In the triangle ABC, the median BM is drawn, on the side AB, the point K is taken so that AK = 1 / 5AB

In the triangle ABC, the median BM is drawn, on the side AB, the point K is taken so that AK = 1 / 5AB, the area of the triangle AMK is 3, find the area of the triangle ABC.

Let side AB = a, side AC = c, the angle between AB and AC will be b;
The area of a triangle ABC is calculated as half the product of these sides multiplied by the sine of the angle between them, i.e. a * b * sin (b) / 2;
The median of a triangle is a segment connecting the apex of the triangle with the middle of the opposite side, i.e. point K divides side AC by 2, i.e. AK = a / 2;
The area of the triangle KAM = 3 and is calculated as (a / 5) * (b / 2) * sin (b) / 2 = 3;
From (a / 5) * (b / 2) * sin (b) / 2 = 3 we find a * b * sin (b) / 2 by multiplying by 5 and 2;
Area ABC = 3 * 5 * 2 = 30.