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
Explain TypeExplanation Content
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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,

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

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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