1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area of the triangle math

  1. Jan 10, 2005 #1
    The shorter sides of a right-angled triangle are of length (x+y)cm, (x-y)cm respectively. Given that the length of the hypotenuse is [tex]\sqrt68[/tex]cm and that the area of the triangle is [tex]8cm^2[/tex], find x and y.

    [tex]\displaystyle{(x+y)^2 + (x-y)^2 = (\sqrt68)^2}[/tex]
    [tex]\displaystyle{x^2+y^2=34}[/tex] ----------- (1)
    [tex]\displaystyle{\frac{1}{2}(x+y)(x-y)=8}[/tex]
    [tex]\displaystyle{x^2+y^2=16}[/tex] ----------- (2)
    I can't seem to solve it. Did I do something wrong?
     
  2. jcsd
  3. Jan 10, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Equation (2) should be [itex]x^2-y^2=16[/itex].

    Can you solve it now?
     
  4. Jan 10, 2005 #3
    Why?
    This is how I did it.
    [tex]\displaystyle{\frac{1}{2}(x+y)(x-y)=8}[/tex]
    [tex]x^2-xy+xy+y^2=16[/tex]
    [tex]\displaystyle{x^2+y^2=16}[/tex]
    Does that mean that -y multiply by y equal to -y^2? Isn't it y^2?
     
  5. Jan 10, 2005 #4
    [tex]-y \times y = -y^2[/tex] is correct. Maybe you're thinking of [tex]-y \times -y = y^2[/tex]
     
  6. Jan 10, 2005 #5
    I was thinking of the first one. I always thought that because there is a square, so it would be positive.
     
  7. Jan 10, 2005 #6
    No, it's negative!
     
  8. Jan 10, 2005 #7
    -y*y = -1*y*y = -1*(y*y) = -1*[tex]y^2[/tex] = -[tex]y^2[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?