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

QUESTION

In the given figure, $\dpi{120} \fn_jvn \large AB || DC || EF, BE ||AD, \;\;and\;\;AF||DE$. Show that ar (ABCD) = ar(DEFH).

EXPLANATION
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Text

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 )       ..........................(1)

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 )      ........................(2)

From Eq (1) and (2)

ar(|| gm ABCD ) =  ar (|| gm DEFH )

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

Hence proved.

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