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Math Help - Triangle and Perpendicular medians

  1. #1
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    Thumbs up Triangle and Perpendicular medians

    Hi Forum!
    I found an interesting question.
    In triangle ABC, the median from vertex A is perpendicular to the other median from vertex B. If the lengths of sides A and B are 6 and 7 respectively, then the length of AB is?

    I managed to do some of it. I got impressed with the solution, since it was somehow complex.
    It says, that the ratio of the medians will be 2:1 because they are perpendicular.
    Can this be proved? I don't understand the line of thought here.
    What median is mentioned? We can have m_a, m_b or m_c.
    Calculating m_c it gets easy to find AB.
    So, this involves a little bit of Pythagorean Theorem and Apollonius' Theorem.
    But the problem here is the usage.
    It is difficult to find a way to solve it without a graph too.

    Thanks.
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  2. #2
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    earboth's Avatar
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    Re: Triangle and Perpendicular medians

    Quote Originally Posted by Zellator View Post
    Hi Forum!
    I found an interesting question.
    In triangle ABC, the median from vertex A is perpendicular to the other median from vertex B. If the lengths of sides A and B are 6 and 7 respectively, then the length of AB is?

    I managed to do some of it. I got impressed with the solution, since it was somehow complex.
    It says, that the ratio of the medians will be 2:1 because they are perpendicular.
    Can this be proved? I don't understand the line of thought here.
    What median is mentioned? We can have m_a, m_b or m_c.
    Calculating m_c it gets easy to find AB.
    So, this involves a little bit of Pythagorean Theorem and Apollonius' Theorem.
    But the problem here is the usage.
    It is difficult to find a way to solve it without a graph too.

    Thanks.
    1. Draw a sketch. (see attachment)

    2. The medians of a triangle intersect in the centroid, labeled here by S. S divides each median in the ratio 2 : 1 (from the vertex!)

    3. According to the text the median m_A = m and m_B = s are perpendicular. That means you have 2 right triangles.

    4. Use Pythagorean theorem and you'll get a system of equations:

    \left|\begin{array}{rcl}\left(\frac13 m \right)^2+\left(\frac23 s \right)^2 &=& \left(\frac62 \right)^2 \\ \left(\frac23 m \right)^2+\left(\frac13 s \right)^2 &=& \left(\frac72 \right)^2\end{array}\right.

    Solve for m and s.

    5. Use Pythagorean theorem to calculate the length of the side |\overline{AB}| = c :

    \left(\frac23 m \right)^2+\left(\frac23 s\right)^2 = c^2

    You should come out with c = \sqrt{17}
    Attached Thumbnails Attached Thumbnails Triangle and Perpendicular medians-seiteausmedians.png  
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  3. #3
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    Re: Triangle and Perpendicular medians

    Quote Originally Posted by earboth View Post
    1. Draw a sketch. (see attachment)

    2. The medians of a triangle intersect in the centroid, labeled here by S. S divides each median in the ratio 2 : 1 (from the vertex!)

    3. According to the text the median m_A = m and m_B = s are perpendicular. That means you have 2 right triangles.

    4. Use Pythagorean theorem and you'll get a system of equations:

    \left|\begin{array}{rcl}\left(\frac13 m \right)^2+\left(\frac23 s \right)^2 &=& \left(\frac62 \right)^2 \\ \left(\frac23 m \right)^2+\left(\frac13 s \right)^2 &=& \left(\frac72 \right)^2\end{array}\right.

    Solve for m and s.

    5. Use Pythagorean theorem to calculate the length of the side |\overline{AB}| = c :

    \left(\frac23 m \right)^2+\left(\frac23 s\right)^2 = c^2

    You should come out with c = \sqrt{17}
    Hi earboth!
    Thanks for the graph! It is much easier to think with it.
    I also have found easy to understand with your explanation!

    Thanks again!
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