The **Pythagorean theorem** states that if a triangle has one right angle, then the square of the longest side, called the hypotenuse, is equal to the sum of the squares of the lengths of the two shorter sides, called the legs. So if \( a \) and \( b \) are the lengths of the legs, and \( c \) is the length of the hypotenuse, then \(a^2+b^2=c^2\).

The theorem is a fundamental building block of geometry and has numerous applications in physics and other real-world situations. It is also the basis for the distance formula in coordinate geometry.

The converse of the theorem is also true: if \( a^2+b^2=c^2 \), then a triangle with side lengths \(a, b, c\) will be a right triangle, with right angle between the sides of lengths \(a\) and \(b\). One sample application of the converse is in construction: to measure a right angle given various lengths of rope, a surveyor can use ropes of length 3,4,5 (a Pythagorean triple) to make a triangle. The angle between the two shorter sides will be a right angle.

#### Contents

- Examples
- Pythagorean Triples
- Trigonometric Identities
- Proof of the Theorem
- See Also

## Examples

## Given a right triangle with leg lengths of 5 and 12, what is the length of the hypotenuse?

By the Pythagorean theorem, \( 5^2 + 12^2 = h^2 \), so \( h^2 = 169 \), which means \( h = 13\). Thus, the length of the hypotenuse is \(13\). \(_\square\)

Here is an example motivating the distance formula:

What is the distance between the points \( A=(2,3) \) and \( B=(6,11)?\)

Letting point \( C = (6,3) \), we have \( \triangle ABC \) with side lengths \( \overline{AC} = 4 \) and \( \overline{BC} = 8 \). Further, since \( AC \) and \( BC \) are parallel to the \(y\)- and \(x\)-axes, respectively, they are perpendicular and form the legs of a right triangle with \(AB\) as the hypotenuse.

Then, by the Pythagorean theorem, we have \( \overline{AC}^2 + \overline{BC}^2 = \overline{AB}^2 \), which gives us \( 4^2 + 8^2 = 16 + 64 = 80 = \overline{AB}^2 \). So \( \overline{AB} = \sqrt{80} = 4\sqrt{5} \). \(_\square\)

For a full discussion of this technique, see Distance Formula.

Is triangle \( XYZ \) a right triangle?

6-7-9 triangle

If \( XYZ \) is a right triangle, then the side lengths will follow the relationship \( a^2 + b^2 = c^2 \), where \(c\) is the longest side. However, \( 6^2 + 7^2 = 85 \not = 9^2 = 81 \). So \(XYZ\) is not a right triangle. \(_\square\)

15 16 17 18

Right triangle \(ABC\) has side lengths \(N+1, N-1,\) and \( \frac {N}{2}.\)

What is the value of \(N?\)

## Given a triangle with side lengths of 5, 12, and 14, how would you classify the largest angle in the triangle? \[\]

\(\quad\) A) Right

\(\quad\) B) Acute

\(\quad\) C) Obtuse

\(\quad\) D) Not enough informationBy the Pythagorean theorem, we know that a triangle with side lengths 5, 12, and 13 is a right triangle since \( 5^2 + 12^2 = 13^2 \). Thus, this triangle cannot be a right triangle.

Furthermore, since the longest side is greater than the equivalent hypotenuse in a right triangle, i.e. \( 14 > 13 \), we know that the angle which subtends that side must also be greater. Thus, the angle is obtuse and the answer is C). \(_\square\)

The diagram above shows a big rectangular lawn cut into two smaller rectangles, by the line \(BC\), with dimensions \(3 \times 12\) and \(6 \times 12\) each. What is the length of \(AD?\)

Circle \(O\) is inscribed in \(\triangle ABC,\) as shown in the figure above. The points of tangency are at \(D,E,\) and \(F\). Given that \(AD=2\) and \(DC=3\), find the area of \(\triangle ABC\).

Only \([1]\) \([2]\) and \([3]\) \([1]\), \([2]\), and \([3]\) None of them are correct

You have a right triangle that has integer sides. Read the following statements about that triangle:

\([1]\) At least one of the sides of that triangle must be divisible by \(3\).

\([2]\) At least one of the sides of that triangle must be divisible by \(4\).

\([3]\) At least one of the sides of that triangle must be divisible by \(5\).

Which of these statements is(are) correct?

\(\) **Note**: This problem is a part of this set.

Find the area of triangle \(ABC\) with side lengths \(AB = \sqrt{13}\), \(BC = \sqrt{17}\), and \(CA = \sqrt{20}\).

**Bonus**: Can you calculate without using the cosine rule?

Find the radius of the circle in the diagram.

Below is an isosceles triangle. Find the length of the base, \(x.\)

\[a+b+c\] \[\frac{\sqrt{3}}{3}(a+b+c)\] \[a^2+b^2+c^2-a-b-c\] \[\sqrt{ab+bc+ac}\] \[\sqrt{a^2+b^2+c^2}\] \[\sqrt[3]{abc}\] \[\sqrt[3]{a^3+b^3+c^3}\] \[\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\]

\(O,X,Y,Z\) are four points on a 3D Cartesian space, where \(O\) is the origin, \(X\) a point on the \(x\)-axis, \(Y\) a point on the \(y\)-axis, and \(Z\) a point on the \(z\)-axis.

Now, the area of \(\triangle OXY\) is given by \(a\), the area of \(\triangle OYZ\) by \(b\), and the area of \(\triangle OXZ\) by \(c\).

Find the area of \(\triangle XYZ\) in terms of \(a,b,c\).

## Pythagorean Triples

Ordered triples of *positive integers* \( (a,b,c) \) such that \( a^2+b^2=c^2\) are called **Pythagorean triples**. There are infinitely many such triples with \( \text{gcd}(a,b,c) = 1 \), and there is a simple parameterization of these Pythagorean triples which generates them all. This is a fundamental example of a nontrivial, nonlinear Diophantine equation. Fermat's attempt to generalize this equation by studying solutions to \( a^n+b^n = c^n\) led to his famous "last theorem." For more on this subject, see the wiki on Pythagorean triples.

## Trigonometric Identities

Basic trigonometric identities are consequences of the Pythagorean theorem. For instance, in a right triangle with sides \(a,b,c,\) the trigonometric functions are defined for the angle \( \theta \) opposite \( a \) by

\[\sin \theta = \frac{a}{c}, \ \cos \theta = \frac{b}{c}.\]

So

\[\begin{align}\sin^2 \theta+\cos^2 \theta &= \frac{a^2}{c^2} + \frac{b^2}{c^2} \\&= \frac{a^2+b^2}{c^2} \\&= \frac{c^2}{c^2} = 1.\end{align}\]

For more on this and other related identities, see Pythagorean identities.

## Proof of the Theorem

For additional proofs of the Pythagorean theorem, see: Proofs of the Pythagorean Theorem.

There are many unique proofs (more than 350) of the Pythagorean theorem, both algebraic and geometric. The proof presented below is helpful for its clarity and is known as a proof by rearrangement.

In a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse.

Given any right triangle with legs \( a \) and \(b \) and hypotenuse \(c,\) make a square with sides \( a+b\) and inscribe four copies of the given triangle as shown below:

This forms a square in the center with side length \( c \) and thus an area of \( c^2. \)

However, if we rearrange the four triangles as follows, we can see two squares inside the larger square, one that is \( a^2 \) in area and one that is \( b^2 \) in area:

Since the larger square is the same in both cases, i.e. \( (a+b)^2 \), and since the four triangles are the same in both cases, we must conclude that the two squares \( a^2 \) and \( b^2 \) are in fact equal in area to the square \( c^2 \).

Thus, \( a^2 + b^2 = c^2 .\) \( _\square \)

The converse of the Pythagorean theorem is a special case of the cosine rule:

## If the sides of a triangle satisfy \( a^2 + b^2 = c^2 \), then the angle opposite \(c\) is a right angle.

Let \( \angle ACB = \theta \); then the cosine rule gives

\[ \cos \theta = \frac{ a^2 + b^2 - c^2 } { 2ab} = \frac{0}{ 2ab} = 0 . \]

Since \( 0 ^ \circ \leq \theta \leq 180 ^ \circ \), the only solution is \( \theta = 90 ^ \circ \). \(_\square\)

## See Also

- Triangles
- Pythagorean Triples
- Distance Formula
- Pythagorean Identities
- Fermat's Last Theorem