Finding the area of the triangle

There is a point inside a equilateral triangle which is at a distance 1,2 and 3 from the sides then what is the area of the triangle? Please help me.

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Hint: The height of the equilateral triangle must be the sum of the distances between an internal point and the sides.

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Suppose that the triangle ABC has sidelength $a$ and the point in the interior is P. Now calculate the areas of ABP, BCP and CAP (you know their base and height !) and compare it with the formula for the area of an equilateral triangle.

This gives you $a$, so you can calculate the area, too.

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 This is the obvious way to do it. Some other ways are more complicated. – Michael Hardy Oct 12 '11 at 18:44

Let's denote :

$AE=a_1 , BE=a_2 , BG=a_3 . GC=a_4 , FC=a_5 , AF=a_6$ ,$ED=1 ,DF=2 ,DG=3$(see picture bellow)..then:

$A=\frac{a_1}{2}+\frac{a_2}{2}+\frac{3a_3}{2}+\frac{3a_4}{2}+\frac{2a_5}{2}+\frac{2a_6}{2}=\frac{a}{2}+\frac{3a}{2}+a=3a$

You may calculate $a$ from equality $$3a=\frac{a^2\sqrt{3}}{4}$$

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