How to find area of triangle from its medians The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is

a) $48$

b) $144$

c) $24$

d) $72$

I don't want whole solution just give me the hint how can I solve it.Thanks.
Area of a Triangle from the Medians

A triangle is divided in to $6$ equal areas by its medians:

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In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.

In the case where the two red triangles share a common $\frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.

Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:

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The triangle in green has sides $\frac23a$, $\frac23b$, and $\frac23c$, and by Heron's formula has area
$$
\frac49\sqrt{s(s-a)(s-b)(s-c)}\tag{1}
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
\frac43\sqrt{s(s-a)(s-b)(s-c)}\tag{2}
$$
There is a direct formula:

Let
$$s = (m_1+m_2+m_3)/2,$$

Then
$$\text{area} = \frac{4}{3}\sqrt{s(s-m_1)(s-m_2)(s-m_3)}.$$

This gives answer of above question as $72$.