Question 2. Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
(i) Let m² – 1 = 6
[Triplets are in the form 2m, m²– 1, m² + 1]
m² = 6 + 1 = 7
So, the value of m will not be an integer.
Now, let us try for m² + 1 = 6
⇒ m² = 6 – 1 = 5
Also, the value of m will not be an integer.
Now we let 2m = 6 ⇒ m = 3 which is an integer.
Other members are:
m² – 1 = 3² – 1 = 8 and m² + 1 = 3² + 1 = 10
Hence, the required triplets are 6, 8 and 10
(ii) Let m² – 1 = 14 ⇒ m² = 1 + 14 = 15
The value of m will not be an integer.
Now take 2m = 14 ⇒ m = 7 which is an integer.
The member of triplets are 2m = 2 × 7 = 14
m² – 1 = (7)² – 1 = 49 – 1 = 48
and m² + 1 = (7)² + 1 = 49 + 1 = 50
i.e., (14, 48, 50)
(iii) Let 2m = 16 m = 8
The required triplets are 2m = 2 × 8 = 16
m² – 1 = (8)² – 1 = 64 – 1 = 63
m² + 1 = (8)² + 1 = 64 + 1 = 65
i.e., (16, 63, 65)
Class 8 Maths Chapter 6 Ex 6.2
(iv) Let 2m = 18 ⇒ m = 9
Required triplets are:
2m = 2 × 9 = 18
m² – 1 = (9)² – 1 = 81 – 1 = 80
and m² + 1 = (9)² + 1 = 81 + 1 = 82
i.e., (18, 80, 82)