Question

Write a Pythagorean triplet whose one member is:
a) 14
b) 10
c) 16
d) 22​

Answer 1

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(i) There are three numbers $2m,\ m^2-1\ and\ m^2+1$ in a Pythagorean Triplet.

Here, 2m = 6 $\Rightarrow\ m=\frac{6}{2}=3$

Therefore, second number $\left(m^2-1\right)=\left(3\right)^2-1=9-1=8$

Third number $m^2+1=\left(3\right)^2+1=9+1=10m$

Hence Pythagorean triplet is is (6, 8, 10).

(ii) There are three numbers

$2m,\ m^2-1\ and\ m^2+1$ in a Pythagorean Triplet.

Here, $2m = 14 \Rightarrow m=\frac{14}{2}=7$

Therefore, Second number $m^2-1=\left(7\right)^2-1=49-1=48m$

Third number $m^2+1=\left(7\right)^2+1=49+1=50m$

Hence, Pythagorean triplet is (14, 48, 50).

(iii) There are three numbers $2m,m^2-1\ and\ m^2+1$ in a Pythagorean Triplet.

Here, $2m = 16 \Rightarrow m=\frac{16}{2}=8$

Therefore, Second number $\left(m^2-1\right)=\left(8\right)^2+1=64+1=65$

Hence, Pythagorean triplet is (16, 63, 65).

(iv) There are Three numbers $2m,\ m^2-1\ and\ m^2+1$ in a Pythagorean Triplet.

Here, $2m = 18 \Rightarrow m=\frac{18}{2}=9$

therefore, Second number $\left(m^2-1\right)=\left(9\right)^2-1=81-1=80$

Third number $m^2+1=\left(9\right)^2+1=81+1=81$

Hence, Pythagorean triplet is (18, 80, 82).