Write a Pythagorean triplet whose one member is 8.
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Answer:
It is given that smallest number is 8. So, 2m = 8. Now, second number = m² -1 = 4² – 1 = 15 and, third number = m² + 1 = 4² + 1 = 17. Hence, the triplet is 8, 15 and 17.
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Answer:
h^2=p^2+b^2
8^2=x^2+x^2
64=2x^2
32=x^2
√32=x
4√2=x
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