Maths / Areas Of Parallelograms And Triangles / Parallelograms on Same Base and Between Same Parallels

QUESTION
 

Prove that parallelogram on the same base and between the same parallel lines are equal in area.

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

Given : Two parallelograms ABCD and ABEF are on the same base AB and betweem the same parallels.

To prove : ar(ABCD)=ar(ABEF)

Proof : In parallelograms ABEF and ABCD, one common figure can be observed, which is trapezium ABCF.

Both the parallelograms consist of two figures : a triangle and a trapezium. Now, the trapezium is the common figure between these two parallelograms.

therefore             We can write :

ar(ABEF)=ar(ABCF)+ar(BCE): : : : : : : : : : : : ...............(1)

And  ar(ABCD)=ar(ABCF)+ar(AFD): : : : : : : : : : : : ...............(2)

The right hand sides of thew above two equations will be equal only when ar (BCE) = ar (AFD.

For this, we have to prove that BCE and AFD are congruent triangles.

Let us prove it.

In parallelogram ABCD,

AB = AD

Rightarrow AB=CF+FD : : : : : : : : : : : : : : ................(3)

Similarly, in parallelogram ABEF,

AB = EF

Rightarrow AB=CF+CE : : : : : : : : : : : : : : ................(4)

From (3) and (4), we get

CF+FD=CF+CE

Or  FD=CE : : : : : : : : : : : : .................(5)

Now, in Delta AFD ; and ; Delta BEF,

angle 1=angle2                                         ............... (Corresponding angles)

FD=CE                                     ................  (Proved above)

angle 3=angle 4                                        .................  (Corresponding angles)

therefore : : : : : : : : Delta AFD cong Delta BEC            ..............  (ASA)

Rightarrow : : : : : ar(AFD)=ar(BEC)          .......... (6)

Using (1), (2) and (6), we get ar (ABEF) = ar (ABCD).

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