CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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, $$ar ABCD= ar \bigtriangleup ADE+ ar EBCD$$
                          $$= ar \bigtriangleup BFC+ ar EBCD$$
                     $$\therefore ar ABCD= ar EFCD$$

969511_1054862_ans_f0cdea6d53c841e58b08e80f4f9dae88.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image