2.Write a Pythagorean triplet whose one member is
(i) 6

(ii) 14

(iii) 16

(iv) 18


Answer :

For a natural number m greater than 1,Triplets are in the form 2m, m2 – 1, m2 + 1
(i) Let \(m^2 – 1 = 6\)

\(\Rightarrow m^2 = 6 + 1 = 7\)

So,here we observe that value of m is not integer.

Now, let us try for \(m^2 + 1 = 6\)

\(\Rightarrow m^2 = 6 – 1 = 5\)

So, here too we have no integer value of m.

Now, let 2m = 6

\(\Rightarrow m = 3\; which\; is \;an\; integer.\)

Other number are:

\(m^2 – 1 = 32 – 1 = 8\) and \(m^2 + 1 = 32 + 1 = 10\)

Hence, the required triplets are 6, 8 and 10.

(ii) Let \(m^2 – 1 = 14\)

\(\Rightarrow m^2 = 1 + 14 = 15\)

So, we observe that the value of m will not be an integer.

Now take 2m = 14

\(\Rightarrow\) m = 7 which is an integer.

The other numbers of triplets are 2m =\( 2 \times 7\) = 14

\(m^2 – 1 = (7)2 – 1 = 49 – 1 = 48\)

and \(m^2 + 1 = (7)2 + 1 = 49 + 1 = 50\)

Hence the triplets are 14, 48 and 50.

(iii) Let, 2m = 16

\(\Rightarrow\)m = 8

The other numbers are

\(m^2 – 1 = (8)2 – 1 = 64 – 1 = 63\)

\(m^2 + 1 = (8)2 + 1 = 64 + 1 = 65\)

Hence the triplets are 16, 63 and 65

(iv) Let 2m = 18

\(\Rightarrow\) m = 9

The other triplets are:

\(m^2 – 1 = (9)2 – 1 = 81 – 1 = 80\)

and \(m^2 + 1 = (9)2 + 1 = 81 + 1 = 82\)

Hence we have the triplets as 18, 80 and 82.

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