Pythagorean Triples and the Right Triangle Test | Grade 8 Math | LevelKite

Can three side lengths alone tell you that a triangle has a right angle? In this lesson you learn a test that answers that question. It starts from a fact you have already met, the Pythagorean theorem.

A right triangle is a triangle with one right angle ( $90^\circ$ ). The two sides that form the right angle are the legs, and the side opposite the right angle is the hypotenuse. Let $a$ and $b$ be the lengths of the legs and $c$ the length of the hypotenuse. All three are positive lengths measured in the same unit. The theorem says that in every right triangle, the squares of the two leg lengths add up to the square of the hypotenuse length. Here $a^2$ means $a \times a$ . This lesson uses the theorem as a known fact and does not prove it.

a^2 + b^2 = c^2 \quad \text{(right triangle with hypotenuse } c\text{)}

The theorem starts with a right triangle and ends with an equation. In this lesson you go the other way: you start with three numbers, test them in the equation, and decide what the result tells you.

Begin with whole numbers. A Pythagorean triple is a set of three positive whole numbers $a$ , $b$ and $c$ that make $a^2 + b^2 = c^2$ true. Positive whole numbers are the counting numbers $1$ , $2$ , $3$ and so on. A well-known triple is $3$ , $4$ , $5$ . The two smaller squares are $3^2 = 9$ and $4^2 = 16$ , and they add to $9 + 16 = 25$ . The largest square is $5^2 = 25$ . The two smaller squares add up exactly to the largest square, and that exact match is what makes a triple.

Before you test three numbers, you must decide which one takes the place of $c$ . It is always the largest, and here is why. In $a^2 + b^2 = c^2$ with positive numbers $a$ , $b$ and $c$ , the number $c^2$ equals $a^2$ plus the positive number $b^2$ , so $c^2$ is greater than $a^2$ . In the same way, $c^2$ is greater than $b^2$ . Squaring keeps positive numbers in the same order, because a larger positive number times itself gives a larger product. So the number with the greatest square is the greatest number, and that number is $c$ .

The check for a triple therefore has three parts: all three numbers are positive whole numbers, the largest is $c$ , and $a^2 + b^2$ equals $c^2$ exactly. If any part fails, the numbers are not a Pythagorean triple.

Worked example

Checking for a Pythagorean triple

Is $8$ , $15$ , $17$ a Pythagorean triple?

Check the kind of numbers. A Pythagorean triple needs three positive whole numbers, and $8$ , $15$ and $17$ are all positive whole numbers.
Choose $c$ . The largest number is $17$ , so it takes the place of $c$ . Only the largest can, because $c^2$ is the sum of two positive squares, so it is greater than each of them.
$a = 8,\quad b = 15,\quad c = 17$
Square the two smaller numbers and add the results to get $a^2 + b^2$ , the left side of the equation. Here $8 \times 8 = 64$ and $15 \times 15 = 15 \times 10 + 15 \times 5 = 150 + 75 = 225$ .
$8^2 + 15^2 = 64 + 225 = 289$
Square the largest number to get $c^2$ , the right side. Split the product to make it easier: $17 \times 17 = 17 \times 10 + 17 \times 7 = 170 + 119 = 289$ .
$17^2 = 289$
Compare the two results. Both are exactly $289$ , so $a^2 + b^2 = c^2$ is true for these numbers.
$8^2 + 15^2 = 17^2$

Answer

Yes. $8$ , $15$ , $17$ is a Pythagorean triple, because the numbers are positive whole numbers and $8^2 + 15^2 = 17^2$ , with both sides equal to $289$ .

The same check works on any three positive numbers. The table tests each set with its largest number in the place of $c$ , wherever that number sits in the list. Look closely at the last two rows. A set is not a Pythagorean triple when the two smaller squares do not add up to the largest square. It is also not a triple when they do add up to it but the numbers are not all positive whole numbers.

Testing sets of three numbers for a Pythagorean triple

Numbers	Largest number	Sum of the two smaller squares	Largest square	Equal?	Pythagorean triple?
$3$ , $4$ , $5$	$5$	$9 + 16 = 25$	$5^2 = 25$	Yes	Yes
$5$ , $12$ , $13$	$13$	$25 + 144 = 169$	$13^2 = 169$	Yes	Yes
$10$ , $6$ , $8$	$10$	$36 + 64 = 100$	$10^2 = 100$	Yes	Yes
$7$ , $24$ , $25$	$25$	$49 + 576 = 625$	$25^2 = 625$	Yes	Yes
$4$ , $5$ , $6$	$6$	$16 + 25 = 41$	$6^2 = 36$	No	No, because $41 \neq 36$
$1.5$ , $2$ , $2.5$	$2.5$	$2.25 + 4 = 6.25$	$2.5^2 = 6.25$	Yes	No, because $1.5$ and $2.5$ are not whole numbers

Now connect the numbers to triangles. The Pythagorean theorem is an if-then statement: if a triangle is a right triangle with legs $a$ and $b$ and hypotenuse $c$ , the side opposite the right angle, then $a^2 + b^2 = c^2$ .

The converse of an if-then statement swaps the two parts. A converse is a new statement, and it is not automatically true. For example, if a whole number ends in the digit 0, then it is even is true. Its converse, if a whole number is even, then it ends in the digit 0, is false, because $12$ is even and does not end in $0$ .

The converse of the Pythagorean theorem is true: if a triangle has positive side lengths $a$ , $b$ and $c$ , with $c$ the largest, and $a^2 + b^2 = c^2$ , then it is a right triangle, and its right angle is opposite the side of length $c$ . In this lesson you use the converse as a known fact. It tells you that every Pythagorean triple gives the side lengths of a right triangle. For example, a triangle with sides $3$ , $4$ and $5$ units long is a right triangle.

Key concept

The side-length test for a right triangle

Take three positive lengths in the same unit. Call the largest $c$ and the other two $a$ and $b$ . Compute the squares exactly, then compare $a^2 + b^2$ with $c^2$ .

Equal: the lengths form a right triangle, and the right angle is opposite the side of length $c$ . This is the converse of the Pythagorean theorem.
Not equal: the lengths cannot form a right triangle. If they did, the hypotenuse, the side opposite the right angle, would be the longest side $c$ , and the theorem would make the two results equal.

Lengths are never negative, so a length is the nonnegative square root of its square, and the longest side has the greatest square. If the numbers are positive whole numbers and the results are equal, they are also a Pythagorean triple.

a^2 + b^2 = c^2 \quad \text{with } c \text{ the largest length}

Worked example

A near miss

A triangle has side lengths $8$ cm, $9$ cm and $12$ cm. Is it a right triangle?

Check the conditions and choose $c$ . All three lengths are positive and in centimeters. The largest is $12$ , so $c = 12$ . If this triangle had a right angle, the side opposite it, the hypotenuse, would be the longest side, so the $12$ cm side is the only one that could be the hypotenuse.
$a = 8,\quad b = 9,\quad c = 12$
Square the two shorter lengths exactly and add the results to get $a^2 + b^2$ . Each square is a number of square centimeters.
$8^2 + 9^2 = 64 + 81 = 145$
Square the longest length to get $c^2$ , also in square centimeters.
$12^2 = 144$
Compare the two results. They differ by only $1$ , but the test needs exact equality, so close does not count. The Pythagorean theorem says that a right triangle with hypotenuse $12$ cm and legs $8$ cm and $9$ cm would make the results equal. They are not equal, so this triangle is not a right triangle.
$145 \neq 144$

Answer

No. The triangle with sides $8$ cm, $9$ cm and $12$ cm is not a right triangle, because $8^2 + 9^2 = 145$ and $12^2 = 144$ are not equal.

Watch the order. A common mistake is to put the numbers into $a^2 + b^2 = c^2$ in the order they appear. For a triangle with side lengths $5$ , $13$ and $12$ , that gives $5^2 + 13^2 = 25 + 169 = 194$ and $12^2 = 144$ . These are not equal, so it looks as if the triangle is not a right triangle.

This fails because it tests the wrong side. In a right triangle, $c$ is the hypotenuse, the side opposite the right angle, and it is always the longest side. The side of length $12$ is not the longest, so it could never be the hypotenuse. With these positive lengths and the largest as $c$ : $5^2 + 12^2 = 25 + 144 = 169$ and $13^2 = 169$ . The results are equal, so by the converse the triangle is a right triangle.

Worked example

Decimal lengths in a mixed order

A triangle has side lengths $6.5$ cm, $2.5$ cm and $6$ cm. Is it a right triangle? Are the numbers $6.5$ , $2.5$ , $6$ a Pythagorean triple?

The lengths are positive and in the same unit. The largest is $6.5$ , so it is $c$ , even though it comes first in the list.
$a = 2.5,\quad b = 6,\quad c = 6.5$
Square the two shorter lengths exactly, without rounding, and add. Here $2.5 \times 2.5 = 2.5 \times 2 + 2.5 \times 0.5 = 5 + 1.25 = 6.25$ .
$2.5^2 + 6^2 = 6.25 + 36 = 42.25$
Square the longest length exactly. Here $6.5 \times 6.5 = 6.5 \times 6 + 6.5 \times 0.5 = 39 + 3.25 = 42.25$ .
$6.5^2 = 42.25$
Compare. Both results are exactly $42.25$ , so by the converse of the Pythagorean theorem this is a right triangle. Its right angle is opposite the longest side.
$2.5^2 + 6^2 = 6.5^2$
Check for a triple. A Pythagorean triple needs positive whole numbers, and $6.5$ and $2.5$ are not whole numbers.

Answer

Yes, it is a right triangle, with the right angle opposite the $6.5$ cm side. The numbers are not a Pythagorean triple, because $6.5$ and $2.5$ are not whole numbers.

Here is the idea to keep. Take three positive lengths, call the largest $c$ , and compare $a^2 + b^2$ with $c^2$ using exact squares. If the two results are equal, the converse of the Pythagorean theorem tells you the lengths form a right triangle, with the right angle opposite the side of length $c$ . If they are not equal, the lengths cannot form a right triangle. When the three numbers are also positive whole numbers and the results are equal, they are a Pythagorean triple.

So far you have used the converse as a known fact. In the next lesson, you will explain a proof of it.