Question

pythagovean tripet whose one member is 18?​

Answer 1

Answer:

57d5d57d57d57d57dt7dt7f8tf8yciyc8ycy9c

Answer 2

Answer:

We have pythagorean triplet (2n,n^2−1,n^2+1) where n>1 (natural number)

Let n^2−1=18

n^2=19

n=root 19 (not an integer)

n^2+1=18

n^2=18−1

n^2=17

n=root 17 (is not an integer)

So, let us take 2n=18

n=9

∴ The required triplet is (an integer)

(2n,n^2−1,n^2+1)=(18,81−1,81+1)

i.e., (18,80,82)