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The number of right triangles with sides of lengths \[\sqrt{n+1}\], \[\sqrt{n+2}\], and \[\sqrt{n+3}\] is: a) 0 b) 1 c) 2 d) 3 e) unbounded
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\(\sqrt{n+3}\) is the longest side, so its the hypotenuse. now apply pythagoras theorem.
\[\sqrt{n+1}^{2}+\sqrt{n+2}^{2}=\sqrt{n+3}^{2}\]
(n+1)+(n+2)=(n+3)
yeah, how many values of 'n' satisfy that equation ?
2n+3=n+3 n=0
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number of right triangles = number of values of n
0 right triangles?
no, n=0 is one value of n. so there's one right triangle with side, sqrt 1 sqrt 2 and sqrt 3
oh, so there is 1 right triangle, where n = 0
yup.
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