# Thread: Triangle and Perpendicular medians

1. ## 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.

2. ## Re: Triangle and Perpendicular medians

Originally Posted by Zellator
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 $c = \sqrt{17}$

3. ## Re: Triangle and Perpendicular medians

Originally Posted by earboth
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!