$\begin{matrix} In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that \end{matrix}$
(i) $ar (A C B) = ar (A C F)$
(ii) $ar (A E D F) = ar (A B C D E)$

Open in App

Solution

Given:A pentagon $A B C D E$ where $B F ∥ A C$
$(i) △ A C B$ and $△ A C F$ lie on the same base $A C$ and are between the same parallels $A C$ and $B F$ .

$∴$ area $(△ A C B) =$ area $(△ A C F)$ since triangles with same base and between the same parallels are equal in area.

$(i i)$ Both $A E D F$ and $A B C D E$ have $A E D C$ common.

In part $(1)$ , we proved that

area $(△ A C B) =$ area $(△ A C F)$

Adding area $(A E D C)$ both sides,

$\Rightarrow$ area $(△ A C B) +$ area $(A E D C) =$ area $(△ A C F) +$ area $(A E D C)$

$\Rightarrow$ Area $(A B C D E) =$ Area $(A E D F)$

Hence proved.

Suggest Corrections

Similar questions

Q. In fig,

A B C D E

is a pentagon. A line through

B

parallel to

A C

meets

D C

produced at

F

. Show that
(i)

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

(ii)

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

a r (△ A C B) = a r (△ A C F)

Question 11 (ii)
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:

(ii) ar(AEDF) = ar(ABCDE)

Q. ABCD is parallelogram and ABEF is a rectangle and DG is perpendicular on AB.
Prove that (i)

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

(ii)

a r (A B C D) = A B \times D G

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

Join BYJU'S Learning Program

In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar(ACB)=ar(ACF)(ii) ar(AEDF)=ar(ABCDE)

$\begin{matrix} In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that \end{matrix}$
(i) $ar (A C B) = ar (A C F)$
(ii) $ar (A E D F) = ar (A B C D E)$