Write a pythagorean triplet whose smallest member is
Question:

Write a pythagorean triplet whose smallest member is

(i) 6

(ii) 14

(iii) 16

(iv) 20

Solution:

For every number $m>1$, the Pythagorean triplet is $\left(2 m, m^{2}-1, m^{2}+1\right)$.

Using the above result:

(i) $2 m=6$

$m=3, m^{2}=9$

$m^{2}-1=9-1=8$

$m^{2}+1=9+1=10$

Thus, the Pythagorean triplet is $[6,8,10]$

(ii) $2 m=14$

$m=7, m^{2}=49$

$m^{2}-1=49-1=48$

$m^{2}+1=49+1=50$

Thus, the Pythagorean triplet is $[14,48,50]$.

(iii) $2 m=16$

$m=8, m^{2}=64$

$m^{2}-1=64-1=63$

$m^{2}+1=64+1=65$

Thus, the Pythagorean triplet is: $[16,63,65]$

(iv) $2 m=20$

$m=10, m^{2}=100$

$m^{2}-1=100-1=99$

$m^{2}+1=100+1=101$

Thus, the Pythagorean triplet is $[20,99,101]$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.