Question

How is this possible ? Even though the figure satisfies Pythagoras Theorem, How can the measurement of a length (Hypotenuse) be 0 ?
Refer The Attachment for the Figure !

Hypotenuse = 0,
Adjacent side = - 1 (or) can be taken as 1
Opposite side = i (√-1)





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Answer 1

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$\sf \: by \: pythagorean \: theorem :$

$\sqrt{ {1}^{2} + { (\sqrt{ - 1}) }^{2} }$

$\sqrt{1 + ( - 1)}$

$\sqrt{1 - 1}$

$\sqrt{0}$

$0$

But, here's the problem :

There no numer existing whose square is - 1, as we take i (√-1) so we cannot take i as the side

Answer 2

Answer:

$by \: \: \: \: pythagorean \: \: \: \: theorem:</p><p></p><p>\sqrt{ {1}^{2} + { (\sqrt{ - 1}) }^{2} }12+{(−1) }^{2} \\ </p><p> \\ </p><p>\sqrt{1 + ( - 1)}1+(−1) \\ </p><p></p><p>\sqrt{1 - 1}1−1 \\ </p><p></p><p>\sqrt{0}\\ </p><p></p><p>0 \\ 0</p><p></p><p>$

But, here's the problem :

There no numer existing whose square is - 1, as we take i (√-1) so we cannot take i as the side