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Standard VIII

Mathematics

Elements of a Parallelogram

In Fig., ABCD...

Question

In Fig., $ ABCDE$ is a pentagon. A line through $ B$ parallel to $ AC$ meets $ DC$ produced at $ F$. Show that $ ar.\left(AEDF\right)$$ =$$ ar.\left(ABCDE\right)$

D

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Solution

Prove the area of given quadrilaterals are equal.
$∆ A C B$ and $∆ A C F$ lie on the same base $A C$ and between the same parallel $A C$ and $B F$
Therefore,
$a r . (∆ A C B)$ $=$ $a r . (∆ A C F)$ .
Now, on adding $a r . (A C D E)$ both sides, we get;
$\Rightarrow a r . (∆ A C B) + a r . (A C D E) = a r . (∆ A C F) + a r . (A C D E) \Rightarrow a r . (A B C D E) = a r . (A E D F)$
Hence proved, $a r . (A E D F)$ $=$ $a r . (A B C D E)$

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Similar questions

Q. In given figure the side

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of a parallelogram

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P

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and then parallelogram

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a r e a (A B C D) = a r e a (P B Q R)

Q. In fig,

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(ii)

a r (A E D F) = a r (A B C D E)

The given figure shows a pentagon ABCDE. EG,drawn parallel to DA,meets BA produced at G,and CF,drawn parallel to DB,meets AB produced at F.Show that ar(pentagon ABCDE)=ar( $△ D G F$ ).

Q. The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar(pentagon ABCDE) = ar(∆DGF).

Q. The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.
Show that ar(pentagon ABCDE) = ar(∆DGF).

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In Fig., $ ABCDE$ is a pentagon. A line through $ B$ parallel to $ AC$ meets $ DC$ produced at $ F$. Show that $ ar.\left(AEDF\right)$$ =$$ ar.\left(ABCDE\right)$D

Prove the area of given quadrilaterals are equal.∆ACB and ∆ACF lie on the same base AC and between the same parallel AC and BFTherefore, ar.∆ACB=ar.∆ACF.Now, on adding ar.ACDE both sides, we get;⇒ar.∆ACB+ar.ACDE=ar.∆ACF+ar.ACDE⇒ar.ABCDE=ar.AEDFHence proved, ar.AEDF=ar.ABCDE

In Fig., $ ABCDE$ is a pentagon. A line through $ B$ parallel to $ AC$ meets $ DC$ produced at $ F$. Show that $ ar.\left(AEDF\right)$$ =$$ ar.\left(ABCDE\right)$

D