The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, mee

1.	The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced to G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar (pentagon ABCDE) = ar (△ DGF).
Answer» Consider △ DGA and △ AED We know that both the triangles have the same base AD and lie between the parallel lines AD and EG. So we get Area of △ DGA = Area of △ AED ……. (1) Consider △ DBC and △ BFD We know that both the triangles have the same base DB and lie between the parallel lines BD and CF. So we get Area of △ DBF = Area of △ BCD ……. (2) By adding both the equations Area of △ DGA + Area of △ DBF = Area of △ AED + Area of △ BCD By adding △ ABD both sides Area of △ DGA + Area of △ DBF + Area of △ ABD = Area of △ AED + Area of △ BCD + Area of △ ABD So we get Area of △ DGF = Area of pentagon ABCDE Therefore, it is proved that ar (pentagon ABCDE) = ar (△ DGF).

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

Answer»

Consider △ DGA and △ AED

We know that both the triangles have the same base AD and lie between the parallel lines AD and EG.

So we get

Area of △ DGA = Area of △ AED ……. (1)

Consider △ DBC and △ BFD

We know that both the triangles have the same base DB and lie between the parallel lines BD and CF.

So we get

Area of △ DBF = Area of △ BCD ……. (2)

By adding both the equations

Area of △ DGA + Area of △ DBF = Area of △ AED + Area of △ BCD

By adding △ ABD both sides

Area of △ DGA + Area of △ DBF + Area of △ ABD = Area of △ AED + Area of △ BCD + Area of △ ABD

So we get

Area of △ DGF = Area of pentagon ABCDE

Therefore, it is proved that ar (pentagon ABCDE) = ar (△ DGF).