Pythagoras theorem

Draw a right-angled triangle, build a square outward on each of its three sides, and something remarkable happens: the big square sitting on the hypotenuse has exactly the same area as the other two squares put together. That single fact — algebra hiding inside geometry — is Pythagoras theorem, and it is arguably the most-used result in all of school mathematics. It lets you find a missing side of a right triangle from the other two, check whether a triangle is right-angled at all using its converse, and it underpins mensuration problems for years afterwards: the diagonal of a rectangle, the height of an isosceles triangle, the distance between two points. This article builds the theorem from an area-based proof, works two numerical examples end to end, and flags the traps that lose the most marks.

The core insight

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides, $a^2+b^2=c^2$ , and this follows from comparing areas of squares built on each side.

The core idea

The statement. Let $\triangle ABC$ have a right angle at $C$ . Call the two sides meeting at $C$ the legs, and the side opposite $C$ — the longest side — the hypotenuse. Pythagoras theorem says

a^2 + b^2 = c^2,

where: $a$ and $b$ are the lengths of the two legs and $c$ is the length of the hypotenuse.

The area picture. The same statement can be read as a fact about areas. If a square is drawn outward on each of the three sides, then

\text{Area(square on hypotenuse)} = \text{Area(square on leg 1)} + \text{Area(square on leg 2)}.

Squaring a length gives the area of a square with that length as its side, so the two readings — numbers squared, and areas of squares — are the same statement in different clothing. This is exactly why the ICSE syllabus insists on an area-based proof: it is not enough to just quote the formula, one should be able to see it in the picture below.

Figure

Right triangle with squares drawn outward on all three sides showing hypotenuse square area equals the sum of the two leg square areas

Converse. If a triangle's sides satisfy $a^2+b^2=c^2$ for its longest side $c$ , the angle opposite $c$ must be a right angle. This runs the theorem backwards and turns it into a test: given three lengths, square the longest and compare it with the sum of the squares of the other two.

Derivation AArea-based proof of Pythagoras theorem

The goal is to show, using only the triangle-area theorems already known from the ICSE syllabus (triangles on the same base between the same parallels have equal area, and a triangle's area is half that of a parallelogram on the same base between the same parallels), that the square drawn outward on the hypotenuse of a right triangle has the same area as the two squares drawn outward on the legs combined, without assuming any trigonometry or coordinate algebra.

Step 1. Set up the figure. Let $\triangle ABC$ have its right angle at $C$ . Draw squares outward on all three sides: square $BCED$ on leg $BC$ , square $ACFG$ on leg $AC$ , and square $ABML$ on hypotenuse $AB$ . From $C$ , drop the altitude to $AB$ , meeting it at $N$ , and extend $CN$ to meet $LM$ at a point $P$ . This line splits the hypotenuse-square $ABML$ into two rectangles: $ANPL$ (nearer $A$ ) and $NBMP$ (nearer $B$ ).

Step 2. Compare the rectangle near $A$ with the square on $AC$ . Consider triangles $ACL$ and $ABG$ . Both sit on sides of equal length coming out of the square on $AC$ and the square on $AB$ respectively, and a careful angle-chase (using that $\angle CAB$ and $\angle GAL$ are both right angles plus a shared angle) shows $\triangle ACL \cong \triangle AGB$ by SAS, so they have equal area. But $\triangle ACL$ has the same area as half of rectangle $ANPL$ (same base $AL$ , between the same parallels), and $\triangle AGB$ has the same area as half of square $ACFG$ (same base $AG$ , between the same parallels). Doubling both equal halves gives $\text{Area}(ANPL) = \text{Area}(ACFG)$ , that is, the rectangle near $A$ equals the square on leg $AC$ .

Step 3. Repeat the same argument on the other side, comparing rectangle $NBMP$ with the square on leg $BC$ using triangles $BCD$ and $BAE$ ; congruence again gives $\text{Area}(NBMP) = \text{Area}(BCED)$ , the rectangle near $B$ equals the square on leg $BC$ .

Step 4. Add the two rectangle areas. Since the two rectangles together make up the whole square on the hypotenuse, $\text{Area}(ANPL) + \text{Area}(NBMP) = \text{Area}(ABML)$ . Substituting Step 2 and Step 3, $\text{Area}(ACFG) + \text{Area}(BCED) = \text{Area}(ABML)$ , which is exactly $AC^2 + BC^2 = AB^2$ , or in the usual letters $a^2+b^2=c^2$ . No coordinates, no trigonometry — only area theorems already available at this stage were used.

Worked example

A right-angled triangle has legs $9\text{ cm}$ and $12\text{ cm}$ . Find its hypotenuse, and then check separately whether a second triangle with sides $7\text{ cm}$ , $24\text{ cm}$ and $25\text{ cm}$ is right-angled.

Step 1. Apply the theorem to the first triangle. With legs $a=9$ and $b=12$ , the hypotenuse $c$ satisfies $c^2 = a^2+b^2 = 9^2+12^2 = 81+144=225$ .

Step 2. Take the square root: $c=\sqrt{225}=15\text{ cm}$ . This is the $9$ – $12$ – $15$ triple, a scaled copy of the basic $3$ – $4$ – $5$ triple (multiply every side by $3$ ); any such scaled copy of a Pythagorean triple automatically satisfies $a^2+b^2=c^2$ again.

Step 3. Now test the second triangle with the converse. Its longest side is $25\text{ cm}$ , so check whether $25^2$ equals the sum of the squares of the other two sides: $7^2+24^2 = 49+576=625$ , and $25^2=625$ too.

Step 4. Since $7^2+24^2=25^2$ , the converse of Pythagoras theorem confirms the second triangle is right-angled, with the right angle opposite the $25\text{ cm}$ side — that is, at the vertex between the sides of length $7\text{ cm}$ and $24\text{ cm}$ . Final answers: hypotenuse $=15\text{ cm}$ for the first triangle; the second triangle is right-angled, confirmed by the converse.

Competitive stretch

If instead the hypotenuse ( $15\text{ cm}$ ) and one leg ( $9\text{ cm}$ ) of the first triangle were given, the missing leg follows by rearranging: $a^2=c^2-b^2 = 15^2-9^2=225-81=144$ , so $a=\sqrt{144}=12\text{ cm}$ — matching the original data and showing the theorem works equally well to recover either a leg or the hypotenuse.

Common traps

Squaring the wrong side as the hypotenuse. The hypotenuse is always opposite the right angle (and the longest side); plugging a leg into the " $c^2$ " slot gives a wrong equation.
Adding lengths instead of squares. Pythagoras theorem relates squares of lengths, not the lengths themselves; $a+b \ne c$ in general.
Forgetting the final square root. Solving for $c^2$ is not the answer — the length needs $\sqrt{c^2}$ .
Testing the wrong pair with the converse. Always square the largest of the three given sides and compare it with the sum of squares of the other two; testing a different pair gives a false negative.

Ask the tutor: Want the full area-rearrangement proof walked through step by step with a hands-on diagram, or more practice problems mixing the theorem and its converse? Ask the AI tutor.

Why does squaring the sides matter — why not just add the two legs to get the hypotenuse?

Because the relationship is genuinely about areas, not lengths. Picture the squares drawn outward on the three sides: the theorem says the area of the hypotenuse-square equals the combined area of the two leg-squares, and area scales with the square of a length, not the length itself. Try it numerically with legs 9 and 12: adding gives 21, but the true hypotenuse is 15, since $9^2+12^2=225=15^2$ . Only the squared version balances.

How do I quickly tell if three given numbers can be the sides of a right triangle without drawing anything

Square all three numbers, then check whether the square of the largest equals the sum of the squares of the other two. For 7, 24, 25: $7^2+24^2=49+576=625$ , and $25^2=625$ , so yes. For 7, 10, 13: $7^2+10^2=49+100=149$ , but $13^2=169$ , so no — this triangle is not right-angled. This converse test is exactly Pythagoras theorem run backwards, and it needs no diagram at all.

Is it worth memorising triples like 3-4-5 and 5-12-13, or should I always calculate from scratch

Both skills matter, but memorising a handful of small triples (3–4–5, 5–12–13, 7–24–25, and their multiples like 6–8–10 or 9–12–15) is a genuine time-saver, since these appear constantly in exam questions and mental recognition is faster than squaring and adding under time pressure. That said, always be ready to calculate from scratch for numbers that are not a known triple — most right triangles in real problems will not have neat integer sides, and $c=\sqrt{a^2+b^2}$ then gives an irrational answer, which is perfectly normal.

Pythagoras theorem

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads