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Home›Geometry and drawings›Triangles

The Pythagorean theorem, two proofs and the converse

On squared paper, a right triangle with legs of 3 and 4 squares gets squares of 9 and 16 on its legs, and the square on its longest side covers exactly 25.

Count the squares first

Take squared paper and draw a right triangle whose two shorter sides run 3 squares across and 4 squares up. On each side, draw a square that sits outside the triangle. The square on the side of 3 holds 9 small squares and the one on the side of 4 holds 16. The square on the long, slanted side is tilted, so counting is awkward, but measure its side with a ruler against the grid and you get 5, which makes 25 squares. And 9 + 16 = 25. Try legs of 6 and 8: the squares hold 36 and 64, together 100, and the long side measures 10.

Squares on the three sides of a right triangle with legs 3 and 4. The two small squares hold 9 and 16 unit squares, and the tilted square on the hypotenuse holds 25.

In a right triangle the side across from the right angle is the hypotenuse. It is always the longest side, because it faces the largest angle. The two sides that form the right angle are the legs. The pattern you counted is the Pythagorean theorem: in any right triangle, the square of the hypotenuse equals the sum of the squares of the legs, a² + b² = c², where c is the hypotenuse. OpenStax Prealgebra states it in exactly that form. Squaring does not mean the drawing has to show squares; you can use the formula on the lengths alone.

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With the angle at 37° and the hypotenuse at 5, the legs come out at about 3.009 and 3.993, close to 3 and 4. Change either slider and square the two legs on the panel: their sum always equals the hypotenuse squared.

A proof you can cut out of paper

Counting squares works for 3 and 4, but a proof must work for every right triangle. Cut four copies of any right triangle with legs a and b and hypotenuse c. Arrange them inside a square of side a + b in two ways. In the first, put one triangle in each corner so that their hypotenuses enclose a tilted square in the middle; that square has area c². In the second, pair the triangles into two rectangles a by b and push them into opposite corners; the empty space is a square of side a and a square of side b. The big square and the four triangles are the same in both pictures, so the empty space must be the same too: c² = a² + b².

The same square of side a + b holds four copies of the triangle in two ways. On the left the leftover space is c²; on the right it is a² and b².

The same picture gives an algebra proof. The big square has area (a + b)² = a² + 2ab + b². In the left picture that area is the tilted square plus four triangles: c² + 4 × (ab ÷ 2) = c² + 2ab. Set the two equal and the 2ab cancels, leaving a² + b² = c².

BackgroundThe theorem is Proposition 47 of Book I of Euclid’s Elements. In the D. E. Joyce edition, the notes explain that the name Pythagorean theorem was given by Proclus and others centuries after Pythagoras, who OpenStax says lived around 500 BCE. The relation itself is far older. Old Babylonian scribes, around 1900 to 1600 BCE, used it to solve problems, and the clay tablet known as Plimpton 322 shows they could construct Pythagorean triples. In China, the Zhou bi suan jing contains a hypotenuse diagram, probably added by the commentator Zhao Shuang in the third century: four 3 by 4 rectangles placed around a 1 by 1 square make a 7 by 7 square, and the tilted square inside has area 49 − 4 × 6 = 25, so its side is 5. Joyce adds that, according to Proclus, the proof in the Elements is Euclid’s own.
  • c. 1900 to 1600 BCE: Old Babylonian scribes use the relation; the tablet Plimpton 322 lists Pythagorean triples.
  • 6th century BCE: Thales measures a pyramid by its shadow (unit 5).
  • c. 500 BCE: Pythagoras lives in the Greek world.
  • about 2,400 years ago: Euclid proves the theorem (Prop. 47) and its converse (Prop. 48).
  • 3rd century CE: Zhao Shuang’s hypotenuse diagram in the Zhou bi suan jing.

The converse tells you when a corner is square

Proposition 48 runs the theorem backwards. If the square on one side of a triangle equals the sum of the squares on the other two, the angle between those two sides is a right angle. This converse is what builders use with a tape measure: mark 3 m along one wall line and 4 m along the other, and if the marks are exactly 5 m apart, the corner is square. Sides of 5, 12 and 13 work the same way, since 25 + 144 = 169.

The comparison also sorts triangles that are not right. Let c be the longest side. If c² is less than a² + b², the angle across from c is smaller than a right angle and the triangle is acute. If c² is more, the angle is obtuse. Sides 6, 7, 9: 81 is less than 36 + 49 = 85, so acute. Sides 5, 6, 9: 81 is more than 25 + 36 = 61, so obtuse.

Ladders, banners and shortcuts

Most uses are about finding one side from the other two. OpenStax sets a 13 ft ladder 5 ft from a wall: the top reaches √(13² − 5²) = √144 = 12 ft up. A 20 ft ladder 6 ft out reaches √(400 − 36) = √364, about 19.1 ft. For a shortcut closer to home, picture a playground 80 m long and 60 m wide. Walking round two edges is 140 m; the diagonal is √(6400 + 3600) = √10000 = 100 m, so cutting across saves 40 m.

  1. The legs of a right triangle are 3 and 4. Find the hypotenuse.
  2. The legs of a right triangle are 5 and 12. Find the hypotenuse.
  3. A 13 ft ladder stands 5 ft from a wall. How high up the wall does it reach?
  4. A 20 ft ladder stands 6 ft from a house. How high does it reach, to one decimal place?
  5. A garage door is 12 ft high and 16 ft wide. How long is a banner stretched corner to corner?
  6. A playground is 80 m by 60 m. How many metres do you save by walking the diagonal instead of two edges?
  7. A bracket of 10 inches is fixed across a square corner with both ends the same distance from the corner. How far from the corner is each end, to one decimal place?
  8. Sides 7, 24 and 25. The triangle is:
    • right
    • acute
    • obtuse
    • impossible
  9. Sides 6, 7 and 9. The triangle is:
    • acute
    • right
    • obtuse
    • impossible
  10. Sides 5, 6 and 9. The triangle is:
    • obtuse
    • acute
    • right
    • impossible
  11. In the Zhou bi diagram, four 3 by 4 rectangles surround a 1 by 1 square to make a 7 by 7 square. What is the area of the tilted square inside?
  12. To check a corner with the 3, 4, 5 rule, you mark 6 m and 8 m along the two lines. What distance between the marks means the corner is square?
Common mistakes
  • “The legs are 6 and 8, so the hypotenuse is 6 + 8 = 14.” Add the squares, not the lengths: 36 + 64 = 100, so the hypotenuse is 10. The triangle inequality says it must be less than 14 anyway.
  • “A ladder 13 long, 5 from the wall: √(13² + 5²) ≈ 13.9 high.” The ladder is the hypotenuse, so subtract: √(169 − 25) = 12. A leg can never be longer than the hypotenuse.
  • “5² + 6² ≠ 9², so the theorem fails here.” The theorem is only about right triangles. For 5, 6, 9 the inequality 81 > 61 tells you the triangle is obtuse, which is the converse at work.
Teacher's noteIn the 2022 Korean curriculum the Pythagorean theorem is in the middle school band. The explanation asks teachers to have learners conjecture it through varied activities and then justify it, while the converse is to be understood intuitively and used to decide whether three given sides make a right triangle. The curriculum also suggests using its history to build interest. Counting squares on graph paper suits the conjecture stage well. For justification, the four-triangle rearrangement can be cut from card in a few minutes, and learners who see both arrangements side by side rarely need the algebra explained twice.
Key points
  • Right triangle with legs a, b and hypotenuse c: a² + b² = c².
  • Proof: four copies in a square of side a + b, arranged two ways.
  • Converse: if c² = a² + b², the angle opposite c is 90°.
  • c² < a² + b² means acute; c² > a² + b² means obtuse (c the longest side).
Sources for this course (7)
  • Curriculum: 교육부 고시 제2022-33호 [별책 8] 수학과 교육과정 (ncic.re.kr/inv/org/download.do?year=2022&seq=10003559&orgType=ogi4, PDF read 2026-10-19: 초등학교 3~4학년군 [4수03-08] 이등변삼각형·정삼각형, [4수03-09] 직각삼각형·예각삼각형·둔각삼각형, [4수03-25] 삼각형과 사각형의 내각의 크기의 합을 여러 가지 방법으로 추론; 초등학교 5~6학년군 [6수03-01] 합동·대응점·대응변·대응각, [6수03-14] 삼각형의 넓이(높이가 삼각형의 외부에 있는 것도 다룸); 중학교 1~3학년군 [9수03-03] 삼각형의 작도, [9수03-04] 삼각형의 합동 조건, [9수03-05] 삼각형의 내각과 외각, [9수03-09] 이등변삼각형의 성질을 합동 조건으로 정당화하고 증명이라는 용어 도입, [9수03-12]·[9수03-13] 닮음·닮음비·삼각형의 닮음 조건, [9수03-15] 피타고라스 정리를 추측하고 정당화, 역은 직관적으로 이해하고 직각삼각형 판별에 사용; 용어와 기호 대변, 대각, 내각, 외각, 증명, 닮음비, △ABC, ≡, ∽)
  • Textbook: OpenStax, Contemporary Mathematics, 10.3 Triangles (triangle from three noncollinear points; acute, right, obtuse; side a opposite angle A; hypotenuse and legs; angle sum 180° by a parallel line and alternate interior angles; x, x + 17, 3x − 62 gives 45°, 62°, 73°; SSS, SAS, ASA, AAS; similar triangles, scaling factor; tower shadow 180 ft with a 9 ft truck and 15 ft shadow gives 108 ft; Thales and the pyramid: openstax.org/books/contemporary-mathematics/pages/10-3-triangles)
  • OpenStax Prealgebra 2e, 9.3 Use Properties of Angles, Triangles, and the Pythagorean Theorem (a² + b² = c²; Pythagoras around 500 BCE; 13 ft ladder 5 ft from the wall; 10 inch bracket about 7.1 inches; similar triangles: openstax.org/books/prealgebra-2e/pages/9-3-use-properties-of-angles-triangles-and-the-pythagorean-theorem)
  • 9.4 Use Properties of Rectangles, Triangles, and Trapezoids (A = ½bh from half a rectangle; isosceles and equilateral; equilateral perimeter 93 in: .../pages/9-4-use-properties-of-rectangles-triangles-and-trapezoids)
  • OpenStax Precalculus 2e, 8.2 Non-right Triangles: Law of Cosines (Heron’s formula with semi-perimeter s; sides 10, 15, 7 give s = 16 and area about 29.4; Chicago lot 62.4, 43.5, 34.1 m gives about 711.4 m²: openstax.org/books/precalculus-2e/pages/8-2-non-right-triangles-law-of-cosines)
  • History for story blocks only: Euclid’s Elements, D. E. Joyce edition, Clark University (Book I Prop. 20 triangle inequality and Heron’s shortest-path problem; Prop. 26 and its guide on congruence theorems, SSA ambiguity and side-side-right angle; Prop. 32 exterior angle and angle sum, Proclus corollaries; Prop. 47 and its guide on Old Babylonians ca. 1900 to 1600 BCE, Plimpton 322, the Zhou bi hypotenuse diagram, Proclus; Prop. 48 converse; Book VI Prop. 4 equiangular triangles: mathcs.clarku.edu/~djoyce/elements/bookI/propI47.html)
  • MacTutor History of Mathematics, University of St Andrews, Heron of Alexandria (formula proved in Metrica Book I; eclipse of 13 March 62; side of 720 about 26 5/6: mathshistory.st-andrews.ac.uk/Biographies/Heron/) MacTutor, "Euclid of Alexandria" (born about 325 BC, died about 265 BC; dating for the Elements, about 2,300 years ago: mathshistory.st-andrews.ac.uk/Biographies/Euclid/, checked 2026-10-19).

Checked: 2026-10-19