Question

Find a Pythagorean triplet in which one member is 6

Answer 1

Answer:

$\sqrt{11}$

Step-by-step explanation:

a+b=c

a+1=6

a=6-1

a=5

Pythagoras thoerom

${5}^{2} + {x}^{2} = {6 }^{2}$

$25 + {x}^{2} = 36 \\ {x}^{2} = 36 - 25 \\ {x}^{2} = 11 \\ x = \sqrt{11}$

Answer 2

Answer:

the required triplet is 6, 8, and 10

Step-by-step explanation:

Take ${m}^{2} - 1 = 6$

Then, ${m}^{2} = 6 + 1 = 7$

here the value of m will not be an integer.so we try again take $2m = 6$

Then, m = 3

Thus ${m}^{2} - 1 = 9 - 1 = 8 \: and \: {m}^{2} + 1 = 9 + 1 = 10$therefore the required triplet is 6 ,8 and 10

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