The Pythagorean Theorem

Ontario 09 Applied (MFM1P)

Pythagorean Triads

Lesson

A Pythagorean triple (sometimes called a Pythagorean triple) is an ordered triple $\left(a,b,c\right)$(`a`,`b`,`c`) of three positive integers such that $a^2+b^2=c^2$`a`2+`b`2=`c`2.

If $\left(a,b,c\right)$(`a`,`b`,`c`) is a triple then $\left(b,a,c\right)$(`b`,`a`,`c`) is also a triple, since $b^2+a^2$`b`2+`a`2 is the same as $a^2+b^2$`a`2+`b`2. So the order of the first two numbers in the triple doesn't matter.

If $a$`a`, $b$`b` and $c$`c` are relatively prime (that is, they have no common factors), then the triple is called primitive. There are $16$16 primitive Pythagorean triples with $c\le100$`c`≤100, including $\left(3,4,5\right)$(3,4,5), $\left(5,12,13\right)$(5,12,13), $\left(8,15,17\right)$(8,15,17), and $\left(7,24,25\right)$(7,24,25).

$\left(6,8,10\right)$(6,8,10) is also a Pythagorean triple, but it is **not** a primitive Pythagorean triple, since $6$6, $8$8 and $10$10 have a common factor of $2$2. If we divide each number in the triple by this common factor, we get the primitive triple $\left(3,4,5\right)$(3,4,5).

A triangle whose sides form a Pythagorean triple is called a Pythagorean triangle, and will always be a right-angled triangle.

The two smallest numbers in a Pythagorean triple are $20$20 and $21$21. What number, $c$`c`, will complete the triple?

**Think**: Here we have $a=20$`a`=20 and $b=21$`b`=21 We can use the Pythagorean formula and solve for $c$`c`.

**Do**: Using the pythagorean formula,

$c^2$c2 |
$=$= | $a^2+b^2$a2+b2 |

$=$= | $20^2+21^2$202+212 | |

$=$= | $400+441$400+441 | |

$=$= | $841$841 | |

so, $c$c |
$=$= | $\sqrt{841}$√841 |

$=$= | $29$29 |

The missing value is $c=29$`c`=29, forming the triple $\left(20,21,29\right)$(20,21,29).

Sean knows the two largest numbers in a Pythagorean Triple, which are $41$41 and $40$40. What number, $a$`a`, does Sean need to complete the triple?

We would like to find the hypotenuse in a right-angled triangle with shorter side lengths $6$6 and $8$8, using our knowledge of common Pythagorean triples.

Below are some common Pythagorean triples. The two shorter sides $6$6, $8$8 and its hypotenuse will be multiples of the sides in which of the triples?

A $\left(3,4,5\right)$(3,4,5) B $\left(5,12,13\right)$(5,12,13) C $\left(8,15,17\right)$(8,15,17) D $\left(7,24,25\right)$(7,24,25) They will be multiples of the Pythagorean triple:

($\editable{}$,$\editable{}$,$\editable{}$)

What number when multiplied by $3$3 and $4$4 gives $6$6 and $8$8 respectively?

$\editable{}$

Hence, what is the length of the hypotenuse in the triangle with two shorter sides $6$6 and $8$8?

$\editable{}$

Is the Pythagorean triple $\left(66,112,130\right)$(66,112,130) a primitive Pythagorean triple?

**Think**: A primitive triple has no common factors between its elements. So we will need to check if $66$66, $112$112 and $130$130 have any common factors.

**Do**: $66$66, $112$112 and $130$130 are all even, so they have a common factor of at least $2$2. This means that $\left(66,112,130\right)$(66,112,130) is not a primitive triple.

Solve problems using the Pythagorean theorem, as required in applications (e.g., calculate the height of a cone, given the radius and the slant height, in order to determine the volume of the cone)