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Parallelograms on Same Base and Between Same Parallels

Areas Of Parallelograms And Triangles I - Concepts

Congruent Figures

Figures on the Same Base and Between the Same Parallels

Parallelograms on Same Base and Between Same Parallels

Area of Parallelograms

Area of Triangle

Class - 9th Foundation NTSE Subjects

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

In the figure , ABCDis a teapezium. If DP $left | right |$ BC, then which of the following relationship is true?

AD=BC

2ar(PQB) = 1 ar(ABCD)

ar(PBCD)=2 ar(PQD)

$Delta$ ADP = $Delta$ PBQ

Right Option : C

View Explanation

Question : 2

In the given figure, ABCD and PQCD are two parallelogram on the same base CD. If ar(ABCD) =X sq. units and ar(PQCD)=y sq. units,which of the following realtionships is true?