Side lengths of an right triangle are #sqrtn#, #sqrt(n+1)#, and #sqrt(n+2)#. How do you find #n#?

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Answer 1 · 2018-03-16T07:59:31

#n=1#

Pythagoras' theorem tells us that the sides of a right angled triangle with legs of length #a, b# and hypotenuse of length #c# satisfy:

#c^2 = a^2 + b^2#

So in our example, we require:

#(sqrt(n+2))^2 = (sqrt(n))^2+(sqrt(n+1))^2#

Assuming #n >= 0#, this simplifies to:

#n+2 = n+(n+1)#

Subtracting #n+1# from both sides, this becomes:

#1 = n#

So the only solution is #n = 1#, giving us a triangle with sides #1#, #sqrt(2)# and #sqrt(3)#.