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

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Solution

(i) Since triangles $A C B$ and $A C F$ are on the same base $A C$ and between the same parallels $A C$ and $B F$ .
Therefore $a r (A C B) = a r (A C F)$

(ii) Adding $a r (A C D E)$ on both sides, we get
$a r (A C F) + a r (A C D E) = a r (A C B) + a r (A C D E)$
$\Rightarrow a r (A E D F) = a r (A B C D E)$

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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)

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)

Q. Question 11 (i)
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)

\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)

Q. 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)

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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)

(i) Since triangles ACB and ACF are on the same base AC and between the same parallels AC and BF. Therefore ar(ACB)=ar(ACF) (ii) Adding ar(ACDE) on both sides, we getar(ACF)+ar(ACDE)=ar(ACB)+ar(ACDE)⇒ar(AEDF)=ar(ABCDE)

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