+0

# Triangle Problem

-1
50
2

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

Guest Nov 17, 2017

#1
+79894
+2

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 ]

CPhill  Nov 17, 2017
Sort:

#1
+79894
+2

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 ]

CPhill  Nov 17, 2017
#2
0

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

Guest Nov 17, 2017

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