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$A B C D$ is a parallelogram in which $B C$ is a parallelogram in which is produced to $E$ is produced to such that $C E = B C$ such that . $A E$ . intersects $C D$ intersects at $F$ at . If the area of $Δ D F B = 3 cm 2$ . If the area of , then the area of $∥ g m A B C D$ , then the area of is:

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6 cm 2

12 cm 2

18 cm 2

3 cm 2

answer is B.

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Detailed Solution

Given that,

A B C D

is a parallelogram in which

B C

is produced to

E

such that

C E = B C

A E

intersects

C D

F

.
The figure is as follows,
Question Image

We have to find the area of the parallelogram when

Δ D F B = 3 cm 2

.
Here AD is parallel to BE. In

Δ A D F

and

Δ E C F

∠ A D F

= ∠ E C F

. (Alternate interior angles are equal.)
As

A D = B C = C E

.[parallel sides of a parallelogram are equal]
So,

A D = C E

.
And

∠ D F A = ∠ C F A

(vertically opposite angles are equal)
Therefore, by AAS theorem,

Δ A D F ≅ Δ E C F

.
Therefore,

a r Δ A D F = a r Δ E C F

.
Since

a r (Δ A D F) = a r (Δ E C F)

.
By CPCT, DF=FC.
So,

B F

is a median in

Δ B D C

Area of parallelogram

A B C D = 2 ×

Area of

Δ B D C

= 2 × 6 = 12

Hence, the area is

12 cm 2

.
Hence, option 2) is correct.

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