The Area of a Triangle

from the Lengths of the Medians

__The problem__: Given the lengths of the three medians, find the area of the triangle.

The question can be, and often is, solved by means of vectors, but the following, while somewhat longer, has a simple elegance that is very attractive to me.

We begin with an illustration of the problem (left below).

I will accept it as understood that AG:AD = 2:3 and simlarly for the other medians. We will use K{ABC} to mean the area of triangle ABC. To the right below we have made a subtle addition to the original illustration, we have added the 180 degree rotation of point G about point D.

This creates the line segment GG’ which is congruent to AG, and thus is 2/3 of AD. The dotted line G’B is included to help you see that CGBG’ is a parallelogram, and thus CG’ is congruent to GB which is 2/3 of BE.

Now observe the triangle G’CG, formed by segments G’C, G’G, and CG. This triangle has three sides that are each 2/3 the length of a median. If we had a triangle with lengths the sides of the medians it would have an area we shall call K(Median). The triangle formed GCG’, by the properties of similar triangles, must have an area that is 4/9 K(Median).

Note now that CDG’ is congruent to BDG and thus has the same area. This means that triangles CGG’ and CGB have the same area. It is easily seen however that K(CGB) = 1/3 K(ABC).

Lets put these last few statements together. We have:

K(GCG’)= 4/9 K(Median) and

K(GCG’)= K(CGB)=1/3 K(ABC)

From this we can establish that 4/9 K(Median) = 1/3 K(ABC) leading to **K(ABC)= 4/3 K(Median).**

In summary then, we have shown that if we have the three medians of a triangle, we can find the area of the triangle that would be contained by three such sides, and the area of the original triangle is 4/3 that value.