In a right triangle, the two sides meeting at the right angle are the legs and the side opposite that angle is the hypotenuse. If the legs have lengths $a$ and $b$ and the hypotenuse has length $c$, then
$$a^2+b^2=c^2.$$
The statement is about areas as well as lengths: the square on the hypotenuse has the same area as the two squares on the legs combined. This editable lesson develops that idea from an area proof, then uses it for unknown sides, triangle classification, coordinate distance, and realistic applications. A separate teacher key gives complete calculations and misconception guidance.
A proof that explains the formula
Place four congruent right triangles, each with legs $a$ and $b$, inside a square of side $a+b$. Their four hypotenuses form a central quadrilateral with four equal sides of length $c$. At each central corner, the two acute angles of a right triangle meet and sum to $90^\circ$, so the central quadrilateral is a square. The large square has area $(a+b)^2$; the same region consists of four triangles of area $ab/2$ and the central square of area $c^2$. Therefore
$$ (a+b)^2=4\left(\frac{ab}{2}\right)+c^2. $$
Expanding both sides gives $a^2+2ab+b^2=2ab+c^2$. Subtracting $2ab$ proves $a^2+b^2=c^2$. This reasoning proves the relation for every right triangle represented by the construction; it does not merely confirm a few numerical triples.
Finding an unknown side
The hypotenuse is the longest side, so it must be identified before any substitution.
- Missing hypotenuse: $c=\sqrt{a^2+b^2}$.
- Missing leg: $a=\sqrt{c^2-b^2}$ or $b=\sqrt{c^2-a^2}$.
For legs $9$ cm and $12$ cm,
$$c=\sqrt{9^2+12^2}=\sqrt{225}=15\text{ cm}.$$
For a hypotenuse of $13$ m and a known leg of $5$ m,
$$b=\sqrt{13^2-5^2}=\sqrt{144}=12\text{ m}.$$
Not every answer is an integer. Legs $7$ and $11$ give $c=\sqrt{170}\approx13.0$. The radical is exact; the decimal is an approximation. Retaining the exact value until the final line prevents cumulative rounding error.
The converse and all three classifications
The converse runs in the opposite logical direction: if three side lengths form a triangle and satisfy $a^2+b^2=c^2$, with $c$ the longest side, then the triangle is right.
Sorting the lengths as $a\le b\le c$ gives a complete classification:
| Comparison | Triangle |
|---|---|
| $a^2+b^2=c^2$ | right |
| $a^2+b^2>c^2$ | acute |
| $a^2+b^2<c^2$ | obtuse |
The triangle inequality must come first. The lengths $3,4,8$ do not form a triangle because $3+4<8$, so they are not classified as obtuse.
For $9,12,15$, $9^2+12^2=225=15^2$, so the triangle is right. For $7,8,10$, $7^2+8^2=113>100$, so it is acute. For $5,7,10$, $5^2+7^2=74<100$, so it is obtuse.
Coordinate distance is the same theorem
The horizontal and vertical changes between $P(x_1,y_1)$ and $Q(x_2,y_2)$ are perpendicular legs. Their connecting segment is the hypotenuse, giving
$$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$
Between $P(-2,1)$ and $Q(4,9)$, the changes are $6$ and $8$, so $PQ=\sqrt{6^2+8^2}=10$. The distance formula is therefore not a disconnected rule to memorize; it is the Pythagorean theorem on a coordinate grid.
Representative practice
The editable student file contains twenty-four questions. These samples show the range.
- Find the exact and approximate hypotenuse of a right triangle with legs $7$ and $9$.
- Decide whether $8,10,13$ form an acute, right, or obtuse triangle, showing the comparison.
- Find the distance between $(-3,4)$ and $(5,-11)$.
- A $17$ m guy wire is attached $15$ m above level ground. How far is its anchor from the pole?
- Correct the claim $b=\sqrt{17-8}=3$ for a right triangle with hypotenuse $17$ and one leg $8$.
- A triangle has sides $12,35,37$. Classify it, then use the same calculation to find the distance between $(-5,2)$ and $(7,37)$.
The sequence begins with identifying the hypotenuse, moves through exact and approximate lengths, then tests the converse, coordinate distance, applications, and error analysis. The final item asks students to transfer one calculation between geometric and coordinate settings.
Classroom use and adaptation
The lesson is aimed at learners around ages 13-15 who already know squares, square roots, and basic triangle vocabulary. It can serve as a compact lesson with guided practice, a two-session handout, or a review assignment. Teachers can shorten it by selecting one worked example and Questions 1-13, or extend it by asking students to reconstruct the area proof without the final algebra.
The source is curriculum-neutral. Its metadata records relevant mappings to Common Core Grade 8 geometry, England Key Stage 3 geometry and measures, and New Zealand Years 9-10 right-triangle content. Local vocabulary, calculator policy, measurement units, and assessment expectations should still be adapted.
A careful historical note
The theorem bears Pythagoras’s name in much modern English usage, but right-triangle calculations appear in Old Babylonian mathematics more than a millennium before his lifetime, and a corresponding gougu rule is part of ancient Chinese mathematics. The surviving record does not support the simple claim that Pythagoras personally discovered the theorem. The student resource keeps this note brief so that history adds accuracy without replacing the mathematics.
Sources
- Common Core State Standards for Mathematics, Grade 8 geometry standards 8.G.B.6-8.
- National curriculum in England: mathematics programmes of study, Key Stage 3 geometry and measures.
- New Zealand mathematics and statistics curriculum, Years 9-10 right-triangle content.
- Euclid’s Elements, Book I, Proposition 47, including historical commentary and the connection to the converse.
- MacTutor biography of Pythagoras, for the caution required when attributing the result.
The diagrams were drawn for this package. The mathematical exposition and exercises are newly written from standard results and have been recomputed during drafting. Independent mathematics and classroom review remain required before publication.