Gujarat BoardEnglish MediumSTD 8MATHSSquare and Square Roots3 Marks
Question
Write a Pythagorean triplet whose smallest member is $8.$
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Answer
We can get Pythagorean triplets by using general form $2m, m^2– 1, m^2+ 1.$
Let us first take $m^2– 1 = 8$
So, $m^2= 8 + 1 = 9$
which gives $m = 3$
Therefore, $2m = 6$ and $m^2+ 1 = 10$
The triplet is thus $6, 8, 10.$ But $8$ is not the smallest member of this.
So, let us try $2m = 8$
then $m = 4$
We get $m^2– 1 = 16 – 1 = 15$
and $m^2+ 1 = 16 + 1 = 17$
The triplet is $8, 15, 17$ with $8$ as the smallest member.
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