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Area of a right triangle with little data

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data
    9ld4qx.png

    2. Relevant equations
    x^2 + y^2 = 9
    A = 0.5xy
    x ≠ y

    3. The attempt at a solution
    x^2 + y^2 = 9
    A = xy/2

    (x + y)^2 = x^2 + 2xy + y^2 = 9 + 2xy = 9 + 4A
    A = ((x+y)^2 - 9)/4

    Then I am lost. I need to find the area.
     
    Last edited: May 6, 2014
  2. jcsd
  3. May 6, 2014 #2

    BruceW

    User Avatar
    Homework Helper

    it looks like you're trying to use some combination of x and y to find the answer. But why not try to find x and y individually, then find the answer?
     
  4. May 6, 2014 #3
    Because I simply couldn't? The pdf that contained the problem is dealing with the Pythagorean Theorem. I do not think that I need more trigonometry for it.

    If we call the point on the hypotenuse (that divides it in 1 and 2) P:
    As x != y, the two acute angles are different from each other and, subsequently, I do not have right angles at the intersection with the hypotenuse.

    To where?
     
  5. May 6, 2014 #4
    Name the triangle as ##ABC## right-angled at ##B## with ##AB=x##. Let D be the point on AC such that AD=1.

    Also, let ##\angle BAC=\theta##, then you have ##\angle ADB=\frac{3\pi}{4}-\theta## and ##\angle BDC=\frac{\pi}{4}+\theta##. Apply law of sines separately in triangles ADB and BDC to obtain x and y in terms of ##\sin(\pi/4+\theta)##, you should then be able to find x and y from ##x^2+y^2=9##.
     
    Last edited: May 6, 2014
  6. May 6, 2014 #5
    Thank you SO much. I remade the image, made the problem clearer with your info.
    Right to the point:

    [tex]\frac{x}{\sin{\left(\frac{3\pi}{4}-\theta\right)}}=\frac{2}{\frac{\sqrt{2}}{2}}\Rightarrow x=\frac{2}{\sqrt{2}}\sin{\left(\frac{3\pi}{4}-\theta\right)}=\left(\frac{2}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{2}\cos{\theta}+\frac{\sqrt{2}}{2}\sin{\theta}\right)\Rightarrow x=\cos{\theta}+\sin{\theta}[/tex]
    [tex]\frac{y}{\sin{\left(\frac{\pi}{4}+\theta\right)}}=\frac{2}{\frac{\sqrt{2}}{2}}\Rightarrow y=\frac{4}{\sqrt{2}}\sin{\left(\frac{\pi}{4}+\theta\right)}=\left(\frac{4}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{2}\cos{\theta}+\frac{\sqrt{2}}{2}\sin{\theta}\right)\Rightarrow y=2\left(\cos{\theta}+\sin{\theta}\right)[/tex]
    [tex]y=2x[/tex]
    [tex]y^2+x^2=5x^2=9[/tex]
    [tex]x=\sqrt{\frac{9}{5}}\Rightarrow A=\frac{x\cdot y}{2}=\frac{9}{5}[/tex]

    Can anyone, please, confirm this?
     
  7. May 6, 2014 #6
    Looks right to me! :)
     
  8. May 6, 2014 #7
    Is the law of sines derived from the Pythagorean Theorem? If not, do you think it is doable without the law of sines?
     
  9. May 6, 2014 #8
    Yes, there is an alternative way, I missed it before.

    Do you see that AD is an angle bisector? Can you recall some property related to it (hint, it is related to ratio of sides)?
     
  10. May 6, 2014 #9
    [tex]\frac{x}{y}=\frac{1}{2}[/tex]
    Holy ****. I went all over the mountain instead of taking the tunnel!
     
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