find a Pythagorean triplet with 41 as one of the numbers
Please provide with full step by step solution
Answers
Answered by
0
Answer: General Pythagoras triplet is (m, m²+1, m²-1)
∴m = 41
Genaral pythagoras triplet = (41 , 41²+1 ,41²-1)
= (41 , 1681+1 , 1681-1)
= (41 , 1682 , 1680)
Answered by
0
Genral pythagoras triplet is (m, m ^ 2 + 1 , m^ 2 . 1)
m = 41
Genaral pythagoras triplet =(41,41^ 2 +1 , 41 ^ 2 1)
1681-1)
=(41,1681 +1,
=(41,1682, 1680). please comment
Similar questions