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 $(1)$: $$ \frac43\sqrt{s(s-a)(s-b)(s-c)}\tag{2} $$

You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:

I guess you saw the right triangle.

There exists a formulae giving the area of a triangle in function of its medians. It is $$A=\frac13\sqrt{2\alpha^2\beta^2+2\beta^2\gamma^2+2\gamma^2\alpha^2-\alpha^4-\beta^4-\gamma^4}$$ where it is clear what are $\alpha,\beta$ and $\gamma$.