Write a Pythagorean triplet whose one member is 14.
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Answer:
Hence the triplet is 14,48 and 50 Answer.
Step-by-step explanation:
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Verified answer
For any natural number m > 1, 2m, m² - 1, m² + 1 forms a Pythagorean triplet.
If we take,
[tex] {m}^{2} + 1 = 14[/tex]
Then,
[tex] {m}^{2} = 13[/tex]
The value of m will not be an integer.
If we take ,
[tex] {m }^{2} - 1 = 14[/tex]
Then ,
[tex] {m}^{2} = 15[/tex]
Again the value of m is not an integer.
Let
[tex]2m = 14[/tex]
[tex]m = 7[/tex]
Thus,
[tex] {m}^{2} - 1 = 49 - 1 = 48[/tex]
And,
[tex] {m + 1}^{2} = 49 + 1 = 50[/tex]
Therefore, the required triplet is 14, 48, and
50.