PYTHAGOREAN TRIPLES

The set of positive integers {a, b, c} is a Pythagorean triple if it obeys the rule

a2 + b2 = c2

Example 1 :

Show that {5, 12, 13} is a Pythagorean triple

Solution :

From the given numbers, the largest number is 13.

132  =  52 + 122

169  =  25 + 144

169  =  169

So, {5, 12, 13} is a Pythagorean triple.

Example 2 :

Find k if {9, k, 15} is a Pythagorean triple.

Solution :

Let 92 + k2 = 152

81 + k2  =  225

k2  =  144

 k  =  √144

 k  =  12

Example 3 :

For what values of n does {n, n+1, n+2} form a Pythagorean triple?

Solution :

Let n2 + (n+1)2 = (n+2)2

n2+n2+2n+1  =  n2+4n+4

2n2-n2+2n-4n+1-4  =  0

n2-2n-3  =  0

(n-3)(n+1)  =  0

n  =  3 and n  =  -1

So, for n  =  3 the given set will create Pythagorean triple.

Example 4 :

Show that {n, n+1, n+3} cannot form a Pythagorean triple.

Solution :

Let n2 + (n+1)2 = (n+3)2

n2+n2+2n+1  =  n2+6n+9

2n2-n2+2n-5n+1-9  =  0

n2-3n-8  =  0

It is not factorable. So, by solving this quadratic equation, we will not get integer value for n. 

Therefore the given set of numbers will not create Pythagorean triple.

Example 5 :

Write a Pythagorean triplet whose one number is 12.

Solution :

For any natural number where, “m” > 1

m2 - 1 = 12

m2 = 12 + 1

m2 = 13 (the value of m is not an integer)

Then, let 2m = 12

m = 6

m2 - 1 = 62 - 1

= 36 - 1

= 35

m2 + 1 = 62 +- 1

= 36 + 1

= 37

So, the required Pythagorean triplets are 12, 36 and 37.

Example 5 :

Write a Pythagorean triplet where the smallest number is 6.

Solution :

For any natural number where, “m” > 1

m2 - 1 = 6

m2 = 6 + 1

m2 = 7

Here m is not an integer. Then,

m2 + 1 = 6

m2 = 6 - 1

m2 = 5

Here m is not an integer. Then,

2m = 6

m = 3 (is an integer)

m2 - 1 = 32 - 1

= 9 - 1

= 8

m2 + 1 = 32 + 1

= 9 + 1

= 10

So, the Pythagorean triples are 3, 8 and 10.

Example 6 :

What is the Pythagorean triples of two positive numbers 5 and 12 ?

Solution :

When two numbers in Pythagorean triples are given, the numbers are

m2 - n2, 2mn and m2 + n2

m2 - n2 = 122 - 52

= 144 - 25

= 119

2mn = 2(12)(5)

= 24(5)

= 120

m2 + n2 = 122 + 52

= 144 + 25

= 169

So, the required Pythagorean Triplet for the given values are 119, 120 and 169.

Example 7 :

Out of the dates given below which date constitutes a Pythagorean triplet ?

A) 15/08/17     B) 16/08/16    C) 3/5/17    D) 4/9/15

Solution :

Option A :

15/08/17 

Let 2m = 8

m = 4

m2 - 1 = 42 - 1

= 16 - 1

= 15

m2 + 1 = 42 + 1

= 16 + 1

= 17

So, date A is Pythagorean Triplet.

Example 8 :

In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?

A) 15      B) 13      C) 5      D) 12

Solution :

The numbers which are in Pythagorean triplets, the sum of the squares of two numbers will be equal to the square of Hypotenuse.

Let h be the square of hypotenuse.

h2 = 169

h = √169

h = 13 cm

So, the length of hypotenuse is 13 cm.

Example 9 :

Out of the following, which is the Pythagorean triplet?

A) (1, 5, 10)     B) (3, 4, 5)    C) (2, 2, 2)      D) (5, 5, 2)

Solution :

Option A :

(1, 5, 10)

2m = 10

m = 5

m2 - 1 = 52 - 1

= 25 - 1

= 24

m2 + 1 = 52 + 1

= 25 + 1

= 26

So, option A is incorrect.

Option B :

(3, 4, 5)

2m = 4

m = 2

m2 - 1 = 22 - 1

= 4 - 1

= 3

m2 + 1 = 22 + 1

= 4 + 1

= 5

So, option B is correct.

Example 10 :

The hypotenuse of right angled triangle with perpendicular sides 4 and 5 is _________.

A) 41       B) √41      C) 6      D) None of the above

Solution :

Let a = 4 and b = 5 and c be the hypotenuse.

Square of hypotenuse = sum of squares of remaining two sides.

42 + 52 = c2

c2 = 16 + 25

c2 = 41

c = √41

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