Question

Write a Pythagorean triplet whose one member is 14.​

Answer 1

Answer:

Hence the triplet is 14,48 and 50 Answer.

Step-by-step explanation:

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Answer 2

For any natural number m > 1, 2m, m² - 1, m² + 1 forms a Pythagorean triplet.

If we take,

${m}^{2} + 1 = 14$

Then,

${m}^{2} = 13$

The value of m will not be an integer.

If we take ,

${m }^{2} - 1 = 14$

Then ,

${m}^{2} = 15$

Again the value of m is not an integer.

Let

$2m = 14$

$m = 7$

Thus,

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

And,

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

Therefore, the required triplet is 14, 48, and

50.