# 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.

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@labbhattacharjee Thanks I got a new formula. –  iostream007 May 19 '13 at 5:40
You can also use Appolonius theorem. –  Aryabhata May 19 '13 at 5:43
@iostream007, welcome. Have you tried proving it? –  lab bhattacharjee May 19 '13 at 5:44
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question? –  iostream007 May 19 '13 at 5:48

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.

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What right triangle? Can you please elaborate? –  Aryabhata May 19 '13 at 6:05
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10. –  nikamed May 19 '13 at 6:06
Thanks! I see it now :-) +1. –  Aryabhata May 19 '13 at 6:08

Area of a Triangle from the Medians

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

$\hspace{2cm}$

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:

$\hspace{3cm}$

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

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The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.

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Are you sure about the $75\%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles. –  N. F. Taussig Mar 21 at 9:47

there is a direct formula. let s be there. s= (m1+m2+m3)/2 area= square_root(s*(s-m1)(s-m2)(s-m3));

this gives answer of above ques as 72.

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There is a missing factor of $\frac43$ missing from your formula. –  robjohn Sep 8 '14 at 2:17

4/3 * 1/2*12*9 = 72 area of triangle of medians = 1/2*12*9=54

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You should provide some text so that your answer is easier to follow. Also, $54$ is not one of the given options... –  user1729 Sep 1 '14 at 11:47