cm}

Calculate · Application

Hypotenuse 25 mm and one leg 7 mm. Find the other leg.

Show hint

Subtract the leg square from the hypotenuse square.

Learning goal: Find a missing leg.

Show solution and calculation

Approach: Rearrange $a^2+b^2=c^2$ for the missing positive length: $a=\sqrt{c^2-b^2}$ .
Substitution gives $a=\sqrt{(25\,\mathrm{mm})^2-(7\,\mathrm{mm})^2}=24\,\mathrm{mm}$ .

Check: $(24\,\mathrm{mm})^2+(7\,\mathrm{mm})^2=(25\,\mathrm{mm})^2$ .

a=\sqrt{(25\,\mathrm{mm})^2-(7\,\mathrm{mm})^2}=24\,\mathrm{mm}

Calculate · Application

A rectangle has sides 90 cm and 1.20 m. Find the diagonal in metres.

Show hint

Convert 90 cm to 0.90 m first.

Learning goal: Calculate rectangle diagonal with unit conversion.

Show solution and calculation

Approach: The rectangle diagonal is the hypotenuse, so $d=\sqrt{a^2+b^2}$ .
First convert $90\,\mathrm{cm}=0.90\,\mathrm m$ .
Then $d=\sqrt{(0.90\,\mathrm m)^2+(1.20\,\mathrm m)^2}=1.50\,\mathrm m$ .

Check: $(1.50\,\mathrm m)^2=2.25\,\mathrm{m^2}$ ; the diagonal is longer than either side.

d=\sqrt{(0.90\,\mathrm m)^2+(1.20\,\mathrm m)^2}=1.50\,\mathrm m

Interpret · Application

A nondegenerate triangle has sides 5, 6 and 8 cm, with 8 cm largest. A sketch makes its opposite angle look right. Determine from lengths whether this is possible and whether that angle is acute or obtuse.

Show hint

Compare the longest-side square with the sum of the other squares.

Learning goal: Test right-angle geometry from side lengths.

Show solution and calculation

Approach: The converse theorem requires equality of squares for a right angle.
Here 64 exceeds 61, so the angle is obtuse rather than right.
The cosine rule confirms a negative cosine.
The lengths also satisfy the triangle inequality.
An unscaled sketch cannot replace the length test.

\begin{aligned}8^2&=64>5^2+6^2=61\\{}\cos\gamma&=\frac{5^2+6^2-8^2}{2\cdot5\cdot6}=-\frac1{20}<0\end{aligned}

[2][4]

Error analysis · Application

A right-angled route runs 9 m east then 12 m north. Someone calls 21 m the direct distance by adding both segments. State path length and direct distance separately and explain the mistake.

Show hint

The hypotenuse joins start and finish rather than following the route.

Learning goal: Reject adding lengths in place of squares.

Show solution and calculation

Approach: Add lengths for the actual route, but add squared perpendicular components for the direct connection.
Path length is 21 m; distance is 15 m.

Check: the direct connection is shorter than the detour and longer than either component. The error confuses two distinct quantities.

\begin{aligned}s_{\mathrm{path}}&=9+12=21\,\mathrm m\\{}d&=\sqrt{9^2+12^2}\,\mathrm m=15\,\mathrm m\end{aligned}

[3]

What to remember for your solution

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added square-pyramid height with setup and check, connected relevant side, angle and volume relations and added a source on right-triangle trigonometry; preserved all prior exercises and figures.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified the introduction and prerequisites; added a third worked example, an original figure and a summary; made supplemental material selectively expandable.

Cite this reference page

Quanta. Pythagorean Theorem. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/pythagorean-theorem

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Frequently asked questions about Pythagorean Theorem

How do you calculate the hypotenuse with the Pythagorean theorem?+

How do you calculate a leg when the hypotenuse and the other leg are known?+

When does the Pythagorean theorem not apply?+

How do you use Pythagoras to calculate the distance between two points?+

What is a Pythagorean triple?+