Question

What will be the value of x in the Pythagorean triplet 41,9,x

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

40

Step-by-step explanation:

first, finding the squares of each number

$41^{2} =1681\\9^{2} =81$

pythagoras' theorem= $a^{2} +b^{2} =c^{2}$

according to the equation if 42=a and b=9

then, the sum of their squares should be a perfect square

verification: 1681+81=1762

which is not a perfect square

thus, a=x, b=9, c=41

$x^{2}$= 1681-81

  = 1600

$x=\sqrt{1600}$

[4x4=16 and 10x10 = 100}

x= 40

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