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Coordinate Geometry(Right Triangle)

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    If y=x+2, y=2x+3 are medians of right angled triangle ABC (angle A=90) through B and C respectively such that |BC|=81 units. Find area of triangle.

    3. The attempt at a solution
    I have been trying this since a long time.

    I am posting my work in the form of 2 images. I hope my writing is legible. I think I have generated enough equations for the unknowns. But to solve them is difficult. Please help me.

    I was also told that there exists a shorter method than this. I shall be glad if you could guide me to a shorter approach.

    Picture 1:
    http://img18.imageshack.us/img18/4679/87911896fu2.jpg [Broken]


    Picture 2:

    http://img25.imageshack.us/img25/694/76694925pt3.jpg [Broken]


    regards,
    Ritwik
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 12, 2009 #2
    Try using www.imageshack.us or create an account on photobucket. They support direct links.
     
  4. Feb 12, 2009 #3
    Please help me with the question!!
     
    Last edited: Feb 12, 2009
  5. Feb 13, 2009 #4

    tiny-tim

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    Hi Ritwik! Thanks for the PM. :smile:

    (I find it very difficult to read the pictures of your handwriting: it would be easier if you would type your work directly into your post)

    Hint: the only importance of the x and y coordinates is that they tell you the angle between the medians.

    Work out that angle, and then forget all about the coordinates, just draw a general right-angled triangle ABC, and work out what the area is, given the length BC and the angle between the medians. :smile:
     
  6. Feb 13, 2009 #5
    Hi Tim,
    Its me who should thank you.
    I am very grateful that you replied.
    Your advice was very-very useful. I have been able to solve the question. Now I am working towards a more compact and smarter solution. I hope you will help :P

    Here is how I got the result.
    Let O be the angle between the medians.
    Then tan O=(1/3)

    Area of triangle EMF=(1/12) *Area of the whole triangle(A)

    0.5*(1/3)*M1*(1/3)*M2=(1/12)*A

    Now in the same triangle EMF;
    I apply the cosine rule
    cos(pi-O)=[tex]\frac{((1/3)*M1) ^2 +((1/3)*M2)^2)-(a/2)^2}{\frac{2M1M2}{9}}[/tex]

    After putting values of cos(pi-O), M1*M2. I still have to find M1^2+ M2^2
    Applying cosine rule;
    CF^2(M1^2)=BC^2+BF^2-2*BC* BF* cos B

    Again putting values of 2*BC* BF* cos B from the original cosine rule equation. for triangle ABC

    Thn I plug values of M1^2+ M2^2 in the original equation. an I get the area=729 sq units.
    Now the question arises. Is there a smaller time saving method???
     
    Last edited: Feb 13, 2009
  7. Feb 13, 2009 #6

    tiny-tim

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    Hi Ritwik! :smile:

    I followed you down to …
    … and then I got a bit lost.

    But you could speed things up by using Pythagoras: M12 + M22 = … ? :wink:
     
  8. Feb 14, 2009 #7
    Then I again used the cosine rule there:
    to replace 2*BC*BF*cos B from this expression,
    2*ac*cos B=a^2+c^2-b^2


    But as you said pythagoras was a better option. Thanks a lot.
    I was wondering whether there could be a method without involving the areas of the smaller triangles as I did???

    By the way thanks a lot for your help :D
    and Happy Valentines Day!!!
     
  9. Feb 15, 2009 #8

    tiny-tim

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    You could use vectors instead …
    put the origin at O = A, so that OB = b, OC = c.

    Then the medians are (b/2 - c) and (c/2 - b), and b.c = 0

    so cos = (b/2 - c).(c/2 - b)/√((b/2 - c)2(c/2 - b)2)

    which you can turn into an equation in bc by using b2 + c2 = 812, and by completing the square. :wink:
     
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