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Byju's Answer

Standard IX

Mathematics

Opposite Sides of a Parallelogram Are Equal

In fig 9.27. ...

Question

In fig 9.27. ABCDE is a pentagon . A line through B parallel to C meets DC produced at F . Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)

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Solution

A pentagon $A B C D E$ , $B F ∥ A C$
To Prove: $A r (A C B) = A r (A C F)$
$A r (A E D F) = A r (A B C D E)$
$△ A C B$ and $△ A C F$ lies on same base $A C$ and are between same parallel $A C$ and $B F$
$∴$ $A r (△ A C B) = A r (△ A C F) . . . . . . (1)$
Hence, Proved
add $A r (A E D C)$ both sides
$A r (△ A C B) + A r (A E D C) = A r (A C F) + A r (A E D C)$
$A r (A B C D E) = A r (A E D F)$
Hence proved

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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)

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)

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. 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)

Q. In figure ,

A B C D E

is any pentagon .

B P

drawn parallel to

A C

meets

D C

produced at

P

and

E Q

drawn parallel to

A D

meets

C D

produced at

Q

.
Prove that

a r (A B C D E) = a r (A P Q)

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In fig 9.27. ABCDE is a pentagon . A line through B parallel to C meets DC produced at F . Show that (i) ar(ACB) = ar(ACF)(ii) ar(AEDF) = ar(ABCDE)

A pentagon ABCDE, BF∥ACTo Prove: Ar(ACB)=Ar(ACF)Ar(AEDF)=Ar(ABCDE)△ACB and △ACF lies on same base AC and are between same parallel AC and BF∴ Ar(△ACB)=Ar(△ACF)......(1)Hence, Provedadd Ar(AEDC) both sidesAr(△ACB)+Ar(AEDC)=Ar(ACF)+Ar(AEDC)Ar(ABCDE)=Ar(AEDF)Hence proved

In fig 9.27. ABCDE is a pentagon . A line through B parallel to C meets DC produced at F . Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)