Question
The hypotenuse (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
Work I've done:
x^2+(m-2)^2=m^2
x^2+m^2-4m+4=m^2
x^2+4=4m
x^2=4m-4
Did i do something wrong?
A)m+2n
B)m+n
C)2(m-n)
D)2(m+n)
E)m-2n
Work I've done:
x^2+(m-2)^2=m^2
x^2+m^2-4m+4=m^2
x^2+4=4m
x^2=4m-4
Did i do something wrong?
Answers
Steve
Ok as far as you go.
Now, 2 = m-n, so
4m-4 = 4m-2(m-n) = 4m-2m+2n = 2(m+n)
Note that your basic 6-8-10 triangle works for this.
6^2 + 8^2 = 10^2
6^2 = 2(8+10)
Now, 2 = m-n, so
4m-4 = 4m-2(m-n) = 4m-2m+2n = 2(m+n)
Note that your basic 6-8-10 triangle works for this.
6^2 + 8^2 = 10^2
6^2 = 2(8+10)
Anonymous
2=m-n