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Q.2. Write a Pythagorean triplet whose one member is.

(i) 6

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For any natural number m > 1, 2m, m^{2} - 1 and m^{2} + 1 forms a Pythagorean triplet.

So if we take,           m^{2} - 1 = 6

                                     m^{2} = 6 + 1 = 7

But value of m will not be an integer.

   Now we take,          m^{2} + 1 = 6

                                     m^{2} = 6 - 1 = 5

  but value of m will not be an integer.

If we take                       2m = 6

 then                                 m = 3

Then              m^{2} -1 = 9 - 1 = 8           and          m^{2} +1 =  9 + 1  = 10.

Therefore the required triplet is 6, 8 and 10

Posted by

Devendra Khairwa

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