The hyponetuse (m) and one leg (n) of a right triangle differ by 2. The square of the other side is

a) m+2n

b) m+n

c) 2(m-n)

d) 2(m+n)

e) m-2n

None of these

We know that m - 2 = n ⇒ m = n + 2

So....using the Pythagorean Theorem....we have that the remaining side =

√ [ m^2 - n^2 ]

√ [ ( n + 2)^2 - n^2 ] simplify

√ ( n^2 + 4n + 4 - n^2 )

√ [4n + 4]

So....the square of this side = 4n + 4 [ in terms of n ]

Thanks. I am also trying to get a solution that corresponds with one of the answers.