Point D, AD: DC = 1: 5 is taken on the AC side of triangle ABC with an area of 36 cm2. Find the area of a triangle ABD.

Let’s build the height of the HV triangle ABC.

The segment BD divides the triangle ABC into two triangles, ABD and CBD.

In triangles ABD and CBD, the height BH is common, then the areas of these triangles are referred to as the lengths of their bases.

Let AD = X cm, then, by condition, CD = 5 * X cm.

Then: Savd / Ssvd = АD / СD = 1/5.

Ssvd = Savd * 5.

Savs = 36 = Savd + Svd = Savd + 5 * Savd = S6 * Savd.

Savd = Savs / 6 = 36/6 = 6 cm2.

Answer: The area of the triangle ABD is 6 cm2.