Question

Two parallelograms $$\text ABCD$$ and $$\text EFCD$$ that have the same base $$\text CD$$ and lie on the same parallels $$\text AF$$ and $$\text CD$$, then

A
Area(ABCD) = Area(AFCD)
B
Area(ABCD) = Area(EFCD)
C
Area(EBDC) = Area(EFCD)
D
None of these

Solution

The correct option is B Area($$\text ABCD$$) = Area($$\text EFCD$$)ABCD is a parallelogram, then $$AD\parallel BC$$& AB is transversal , therefore $$\angle DAB = \angle CBF \rightarrow (1)$$ (corresponding angles)Also $$AD=BC\rightarrow (2)$$ (opposite sides of $$\parallel gm$$)Similarly in $$\parallel gm$$ CDEF,$$ED\parallel CF$$ & EF is transversal , therefore$$\angle DEA= \angle CFE \rightarrow (3)$$     (corresponding angles)In $$\bigtriangleup AED$$ & $$\bigtriangleup BFC$$$$\angle DEA=\angle CFE$$  [from(3)]     $$AD=BC$$       [from(2)] $$\angle DAB=\angle CBF$$   [from (1)]$$\therefore \bigtriangleup AED\cong \bigtriangleup BFC$$   (By ASA congruency)$$ar\bigtriangleup AED=ar\bigtriangleup BFC$$   (areas of congruent part)Now, $$ar ABCD= ar \bigtriangleup ADE+ ar EBCD$$                          $$= ar \bigtriangleup BFC+ ar EBCD$$                     $$\therefore ar ABCD= ar EFCD$$Mathematics

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