ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD at F. If ar $A (Δ D F B) = 3 c m^{2}$ , find the area of the parallelogram ABCD.

6 c m^{2}

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9 c m^{2}

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12 c m^{2}

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15 c m^{2}

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Solution

The correct option is C $12 c m^{2}$
$G i v e n - A B C D i s a p a r a l l e l o g r a m . T h e l i n e s B C E & A F E m e e t a t E . B C = E C a n d a r Δ D F B = {3 c m}^{2} T o f i n d o u t - a r p a r a l l e l o g r a m A B C D = ? S o l u t i o n - B C = E C & A B ∥ F C (g i v e n) ∴ F C = \frac{1}{2} A B (b y m i d p o i n t t h e o r e m) B u t A B = D C (o p p o s i t e s i d e s o f a p a r a l l e l o g r a m) ∴ F C = \frac{1}{2} D C ⟹ F C = D F = \frac{1}{2} D C . . . . . . . (i) N o w Δ D F B h a s t h e b a s e o n t h e l i n e D C i . e b a s e o f t h e p a r a l l e l o g r a m A B C D & b o t h t h e f i g u r e s a r e w i t h i n t h e s a m e p a r a l l e s A B & D C . . . . . . . . . . . . . (i i) ∴ F r o m (i) & (i i) w e h a v e a r Δ D F B = \frac{1}{2} (\frac{1}{2} a r A B C D) ⟹ a r A B C D = a r Δ D F B \times 4 = {3 c m}^{2} \times 4 = {12 c m}^{2} A n s - a r A B C D = {12 c m}^{2}$

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