Pythagorean Theorem

The theorem is an area relationship

A square of side length $a$ has area $a^2$. The equation says that the areas of the squares on the legs add to the area of the square on the hypotenuse. It does not say that the leg lengths themselves add to the hypotenuse.

For a $3$–$4$–$5$ right triangle, the areas are $9$, $16$ and $25$ square units. The relation is $9+16=25$, not $3+4=5$.

An alternative dissection moves pieces of the two leg-squares directly into the hypotenuse-square. The next illustration shows that different arrangement; it is not the four-triangle layout used in the algebra above.

An area proof

Arrange four copies of a right triangle with legs $a,b$ around a central square inside a larger square of side $a+b$. Each central side is a hypotenuse of length $c$. The two acute angles of a right triangle add to $90^\circ$, which makes every corner of the central region a right angle. It is therefore a square, not just a four-sided gap.

Calculate the large area in two ways:

Expanding and subtracting the triangular areas gives

This argument works for arbitrary positive leg lengths. The particular diagram helps explain it; the algebra supplies the general relationship. Euclid I.47 provides a different geometric proof.

Finding the hypotenuse

A right triangle has legs $6\text{ cm}$ and $8\text{ cm}$. Substitute those lengths before taking a square root:

Only the positive square root represents a side length. The negative solution to the algebraic equation $c^2=100$ does not represent another physical length.

If the legs are $2\text{ cm}$ and $5\text{ cm}$, the exact answer is

Keep $\sqrt{29}$ during later calculations. Rounding it early can introduce avoidable error.

Finding a missing leg

If the hypotenuse is known, subtract the known leg’s square from the hypotenuse’s square. A $13\text{ m}$ hypotenuse and a $5\text{ m}$ leg give

A common error is to calculate $\sqrt{13^2+5^2}$. That treats the known hypotenuse as a leg and produces a new length larger than it, contradicting the geometry.

An idealized ladder of length $5\text{ m}$ resting against a vertical wall with its foot $1.4\text{ m}$ from the wall reaches

This calculation assumes a right angle between wall and ground and measures the ladder as a straight segment. It is a geometric model, not guidance on safe ladder placement.

The converse: checking whether an angle is right

The theorem begins with a right triangle and concludes a side equation. Its converse begins with positive side lengths that form a triangle. If the square of the longest side equals the sum of the other two squares, the angle opposite that longest side is right.

For sides $8$, $15$ and $17$,

The triangle is right-angled. For $6$, $7$ and $10$, the triangle inequality holds, but $6^2+7^2=85<100=10^2$. The largest angle is obtuse. For $6$, $7$ and $8$, $85>64$, so the largest angle is acute and the triangle is acute. These comparisons follow more generally from the cosine rule.

The equality test is exact. Rounded measurements that nearly satisfy it support an approximate assessment, not an exact theorem about the underlying object.

Diagonals and coordinates

A rectangle’s diagonal is a hypotenuse because adjacent sides meet at right angles. A rectangle $9\text{ cm}$ by $12\text{ cm}$ has diagonal $15\text{ cm}$.

Coordinate differences provide perpendicular legs as well. Between $(1,-2)$ and $(7,6)$, the horizontal difference is $6$ and the vertical difference is $8$, so the straight-line distance is $10$ coordinate units. Distance Between Two Points derives the general distance formula from this construction.

In a rectangular box with mutually perpendicular edges $2$, $3$ and $6$, apply the theorem twice. A base diagonal has square $2^2+3^2=13$. The space diagonal therefore has square $13+6^2=49$ and length $7$. This works because the height is perpendicular to the base plane.

What the theorem does not allow

For a non-right triangle, $a^2+b^2=c^2$ is not the general side relation. Nor can the theorem be applied to three convenient lengths in a diagram without showing that they are sides of the same right triangle.

Squaring a sum is another frequent trap: $(a+b)^2=a^2+2ab+b^2$, not $a^2+b^2$. The proof above uses the middle term deliberately; omitting it would destroy the area argument.

Before calculating, identify the right angle, label the opposite side, confirm common units and decide whether a sum or difference of squares is needed. After calculating, check that every leg is shorter than the hypotenuse.

Related topics

Use Distance Between Two Points for coordinate distances and Trigonometric Ratios when a right-triangle problem also involves acute angles. Proof in Geometry explains the difference between a theorem and its converse.