find the pathagorean tripets for which one member is 10
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10^2=100
divide by 2=100/2=50
(i)add 2 to 50=50+2=52
(ii)subtract 2 by 50= 50-2=48
(i)52/2 =26and (ii)48/2 =24
ANS. is 10,24,26 are the triples (10^2+24^2+26^2)
and bcuz 3,4,5 are a triple their double must also be 6,8,10 are also a triplet (6^2+8^2+10^2)
divide by 2=100/2=50
(i)add 2 to 50=50+2=52
(ii)subtract 2 by 50= 50-2=48
(i)52/2 =26and (ii)48/2 =24
ANS. is 10,24,26 are the triples (10^2+24^2+26^2)
and bcuz 3,4,5 are a triple their double must also be 6,8,10 are also a triplet (6^2+8^2+10^2)
suvetha07:
thank you
answer is 6,8,10
how
If we multiply 3,4,5 the all the three x2 we get 6,8,10
If nice Mark AS BRAINLIEST
thanks you
wlcme
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