Question

3. Write a Pythagorean triplet if one number is 14.​

Answer 1

Answer:

$\sf{Let\:numbers\:be \:= \:2m, m^{2}+1,m^{2}-1}$

Now, the smallest number is 2m = 14

We now find : m = 7

=》 $\tt{m^{2} + 1 = 7^{2} +1 = 50}$

=》 $\tt{m^{2} - 1 = 7^{2} -1 = 48}$

Triplet : 14, 48, 50

Answer 2

Your answer is 14 , 48 , 50.....

$= > 2m = 14 \\ \\ = > m = \frac{14}{2} \\ \\ = > m = 7$

now ,

the value of m2-1 , m2+1 , 2m

●m^2-1

$= > m ^{2} - 1 \\ \\ = > {7}^{2} - 1 \\ \\ = > 49 - 1 \\ \\ = > 48$

●m^2+1

$= > m ^{2} + 1 \\ \\ = > {7}^{2} + 1 \\ \\ = > 49 + 1 \\ \\ = > 50$

●2m

$= > 2m \\ \\ = > 2 \times 7 \\ \\ = > 14$

Your answer is 14 , 48 , 50.....

Please mark it as brainliest.....