Q. 224.8( 9 Votes )

# 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).

Answer :

Given : ABCDE is a pentagon EG is drawn parallel to DA which meets BA produced at G and CF is drawn parallel to DB which meets AB produced at F

To prove: area(pentagon ABCDE) = area(∆DGF)

Proof:

Consider quadrilateral ADEG. Here,

area(∆AED) = area(∆ADG) ------------- (1)

[since two triangles are on same base AD and lie between parallel line i.e, AD||EG]

Similarly now, Consider quadrilateral BDCF. Here,

area(∆BCD) = area(∆BDF) ---------------- (2)

[since two triangles are on same base AD and lie between parallel line i.e, AD||EG]

Adding Eq (1) and (2) we get

area(∆AED) + area(∆BCD) = area(∆ADG) + area(∆BDF) ------------------ (3)

Now add area(∆ABD) on both sides of Eq (3), we get

area(∆AED) + area(∆BCD) + area(∆ABD) = area(∆ADG) + area(∆BDF) + area(∆ABD)

area(pentagon ABCDE) = area(∆DGF)

Hence proved

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