In a land of triangular shape ∆ABC , a line DE intersects the sides AB and AC in such a way that AB : BD = AC : CE. If the area of the land shaped as ∆DBC is 10 sq. meters, what is the area of ∆BEC?

Given that AB : BD = AC : CE, we can say that the triangles ∆ABC and ∆BDE are similar by the Side-Splitter Theorem.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, the ratio of the areas of ∆ABC and ∆BDE is equal to the square of the ratio of AB to BD or AC to CE.

Let's denote the ratio AB : BD = AC : CE = k : 1. Then, the area of ∆ABC is k^2 times the area of ∆BDE.

The area of ∆ABC is also equal to the sum of the areas of ∆BDE and ∆BEC. Therefore, we can write the equation:

k^2 * Area(∆BDE) = Area(∆BDE) + Area(∆BEC)

Substituting the given Area(∆BDE) = 10 sq. meters, we get:

k^2 * 10 = 10 + Area(∆BEC)

Rearranging the terms, we find:

Area(∆BEC) = k^2 * 10 - 10

Since k is greater than 1 (because AB > BD and AC > CE), the area of ∆BEC is greater than zero. Therefore, the area of ∆BEC is (k^2 - 1) * 10 sq. meters.