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  • Write a Pythagorean triplet whose one member is.(iv) 18

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

 (iv) 18

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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 = 18

                                     m^{2} = 18 + 1 = 19

But the value of m will not be an integer.

   Now we take,          m^{2} + 1 = 18

                                     m^{2} = 18 - 1 = 17

  but the value of m will not be an integer.

If we take                       2m = 18

 then                                 m = 9

Then              m^{2} -1 = 81 - 1 = 80           and          m^{2} +1 =  81 + 1  = 82.


Therefore the required combination is 18, 80 and 82

Posted by

Devendra Khairwa

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