In triangle ABC, D and E are two points on the sides AB and AC respectively so that DE $$\parallel$$ BC and $$\frac{AD}{BD} = \frac{3}{4}$$. The ratio of the area of $$\triangle$$ABC to the area of trapezium DECB is:

So this is our corresponding figure.

Let us assume AD = 3x and BD = 4x ( Since it is given that AD:BD= 3:4)

Therefore AB = AD + BD =3x + 4x = 7x

Now Consider two triangles here,

ABC and ADE

(Since DE parallel to BC)

$$\angle$$ADE = $$\angle$$ABC (Corresponding angle)

$$\angle$$AED= $$\angle$$ACB (Corresponding angle)

$$\angle$$BAC = $$\angle$$DAE (Same angle)

The two triangles ABC and ADE are similar.

Hence we can say,

By the property of similar traingles,

We can say that Area of two similar triangles are proportional to the square of their corresponding sides,

Areas of $$\frac{ADE}{ABC}$$ =$$\left(\frac{AD}{BD}\right)^2 $$=$$\frac{9}{49}$$

Hence, ADE = 9x and ABC = 49x

But we have to find the area of Trapezium DECB, which is nothing but the difference of the areas of the two triangles above,

Area of trapezium DECB is 40x.

Hence , the ratio area of $$\triangle$$ABC to the area of trapezium DECB = 40:49