write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18
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we know no's in Pythagorean triplet are 2m,m^2-1,m^2+1
1) let 2m =6
m=6/2=3 then other no's are
m^2-1=(3)^2-1=8
m^2+1=(3)^2+1=10
you can do other parts by this method...
hope you like it....
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1) let 2m =6
m=6/2=3 then other no's are
m^2-1=(3)^2-1=8
m^2+1=(3)^2+1=10
you can do other parts by this method...
hope you like it....
mark as brainliest
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