# Write a Pythagorean triplet whose one member is:

(i) 6 (ii) 14 (iii) 16 (iv) 18

**Solution:**

For any natural number 'm' where m > 1, we have (2m)^{2} + (m^{2} - 1)^{2} = (m^{2} + 1)^{2}.

So, 2m, m^{2} - 1 and m^{2} + 1 forms a Pythagorean triplet

(i) If we take m^{2} + 1 = 6,

m^{2} = 5

The value of m will not be an integer.

If we take m^{2} - 1 = 6

m^{2} = 7

Again, the value of m will not be an integer.

Let 2m = 6

m = 6/2 = 3

m^{2} - 1 = (3)^{2} - 1

= 9 - 1 = 8

m^{2} + 1 = (3)^{2} + 1

= 9 + 1 = 10

Therefore, Pythagorean triplets are 6, 8 and 10

(ii) If we take m^{2} + 1 = 14,

m^{2} = 13

The value of the m will not be an integer.

If we take m^{2} - 1 = 14

m^{2} = 15

Again, the value of m will not be an integer.

Let 2m = 14

Thus, m = 7

m^{2} - 1 = 7^{2} - 1

= 49 - 1 = 48

m^{2} + 1 = 7^{2} + 1

= 49 + 1 = 50

Therefore, 14, 48, 50 are Pythagorean triplets.

(iii) If we take m^{2} + 1 = 16,

m^{2} = 15

The value of the m will not be an integer.

If we take m^{2} - 1 = 16

m^{2} = 17

The value of the m will not be an integer

Let 2m = 16

Thus, m = 8

m^{2} - 1 = 8^{2} - 1

= 64 - 1 = 63

m^{2} + 1 = 8^{2} + 1

= 64 + 1 = 65

Therefore, 16, 63, 65 are Pythagorean triplets.

(iv) If we take m^{2} + 1 = 18,

m^{2} = 17

The value of the m will not be an integer.

If we take m^{2} - 1 = 18

m^{2} = 19

Again, the value of m will not be an integer.

Let 2m = 18

Thus, m = 9

m^{2} - 1 = 9^{2} - 1

= 81 - 1 = 80

m^{2} + 1 = 9^{2} + 1

= 81 + 1 = 82

Therefore, 18, 80, 82 are Pythagorean triplets.

**Video Solution:**

## Write a Pythagorean triplet whose one member is: (i) 6 (ii) 14 (iii) 16 (iv) 18

### NCERT Solutions for Class 8 Maths - Chapter 6 Exercise 6.2 Question 2

**Summary:**

The Pythagorean triplets for the given questions are (i) 6, 8 and 10, (ii) 14, 48 and 50. (iii) 16, 63, 65 and (iv) 18, 80, 82