Parallelograms on Same Base and Between Same Parallels


 
 
Concept Explanation
 

Parallelograms on Same Base and Between Same Parallels

Parallelograms On The Same (Equal) Base and Between The Same Parallels.  Now we are going to discuss about the are of two parallelograms which lie on the same or equal base and between the same parallels.

 

Theorem 1    Parallelograms on the same base and between the same parallels are equal in area.

Given        Two parallelograms ABCD and ABEF, which have same basae AB and which are between the same parallel lines AB and FC.

To prove    ar(parallel ^{gm}ABDC)=ar(parallel ^{gm}ABEF)

Proof       As ABDC is a parallelogram Rightarrow AB = DC   ................(1) [ Opposite sides of a parallelogram]

Similarly in Parallelogram ABEF Rightarrow AB = EF              ................(2)

From Eq. (1) and (2) We get  DC = EF                  ...................(3)

Subtracting FD from both sides in Eq (3) We get   DC- FD = EF - FD Rightarrow CF = DE    ..............(4)

  In Delta s ;AFC;;and;;BED, ;;we;; have

                     AC = BD                            [ Opposite side of the parallelogram ABDC]

                    AF =  BE                             [ Opposite side of the parallelogram ABEF]

                  CF = DE                               [ From Equation(4)]

               So, Delta ;AFC;;cong ;;Delta BED,    [ By SSS criterion of congruence ]

              Rightarrow ;;;ar(Delta AFC)=ar(Delta BED)                ...(5)   [ Because area of congruent figures is equal]

Now, Adding ar(square ABDF ) ob both sides of Eq. (5) we get

Rightarrow ;;;ar(Delta AFC)+ ar(square ABDF)=ar(Delta BED)+ ar(square ABDF)

    Rightarrow        ar(parallel ^{gm}ABDC)=ar(parallel ^{gm}ABEF)

Hence Proved

Illustration: In the given figure, large AB || DC || EF, BE ||AD, ;;and;;AF||DE. If the ar (ABCD) = 25sq.m find the ar(DEFH)

Solution: In the given figure, we have AB || DC and AD || BC,  Rightarrow  ABCD is a parallelogram.

Again,AD || BC Rightarrow AD || GE and AF || DE Rightarrow  ADEG  is a parallelogram.

Now ||gm ABCD and ||gm ADEG are on the same base AD and between the same parallels AD || BE

Rightarrow ar(|| gm ADEG )= ar(|| gm ABCD ) = 25 sq.m.

Again, AF || DE and EF || DC Rightarrow  DEFH  is a parallelogram.

Now ||gm DEFH and ||gm ADEG are on the same base DE and between the same parallels DE || AF.

Rightarrow ar(|| gm DEFH )= ar(|| gm ADEG ) = 25 sq.m.

Hence ar(|| gm DEFH )= 25 sq. m.

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

Two parallelograms are on equal bases and between the same parallels . The ratio of their areas is _____________________

Right Option : B
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Question : 2

In the Given figure If area of parallelogram ABCD = 345 cm^{2}  then find out the area of parallelogram EDCB .

Right Option : A
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Question : 3

In the figure.ABCD and PQCD  are two parallelograms.

Which of the following relationship is true?

 

 

Right Option : C
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