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My question is:

In triangle ABC ,length of median from vertex A is $13$ , length of median from vertex B is $14$ , length of median from vertex C is $15$. Compute the area of triangle ABC.

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One strategy, probably not optimal, is to find the lengths of the sides, and then use Heron's Formula.

Take a triangle $XYZ$, with sides $x$, $y$, $z$ as usual, and let $m$ be the length of the median from $Z$ to the side $XY$, which has length $z$.

Let $P$ be the midpoint of $XY$. We have divided our triangle into two triangles, $ZPX$ and $ZPY$. Let $\theta=\angle ZPX$ and $\phi=\angle ZPY$. Note that $\theta+\phi=\pi$. By the Cosine Law, we have $$y^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\theta.$$ Also by the Cosine Law, we have $$x^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\phi.$$ But $\cos\phi=-\cos\theta$. Add. The awkward cosine terms cancel, and we get $$x^2+y^2=2m^2+2\left(\frac{z}{2}\right)^2,$$ which yields the result $$m^2=\frac{1}{4}\left(2x^2+2y^2-z^2\right).$$

The rest is downhill. We go back to the triangle of the question. The three lengths of the medians give us three linear equations in the squares of the side lengths. If the sides of our triangle are $a$, $b$, and $c$, we get $$4\cdot 13^2=2a^2+2b^2-c^2;\quad 4\cdot 14^2=2b^2+2c^2-a^2;\quad 4\cdot 15^2=2c^2+2a^2-b^2.$$ Solve. A simple way is to use the symmetry of the equations by first of all "adding" the above three equations to first find $a^2+b^2+c^2$.

Now that we have the sides, we can use Heron's Formula.

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Use formula:

$$A = \frac{4}{3} \sqrt{s \cdot (s_m - m_1) \cdot (s_m - m_2) \cdot (s_m - m_3)} $$

where $m_1$ = 1st median, $m_2$ = 2nd median, $m_3$ = 3rd median, $s_m = (m_1+m_2+m_3)/2$

It is one of simplest method

Regards Jay Sharma

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Area: $$\Delta=\dfrac 43\sqrt {s(s-a)(s-b)(s-c)}$$ Where $a,b$ & $c$ are median lengths & $s$ is the semi-perimeter.

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  • $\begingroup$ The formula is a little off. Correct it. $\endgroup$ – Red Floyd Feb 2 '17 at 11:13

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