In the figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ΔDFB is 3 cm², then area of the parallelogram ABCD is :

1. 9 cm²

2. 10 cm²

3. 12 cm²

4. 15 cm²

We know that,

Area of triangles on the same base and between the same parallel lines are equal.

△ ADF and △ DFB lie on same base DF and between same parallel lines AB and DC.

∴ Area of △ ADF = Area of △ DFB = 3 cm²

By converse of mid-point theorem,

If a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side.

In △ ABE,

C is the mid-point of BE and CF || AB.

∴ F is the mid-point of AE. (By converse of mid-point theorem)

∴ EF = AF.

In △ ADF and △ EFC,

⇒ ∠AFD = ∠EFC (Vertically opposite angles are equal)

⇒ EF = AF (Proved above)

⇒ ∠DAF = ∠CEF (Alternate interior angles are equal)

∴ △ ADF ≅ △ ECF (By A.S.A. axiom)

We know that,

Area of congruent triangles are equal.

∴ Area of △ EFC = Area of △ ADF = 3 cm².

In △ BFE,

Since, C is the mid-point of BE.

∴ CF is the median of triangle BFE.

We know that,

Median of triangle divides it into two triangles of equal areas.

∴ Area of △ BFC = Area of △ EFC = 3 cm².

From figure,

⇒ Area of △ BDC = Area of △ BDF + Area of △ BFC

⇒ Area of △ BDC = 3 + 3 = 6 cm².

We know that,

The area of triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

||gm ABCD and △ BDC lies on same base DC and between same parallel lines AB and DC.

∴ Area of △ BDC = $\dfrac{1}{2}$ Area of ||gm ABCD

⇒ Area of ||gm ABCD = 2 × Area of △ BDC

⇒ Area of ||gm ABCD = 2 × 6 = 12 cm².

Hence, option 3 is the correct option.