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

(i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=6 ⇒ m=6/2=3
Therefore,
Second number (m²-1)=(3)²-1=9-1=8
Third number m²+1=(3)²+1=9+1=10
Hence, Pythagorean triplet is (6,8,10).

(ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=14
⇒ m=14/2=7
Therefore,
Second number (m²-1)=(7)²-1=49-1=48
Third number m²+1=(7)²+1=49+1=50
Hence, Pythagorean triplet is (14,48,50).

(iii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=16
⇒ m=16/2=8
Therefore,
Second number (m²-1)=(8)²-1=64-1=63
Third number m²+1 =(8)²+1=64+1=65
Hence, Pythagorean triplet is (16,63,65)..

(iv) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
Here, 2m=18
⇒ m=18/2=9
Therefore,
Second number (m²-1)=(9)²-1=81-1=80
Third number m²+1 =(9)²+1=81+1=82
Hence, Pythagorean triplet is (18, 80, 82).

Class 8 Maths Chapter 6 Exercise 6.2 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/