If the length of the sides of a right triangle are the square roots of three consecutive integers...?

Question

If the length of the sides of a right triangle are the square roots of three consecutive integers, what is the sum of those three integers? Any help would be appreciated.

Answer 1

a^2 + b^2 = c^2

If b=a+1 and c=a+2, the substitute it all out and do the math for 'a'.

Answer 2

i believe it is 12... a 3,4,5 triangle has a right angle.  

Answer 3

your lenghts are sqrt(n), sqrt(n+1) and sqrt(n+2)

squaring these according to the pythagorean theorem gives you:

n+n+1=n+2

2n+1=n+2

n=1

so your three integers sum to 1+2+3=6

Answer 4

call the first integer x

then sqrt(x), sqrt(x + 1) and sqrt(x + 2) are three sides

by Pythagorean theorem

x + x +1= x + 2

2x + 1= x + 2

x= 1

so sum of those 3 integers are 1 + 2 +3= 6

Answer 5

12, since it's a 3-4-5 right triangle. But that is for the lengths, not their square roots. I don't think the question as you posted it has an answer.

Answer 6

a^2 = x

b^2 = x + 1

c^2 = x + 2

x + x+1 = x + 2

2x +1 = x+2

x = 1

Sum = 3x +3

        = 3(1) + 3

        = 6