question_answer 1)

                Write a Pythagorean triplet whose one number is (i) 6                                        (ii) 14                                     (iii) 16                                    (iv) 18.

Answer:

                (i) 6                 Here,     \[2m=6\,\,\Rightarrow \,\,m=\frac{6}{2}=3\] \[{{m}^{2}}-1={{3}^{2}}-1=9-1=8\] and        \[{{m}^{2}}+1={{3}^{2}}+1=9+1=10\] So, a Pythagorean triplet, whose one member is 6, is 6, 8, 10. (ii) 14 Here, \[2m=14\] \[\Rightarrow \]               \[\,m=\frac{14}{2}=7\] \[\therefore \]  \[\,{{m}^{2}}-1={{7}^{2}}-1=49-1=48\] and        \[{{m}^{2}}+1={{7}^{2}}+1=49+1=50\] So, a Pythagorean triplet, whose one member is 14, is 14, 48, 50. (iii) 16 Here, \[2m=16\] \[\Rightarrow \]               \[\,m=\frac{16}{2}=8\] \[\therefore \]  \[\,{{m}^{2}}-1={{8}^{2}}-1=64-1=63\] and        \[{{m}^{2}}+1={{8}^{2}}+1=64+1=65\] So, a Pythagorean triplet, whose one number is 16, is 16, 63, 65.                 (iv) 18                              Here, \[2m=18~\] \[\Rightarrow \]               \[m=\frac{18}{2}=9\] \[\therefore \]  \[{{m}^{2}}-1={{9}^{2}}-1=81-1=80\] and        \[{{m}^{2}}+1={{9}^{2}}+1=81+1=82\] So, a Pythagorean triplet, whose one number is 18, is 18, 80, 82.