Question

find a Pythagorean triplet whose 1 member is 18​

Answer 1

Step-by-step explanation:

We have pythagorean triplet (2n,n2−1,n2+1) where n>1 (natural number)

Let n2−1=18

n2=19

n=19 (not an integer)

n2+1=18

n2=18−1

n2=17

n=17 (is not an integer)

So, let us take 2n=18

n=9

∴ The required triplet is (an integer)

(18,92−1,92+1)=(18,81−1,81+1)

i.e., (18,80,82)

Answer 2

Answer:

answer is attached

Step-by-step explanation:

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