# Pythagorean Triples

Pythagorean triples are a^{2}+b^{2} = c^{2} where a, b and c are the three positive integers. These triples are represented as (a,b,c). Here, a is the perpendicular, b is the base and c is the hypotenuse of the right-angled triangle. The most known and smallest triplets are (3,4,5). Learn Pythagoras theorem for more details.

Pythagoras who was a mathematician was interested in mathematics, science, and philosophy. He was born in Greece in about 570 BC. He is famous for a property of triangles with a right angle i.e 90^{0 }angles, and the property is known as Pythagoras Theorem. In a right-angled triangle, the hypotenuse is the side ‘r’, the side opposite the right angle. Adjacent to the right angle the shorter of the two sides is the side p. In this article, let us discuss what is Pythagorean triples, its formula, list, steps to find the triples, examples, and proof.

## What are Pythagorean Triples?

The integer solutions to the Pythagorean Theorem**, a ^{2}**

**+ b**

^{2}

**= c**

^{2}are called

**Pythagorean Triples**which contains three positive integers a, b, and c.

Example: (3, 4, 5)

By evaluating we get:

3^{2} + 4^{2} = 5^{2}

9+16 = 25

Hence, 3,4 and 5 are the Pythagorean triples.

You can say “triplets,” but “triples” are the favoured term. Let’s start this topic by an introduction of Pythagoras theorem.

### Table

(3, 4, 5) | (5, 12, 13) | (8, 15, 17) | (7, 24, 25) |

(20, 21, 29) | (12, 35, 37) | (9, 40, 41) | (28, 45, 53) |

(11, 60, 61) | (16, 63, 65) | (33, 56, 65) | (48, 55, 73) |

(13, 84, 85) | (36, 77, 85) | (39, 80, 89) | (65, 72, 97) |

## Pythagoras Triples Formula

If a triangle has one angle which is a right-angle (i.e. 90^{o}), there exists a relationship between the three sides of the triangle.

If the longest side (called the hypotenuse) is *r* and the other two sides (next to the right angle) is called *p* and *q*, then:

*p ^{2} + q^{2} = r^{2}*.

or,

The sum of the squares of the other two sides is the same as the square of the longest side.

## Pythagorean Triples List

The list of Pythagorean triples where the value of c is above 100 is given below:

(20, 99, 101) | (60, 91, 109) | (15, 112, 113) | (44, 117, 125) |

(88, 105, 137) | (17, 144, 145) | (24, 143, 145) | (51, 140, 149) |

(85, 132, 157) | (119, 120, 169) | (52, 165, 173) | (19, 180, 181) |

(57, 176, 185) | (104, 153, 185) | (95, 168, 193) | (28, 195, 197) |

(84, 187, 205) | (133, 156, 205) | (21, 220, 221) | (140, 171, 221) |

(60, 221, 229) | (105, 208, 233) | (120, 209, 241) | (32, 255, 257) |

(23, 264, 265) | (96, 247, 265) | (69, 260, 269) | (115, 252, 277) |

Students can pick any triples from the above list and prove the Pythagoras formula,i.e.,

a^{2}+b^{2}=c^{2}

## How to Form Pythagorean Triples?

How to Form Pythagorean Triples?

As we know, the number can be an odd number or an even number. Now, let us discuss how to create the Pythagorean triples.

**Case 1: If the number is odd:**

Let us assume the number be “x”.

If “x” is odd, then the Pythagorean triple = x, (x^{2}/2) – 0.5, (x^{2}/2) + 0.5.

Consider an example (7, 24, 25). Now, let us discuss how to form this Pythagorean triple.

Hre, x = 7, which is an odd number.

(x^{2}/2) – 0.5 = (49/2) – 0.5 = 24.5 – 0.5 = 24

(x^{2}/2) + 0.5 = (49/2) + 0.5 = 24.5 + 0.5 = 25

Hence, the Pythagorean triple formed is (7, 24, 25).

**Case 2: If the number is even:**

If “x” is even, then the Pythagorean triple = x, (x/2)^{2}-1, (x/2)^{2}+1.

Let us assume an example, (16, 63, 65). Now, we will check how to form the Pythagorean triple.

Here, x = 16, which is an even number.

(x/2)^{2} -1 = (16/2)^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63

(x/2)^{2} +1 = (16/2)^{2} +1 = 8^{2} + 1 = 64 + 1 = 65.

Thus, the Pythagorean triple formed is (16, 63, 65)

**Note: **We can observe that there are infinitely many Pythagorean triples because the approaches give us a triple for every positive whole number. Can these procedures, however, produce all of them? The answer is “No”. For example, the Pythagorean triple (20, 21, 29) cannot be formed using these methods.

## Pythagorean Triples Proof

Proof of Pythagoras theorem:

Look at the figure above

In the figure, at left,

Area of square = (a+b)^{2}

Area of Triangle = 1/2(ab)

Area of the inner square = b^{2}.

The area of the entire square = **4(1/2(ab)) + c ^{2}**

Now we can conclude that

**(a + b) ^{2} = 4(1/2 (ab)) + c^{2}**.

or

**a ^{2} + 2ab + b^{2} = 2ab + c^{2}**.

Simplifying, we get Pythagorean triples formula,

**a ^{2} + b^{2} = c^{2}**

Hence Proved.

### Triangular Numbers

The difference between successive squares is successive odd numbers is a fact and suggests that every square is the sum of two successive **triangular **numbers.

And in this, the triangular numbers are the successive sums of all integers.

- 0 + 1 =
**1**, - 0 + 1 + 2 =
**3**, - 0 + 1 + 2 + 3 =
**6**, etc.

So the triangular numbers are **1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120,** etc.

105 + 120 = **225**; 225 is the square of **15**.

## Common Pythagorean Triples

As we know, the specific set of integers that satisfies the Pythagoras theorem is called Pythagorean triples. It means that the set of integer numbers has a special connection with the Pythagoras theorem. Not only the set satisfies the Pythagoras theorem, but also the multiples of the integer set also satisfy the Pythagoras theorem.

For example, (3, 4, 5) is the most common Pythagorean triples. When each integer number is multiplied by 2, we get the set (6, 8, 10), which also satisfies the Pythagoras theorem.

(i.e.,) 3^{2} + 4^{2 } = 5^{2}

9+16 = 25

25 = 25

Similarly,6^{2} + 8^{2} = 10^{2}

36 + 64 = 100

100 = 100

This can be simply expressed as follows:

If a, b and c are the positive integers, which satisfies the Pythagoras theorem, then ak, bk, ck will also satisfy the Pythagoras theorem, when “k” is a positive integer.

Also, the Pythagorean triples can be found using different methods, such as generalized Fibonacci sequence, quadratic equations, using matrices and linear transformations, and so on. The set of Pythagorean triples is endless. We can prove that we have infinitely many Pythagorean triples with the help of (3, 4, 5)

**Facts: **An interesting fact about Pythagorean triples is that Pythagorean triples always consist of all even numbers or two odd numbers and an even number.

A Pythagorean triple never be made up of all odd numbers or two even numbers and an odd numbers

Pythagorean Triples | x 2 (Times 2) | x 3 (Times 3) | x 4 (Times 4) |

3-4-5 | 6-8-10 | 9-12-15 | 12-16-20 |

5-12-13 | 10-24-26 | 15-36-39 | 20-48-52 |

7-24-25 | 14-48-50 | 21-72-75 | 28-96-100 |

9-40-41 | 18-80-82 | 27-120-123 | 36-160-164 |

11-60-61 | 22-120-122 | 33-180-183 | 44-240-244 |

## Pythagorean Triples Examples (With Answers)

- So, the square of 3, 9, is the difference between 16, the square of 4, and 25 the square of 5, giving us the triplet
**7,24,25**. - Similarly, the square of 5, 25 is the difference between 144, the square of 12, and 169, the square of 13, giving us the triplet
**5, 12, 13.**

### Pythagorean Triples Problems

**Example 1: **

Prove that (5, 12, 13) is a Pythagorean triple?

**Solution:**

To prove: (5, 12, 13) is a Pythagorean Triple

We know that, a^{2} + b^{2 } = c^{2}

(a, b, c) = (5, 12, 13)

Now, substitute the values,

5^{2} + 12^{2} = 13^{2}

25 + 144 = 169

169 = 169

Hence, the given set of integers satisfies the Pythagoras theorem, (5, 12, 13) is a Pythagorean triples.

**Example 2: **

Check if (7, 15, 17) are Pythagorean triples.

**Solution:**

(a, b, c) = (7, 15, 17)

We know that a^{2} + b^{2 } = c^{2}

By substituting the values in the equation, we get

7^{2} + 15^{2} = 17^{2}

49 + 225 = 289

274 ≠ 289

Hence, the given set of integers does not satisfy the Pythagoras theorem, (7, 15, 17) is not a Pythagorean triplet. Also, it proves that the Pythagorean triples are not made up of all odd numbers.

## Frequently Asked Questions – FAQs

### What are Pythagorean triples?

Pythagorean triples are non-negative integers say a,b and c, which satisfies the following equation: a^{2}+b^{2} = c^{2}. Here a, b and c are the sides of a right triangle where a is perpendicular, b is the base and c is the hypotenuse.

### What are the five most common Pythagorean triples?

(3,4,5)

(5,12,13)

(7,24,25)

(9,40,41)

(11,60,61)

### How to find Pythagorean triplets?

To find Pythagorean triplets, remember the rules below:

Every odd number is the p side of a Pythagorean triplet.

The q side of a Pythagorean triplet is simply (p^{2}– 1)/2.

The r side is (q^{2} + 1)/2.

If p=9

q=(9^{2}-1)/2 = (81-1)/2 = 80/2 = 40

r=(9^{2}+1)/2 = (81+1)/2 = 82/2 = 41

Hence, (9,40,41) are the Pythagoras triples.

### How to do scaling of triples?

If (3,4,5) are the Pythagorean triples, then if we scale them by 2, we get;

(6,8,10)

So, 6^{2}+8^{2}=10^{2}

36+64=100

100 = 100

### Is (4,5,8) is Pythagorean triple?

If (4,5,8) is a Pythagorean triple, then it should satisfy: 4^{2}+5^{2}=8^{2}

Let us take LHS first,

4^{2}+5^{2} = 16+25 = 41

RHS = 82 = 64

Clearly, 41 is not equal to 64

Therefore (4,5,8) is not a pythagorean triples.

## Formula for Pythagorean Triples

To find the Pythagorean triples, the following formula is used. If a, b are two sides of the triangle and c is the hypotenuse, then, a, b, and c can be found out using this-

- a = m
^{2}-n^{2} - b = 2mn
- c = m
^{2}+n^{2}

These values result in a right-angled triangle with sides a, b, c.

Also, k.a, k.b and k.c are considered as the Pythagorean triple.

**Notes:**

(i) m, n and k are any two positive integers

(ii) m > n

(iii) m and n are coprime and both should not be odd numbers

### Solves Example

**Question: Check if (7, 24, 25) is a Pythagorean triple.**

**Solution: **Given,

Pythagorean triple = (7, 24, 25)

a = 7, b = 24, c = 25

The Pythagorean triples formula is, c^{2} = a^{2} + b^{2}

LHS: c^{2 }= 25^{2} = 625

RHS: a^{2} + b^{2} = 7^{2} + 24^{2} = 49 + 576 = 625

LHS = RHS

So, (7, 24, 25) is a Pythagorean triple.

**Pythagorean Triples List**

Here is the list of the most commonly used Pythagorean triples:

## Pythagorean Triples

**Pythagorean triples** are any three positive integers that completely satisfy the Pythagorean theorem. The theorem states that in any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two legs of the right triangle. These three sides of the right triangle form the Pythagorean triples. Let us learn how to generate a few Pythagorean triples in this article.

## What are Pythagorean Triples?

Pythagorean triples are a set of 3 positive numbers that fit in the formula of the Pythagoras theorem which is expressed as, a^{2} + b^{2 }= c^{2}, where a, b, and c are positive integers. Here, ‘c’ is the ‘hypotenuse’ or the longest side of the triangle and ‘a’ and ‘b’ are the other two legs of the right-angled triangle. The Pythagorean triples are represented as (a,b, c). The most popular example of Pythagorean triples is (3, 4, 5). We can verify that 3, 4 and 5 satisfy the equation a^{2} + b^{2 }= c^{2}. Let us see how! 3^{2} + 4^{2 }= 5^{2 }⇒ 9 + 16 = 25. Observe the right-angled triangle given below which shows the Pythagorean triples, 3, 4, and 5.

## Examples of Pythagorean Triples

The set of the Pythagorean triples is endless**. **The first known Pythagorean triples is (3, 4, and 5). We can generate a few more triples by scaling them up in the following manner. We can create as many triples as possible by taking values for n.

n | (3n, 4n, 5n) |
---|---|

2 | (6, 8, 10) |

3 | (9, 12, 15) |

4 | (12, 16, 20) |

## Pythagorean Triples Proof

Let us consider Pythagorean triples (9, 40, 41) for which we can verify the Pythagorean formula (Hypotenuse^{2 }= side 1^{2 }+ side 2^{2}). The hypotenuse of the right-angled triangle is the longest side = 41

Hypotenuse^{2 }= 41^{2} = 1681

The other two sides of the right angled triangle = 9 and 40

side 1^{2 }+ side ^{2 }⇒ 9^{2 }+ 40^{2 }⇒ 81 + 1600 = 1681

Thus, for any 3 Pythagorean Triples, we can verify the Pythagorean formula.

**Tips and Tricks**

- If we know one of the 3 triples, we can find the other two. If the given number (n) is an odd number, the Pythagorean triples are of the form, (n, (n
^{2}/2 – 0.5) and (n^{2}/2 + 0.5)). For example, consider 5. The triples are (5, 25/2 – 0.5, 25/2 + 0.5) Finally, we get (5, 12 and 13) - If the given number (n) is an even number, the Pythagorean triples are of the form = n, (n/2)
^{2}-1), ((n/2)^{2}+1). For example, consider 6. The triples are (6, (3)^{2}– 1, (3)^{2}+ 1) Finally, we get (6, 8, and 10)

## List of Pythagorean Triples

The following table shows a list of a few Pythagorean Triples.

(3, 4, 5) | (5,12,13) | (7, 24, 25) |

(8, 15, 17) | (9, 40, 41) | (11, 60, 61) |

(12,35, 37) | (13, 84, 85) | (15, 112, 113) |

(16, 63, 65) | (17,144, 145) | (19, 180, 181) |

(20, 21, 29) | (20, 99 ,101) | (21, 220,221) |

### How to Generate Pythagorean Triples?

In order to generate Pythagorean triples, we use a formula, which is also known as the Pythagorean triple checker. Observe the figure given below and make the following assumptions where:

- a, b = legs of a right-angled triangle
- c = Hypotenuse of the right-angled triangle
- m, n are any two positive integers; and where m > n
- m and n are coprime numbers and both should not be odd numbers.

Let us assume any 2 integers ‘m’ and ‘n’, which will help us in generating the Pythagorean formula. Now, the lengths of our sides are a, b, and c. We will use ‘m’ and ‘n’ in order to find the exact values of the sides.

- The length of side ‘a’ is determined by defining the difference between the squares of ‘m’ and ‘n’ which is expressed as a = m
^{2}– n^{2} - The length of side ‘b’ is determined by doubling the product of ‘m’ and ‘n’. This makes it b = 2mn in the equation form.
- Finally, the length of side ‘c’ is computed by the sum of the squares of m and n. This can be expressed as c = m
^{2}+ n^{2}

Now, we have concluded the following formulas:

- a = m
^{2}– n^{2} - b = 2mn
- c = m
^{2}+ n^{2}

It is to be noted that any one of the formulas given above can be used based on the condition and the requirement of finding the Pythagorean triples. To generate random Pythagorean triples, consider random natural numbers m and n, such that m > n and determine the triples (a,b,c) such that a = m^{2}– n^{2}, b = 2mn, c = m^{2}+ n^{2}

**Example:** Let us generate Pythagorean triples using the two integers 2 and 3.

**Solution:** Since m > n, m = 3, and n = 2. Now that we know the values of m and n, let us substitute those values into the formulas of a, b, and c, to get the sides of the right triangle.

Computing a:

**a = m ^{2} – n^{2}**

a = 3^{2} – 2^{2}

9 – 4 = 5

a = 5

Computing b:

**b = 2mn**

b = 2 × 3 × 2

b = 12

Computing c:

**c = m ^{2} + n^{2}**

c = 3^{2} + 2^{2}

c = 13

Let us see if our values for a = 5, b = 12, and c = 13 satisfy the Pythagorean theorem, which is a^{2} + b^{2} = c^{2}

LHS: 5^{2} + 12^{2 }= 25 +144 = 169

RHS : 13^{2 }= 169

Yes, it does! Therefore, (5, 12, 13) are Pythagorean triples.

**Important notes**

- Any three positive integers which satisfy the Pythagorean theorem formula of a
^{2}+ b^{2}= c^{2}are known as Pythagorean triples. - Pythagorean triples cannot be expressed in decimals.

## What Is Pythagorean Triples Formula?

Pythagorean triples formula is used to find the triples or group of three terms that satisfy the Pythagoras theorem. We know that when a, b c are the base, perpendicular and the hypotenuse of a right-angled triangle, then by Pythagoras’ theorem we have: c^{2} = a^{2}+b^{2}. The Pythagorean triples formula can be given as,

- a = m
^{2}– n^{2} - b = 2mn
- c = m
^{2}+ n^{2}

where,

- a, b = Base, and perpendicular of a right-angled triangle
- c = Hypotenuse of a right-angled triangle
- m, n are any two positive integers; m > n
- m and n are coprime and both should not be odd numbers

### Pythagorean Triples Formulas

Any one of the following formulas is used based on the condition and the requirement of finding the Pythagorean triples.

- If any number “m” of a Pythagorean triple is given, then the other two numbers can be evaluated using the trick by taking the triples as (2mn, m
^{2}– 1, m^{2}+ 1) - To generate random Pythagorean triples, consider random natural numbers m and n, such that m >n and determine the triples (a,b,c) in such a condition that a = m
^{2}– n^{2}, b = 2mn, c= m^{2}+ n^{2} - If a number (n) odd is given, then the Pythagorean triples are of the form, (n, (n
^{2}/2 – 0.5) and (n^{2}/2 + 0.5)). - If a number (n) even is given, then the Pythagorean triples are of the form = n, (n/2)
^{2}-1), ((n/2)^{2}+1).

### Pythagorean Triples Formula Verification

We can randomly generate Pythagorean triples using the Pythagorean triples formula. When a = m^{2}−n^{2}, b = 2mn, and c = m^{2}+n^{2}. When a = m^{2}– n^{2}, b = 2mn and c = m^{2}+n^{2},we can verify the Pythagorean triples formula using the Pythagorean theorem: c^{2} = a^{2 }+ b^{2 }

Consider LHS: c^{2} = (m^{2}+n^{2})^{2}

c^{2} = m^{4 }+n^{4 }+2m^{2}n^{2}

Consider RHS: a^{2}+b^{2 }= (m^{2}– n^{2} )^{2 }+ (2mn)^{2}

= m^{4 }+n^{4 }-2m^{2}n^{2 }+ 4m^{2}n^{2 }

a^{2}+b^{2 }= m^{4 }+n^{4 }+2m^{2}n^{2}

LHS= RHS

Thus c^{2} = a^{2 }+ b^{2}

## Examples on Pythagorean Triples Formula

**Example 1: **Check if (5, 12, 13) is a Pythagorean triple.

**Solution:**

To find: Check whether (5, 12, 13) is a Pythagorean triple.

Given: a = 5; b = 12; c = 13

(13 is the longest side and is considered to be the hypotenuse.)

Using the Pythagorean triples formula, we know that a Pythagorean triple satisfies Pythagoras’ theorem: c^{2} = a^{2}+b^{2}

L. H. S. = c^{2} = 13^{2} = 169

R. H. S. = a^{2 }+ b^{2 }= 5^{2 }+ 12^{2} = 25 + 144 = 169

Since the given values satisfy the Pythagoras’ theorem,(5, 12, 13) is a Pythagorean triple.

**Answer:** **(5, 12, 13) is a Pythagorean triple.**

**Example 2:** Evaluate and find why 40, 76, 86 is not a Pythagorean triple.

**Solution:**

Using Pythagorean Triples formula, a^{2} + b^{2} = c^{2}

= 40^{2} + 76^{2} = 1,600 + 5,776

= 7,376 ≠ 86^{2}

**Answer: 40, 76, 86 is not a Pythagorean triple.**

**Example 3:** If (x, 40, 41) is a Pythagorean triple, determine the value of x using the Pythagorean triple formula?

Solution:

Using the Pythagorean Triples formula:

a^{2} + b^{2} = c^{2}

Replacing a by x, b by 40, and c by 41 in the formula we have,

⇒ x^{2} + 40^{2} = 41^{2}

⇒ x^{2} + 1,600 = 1,681

⇒ x^{2} = 1,681 – 1,600 = 81

⇒ x = √81 = 9

## FAQ’s on Pythagorean Triples Formula

### What Is The Pythagorean Triples Formula When one Side Is Given?

The Pythagorean triples formula is applied based on the requirement and the conditions. If one side of a right triangle is given, then the other two sides are determined on the basis of the given number odd or even.

- If a number (n) odd is given, then the Pythagorean triples are of the form, (n, (n
^{2}/2 – 0.5) and (n^{2}/2 + 0.5)). For example, consider 3. The triples are (3, (9/2 – 0.5), (9/2 + 0.5)). Finally, we get (3, 4 and 5). - If a number (n) even is given, then the Pythagorean triples are of the form, (n, (n/2)
^{2}-1), ((n/2)^{2}+1). For example, consider 8. The triples are (8, ((4)^{2}– 1), ((4)^{2}+ 1)). Finally, we get (8, 15, and 17)

### What Is Pythagorean Triples Formula Using random numbers?

Let us take two random numbers m = 4 and n = 3, where m> n. According to the Pythagorean triples formula, a = m^{2}– n^{2}, b = 2mn and c = m^{2}+n^{2}.

a = 4^{2}– 3^{2 }= 16 – 9 = 7,

b = 2(4)(3)= 24

and c = 4^{2}+3^{2 }= 16 +9 = 25

Thus the Pythagorean triples are (7, 24, 25) using random numbers.

### How Do you Find the Pythagorean Triplet of 10?

We find that 10 is even. Thus by the Pythagorean triples formula, we get (10, (10/2)^{2} + 1, (10/2)^{2} – 1) = (10, (5)^{2} – 1, (5)^{2} +1) = (10, 24, 26)

### What Are The Pythagorean Triples Whose Smallest Number is 8?

Let us consider 8 as the base side. Thus using the Pythagorean triples formula, we find 8 = 2m. This implies m = 4. The other two triples are m^{2} + 1 and m^{2} – 1.

Thus we evaluate them as 4^{2} + 1 and 4^{2} – 1 = 17 and 15.

Thus the triples are (8,15,17)

**Example Problems on Pythagorean triples**

**Example 1: ****Can you prove that (5, 12, 13) is a Pythagorean triple?**

**Solution:**

In order to show that (5, 12, 13) is a Pythagorean triple

In mathematics, a^{2} + b^{2} = c^{2 }

(a, b, c) = (5, 12, 13)

Substitute the values now,

5^{2} + 12^{2} = 13^{2}

25 + 144 = 169

169 = 169

Therefore, the given set of integers satisfied Pythagoras’s theorem: (5, 12, 13) is a Pythagorean triple.

**Example 2:**** Check whether (7, 15, 17) are Pythagorean triples.**

**Solution:**

(a, b, c) = (7, 15, 17)

In mathematics, a^{2} + b^{2} = c^{2 }

When we substitute the values in the equation, we get

7^{2} + 15^{2} = 17^{2}

49 + 225 = 289

274 ≠ 289

The given set of integers does not satisfy the Pythagorean triplet theorem; (7, 15, 17) is not a Pythagorean triplet. In addition, this proves that the Pythagorean triples do not include all odd numbers.

**Example 3: ****Find the Pythagorean triplet that consists of 18 as one of its elements.**

**Solution:**

Let us consider the Pythagorean triplet (a, b, c) in which

a = m^{2} – 1, b = 2m and c = m^{2} + 1

Let us take the value of ‘b’ as 18.

b = 2m

18 = 2m

m = 9

Substituting m = 9 in the formulas for ‘a’ and ‘c’, we get

a = m^{2} – 1 = 9^{2} – 1 = 81 – 1 = 80

c = m^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82

Therefore, the Pythagorean triplet is (80, 18, 82).

**Solved Examples**

Q.1: Check if (16, 63, 65) are Pythagorean triples?

Solution: Given, (16, 63, 65)

a = 16, b = 63, c = 65

Pythagorean triples formula used as,

c^{2} = a^{2} + b^{2}

LHS: c^{2 }= 65^{2} = 4225

RHS: a^{2} + b^{2} = 16^{2} + 63^{2} = 256 + 3969 = 4225

LHS = RHS

So, (16, 63, 65) is a Pythagorean triples.

Q.2. Write a Pythagorean triplet whose one member is 18

Solution: Let 2m = 18 , m = 9

Now, m^{2} – 1 = 9^{2} – 1 =81 – 1 = 80

And, m^{2} + 1 = 9^{2} + 1 = 81 + 1 = 82

So, the Pythagorean triple is 18, 80, 82

**Examples of non-primitive Pythagorean triples include**: (6,8,10), (32,60,68), (16, 30, 34) etc.

- (6,8,10) → GCF of 6, 8 and 10 = 2.

a^{2} + b^{2 }= c^{2}

6^{2} + 8^{2 }= 10^{2}

36 + 64 = 100

- = 100
- (32,60,68) → GCF of 32, 60 and 68 = 4

a^{2} + b^{2 }= c^{2}

32^{2} + 60^{2 }= 68^{2}

1,024 + 3,600 = 4,624

4,624 = 4,624

Other examples of commonly used Pythagorean triples include: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) , (20, 21, 29) , (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (16, 63, 65), (33, 56, 65), (48, 55, 73), etc.

*Example 1*

What is the Pythagorean triple of two positive numbers, 1 and 2?

Solution

Given the Pythagorean triples formula: (a, b, c) = (m^{2} − n^{2}; 2mn; m^{2} + n^{2}), where; m > n.

So, let m = 2 and n = 1.

Substitute the values of m and n into the formula.

⇒ a = 2^{2} − 1^{2} = 4 − 1 = 3

a =3

⇒ b = 2 × 2 × 1 = 4

b = 4

⇒ c = 2^{2} + 1^{2} = 4 + 1 = 5

c = 5

Apply the Pythagorean theorem to verify that (3,4,5) is indeed a Pythagorean triple

⇒ a^{2} + b^{2 }= c^{2}

⇒ 3^{2} + 4^{2 }= 5^{2}

⇒ 9 + 16 = 25

⇒ 25 = 25.

Yes, it worked! Therefore, (3,4,5) is a Pythagorean triple.

*Example 2*

Generate a Pythagorean triple from two integers 5 and 3.

Solution

Since m must be greater than n (m > n), let m= 5 and n = 2.

a = m^{2} − n^{2}

⇒a= (5)^{2 }−(3)^{2} = 25−9

= 16

⇒ b = 2mn = 2 x 5 x 3

= 30

⇒ c = m^{2} + n^{2 }= 3^{2 }+ 5^{2}

= 9 + 25

= 34

Hence, (a, b, c) = (16, 30, 34).

Verify the answer.

⇒ a^{2} + b^{2 }= c^{2}

⇒ 16^{2} + 30^{2 }= 34^{2}

⇒ 256 + 900 = 1,156

1,156 = 1,156 (True)

Therefore, (16, 30, 34) is indeed a Pythagorean triple.

*Example 3*

Check if (17, 59, 65) is a Pythagorean triple.

Solution

Let, a = 17, b = 59, c = 65.

Test if, a^{2} + b^{2 }= c^{2}.

a^{2} + b^{2 }⇒ 17^{2} + 59^{2}

⇒ 289 + 3481 = 3770

c^{2 }= 65^{2}

= 4225

Since 3770 ≠ 4225, then (17, 59, 65) is not a Pythagorean triple.

*Example 4*

Find the possible value of ‘a’ in the following Pythagorean triple:(a, 35, 37).

Solution

Apply the Pythagorean equation a^{2} + b^{2 }= c^{2}.

a^{2} + 35^{2 }= 37^{2}.

a^{2 }= 37^{2}−35^{2}=144.

√a^{2 }= √144

a = 12.

*Example 5*

Find the Pythagorean triple of a right triangle whose hypotenuse is 17 cm.

Solution

(a, b, c) = [ (m^{2}-1), (2m), (m^{2}+1)]

c = 17 = m^{2}+1

17 – 1 = m^{2}

m^{2} = 16

m = 4.

Therefore,

b = 2m = 2 x 4

= 8

a = m^{2} – 1

= 4^{2} – 1

= 15

*Example 6*

The smallest side of a right triangle is 20mm. Find the Pythagorean triple of the triangle.

Solution

(a, b, c) =[(2m), (m^{2}-1), (m^{2}+1)]

20 =a = 2m

2m = 20

m =10

Substitute m = 10 into the equation.

b = m^{2} – 1

= 10^{2} – 1= 100 – 1

b = 99

c = m^{2}+1

= 10^{2} + 1

= 100 + 1 = 101

PT = (20, 99, 101)

*Example 7*

Generate a Pythagorean triple from two integers 3 and 10.

Solution

(a, b, c) = (m^{2} − n^{2}; 2mn; m^{2} + n^{2}).

a = m^{2} − n^{2}

= 10^{2} – 3^{2} = 100 – 9

= 91.

b = 2mn = 2 x 10 x 3

= 60

c = m^{2} + n^{2}

= 10^{2 }+ 3^{2} = 100 + 9

= 109.

PT = (91, 60,109)

Verify the answer.

a^{2} + b^{2 }= c^{2}.

91^{2} + 60^{2 }= 109^{2}.

8,281+ 3,600=11,881

11,881=11,881 (True)

*Example 8*

Check whether the set (24, 7, 25) is a Pythagorean triple**.**

Solution

Let a = 24, b = 7 and c = 25.

By Pythagorean theorem: a^{2} + b^{2} = c^{2}

7^{2} + 24^{2} = 625

49 + 576 = 625 (True)

Therefore, (24, 7, 25) is a Pythagorean triple.

*Example 9*

Find the Pythagorean triplet of a right triangle whose one side is 18 yards.

Solution

Given the formula: (a, b, c) = [ (m^{2}-1), (2m), (m^{2}+1)].

Let a or b = 18 yards.

2m = 18

m = 9.

Substitute m = 9 into the formula.

c = m^{2} + 1

= 9^{2 }+ 1 = 81

b or a = m^{2} -1 = 9^{2} -1

= 80

Therefore, the possible triplets are; (80, 18, 81) or (18, 80, 81).

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