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Homework Help: 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

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    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]
     
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