Class 9---areas of parallelograms and triangles----

Dear Student,

Please find below the solution to the asked query:

Given  :  ABCD and AEFD are two parallelograms. 

So,

AB  | |  CD , BC  | | DA 

And

FE | | DA and AE  | | FD  , SO

DA  | | PQ (  As PQ is the part of line FE and we know FE  | | DA ) and DQ  | | AP (  As DQ is the part of line CD and AP is the part of line AB and we know AB  | | CD )

So,

APQD is also a parallelogram .

And

Area of parallelogram APQD  = Area of parallelogram AEFD                 ---- ( 1 )

As both have same base ( AD )  and have same height ( As lies in the same parallel lines DA  | | QE )

And

Area of $∆$ PEA = Area of parallelogram AEFD  -  Area of trapezium APFD                ---- ( 2 )
 

And

Area of $∆$ QFD = Area of parallelogram APQD  -  Area of trapezium APFD , From equation 1 we get 

Area of $∆$ PEA = Area of parallelogram AEFD  -  Area of trapezium APFD                  ---- ( 3 )

From equation 2 and 3 we get

Area of $∆$ PEA = Area of $∆$ QFD                                                                       ( Hence proved )


Hope this information will clear your doubts about Areas of Parallelograms and Triangles .

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Regards,