Question

Write a Pythagorean triplet whose one member is 80

Answer 1

hey here is ur answer

let the other number be x

$\sqrt{ {80}^{2} + {x}^{2} }$

√1600+x^2

I hope this helps you

Answer 2

formulas=2m,m2-1, m2+1

putting value of m= 2m=80
m=80/2=m=40
it's a proper integral value
now,
=m2-1=80
m2=80+1
m2=81
m=underroot 81
m=9
it's also a proper integral value

now
m2+1=80
m2=80-1=71
m2=underroot 71
which is not a proper integral value

now put value of m in all the formulas - 2m, m2-1 and m2+1
put both value 9 and 40 in the value which u get 80 as a member legalise it

hope it helps