In given figure, ABCDE is any pentagon. BP is drawn parallel to AC and meets DC produced at P. EQ is drawn parallel to AD and meets CD produced at Q. Then, which of the following statements is true?

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

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$A r e a (A B C D E) = A r e a (Δ A C Q)$

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$A r e a (A B C D E) = 2 A r e a (Δ A D Q)$

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$N o n e o f t h e s e$

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Solution

The correct option is A
$A r e a (A B C D E) = A r e a (Δ A P Q)$

BP || AC and AD || EQ,
Since triangles on the same base between the same parallels are equal in area
$a r (Δ A B C) = a r (Δ A P C)$ ........(1) (As Triangles that lie between two parallel lines and have same bases are equal in area.)
$a r (Δ A D E) = a r (Δ A D Q)$ ..........(2) (As Triangles that lie between two parallel lines and have same bases are equal in area.)
adding (1) and (2) we get
$a r (Δ A B C) + a r (Δ A D E) = a r (Δ A P C) + a r (Δ A D Q)$
Adding $a r (Δ A C D)$ to both sides, we get
$a r (Δ A B C) + a r (Δ A D E) + a r (A C D) = a r (Δ A P C) + a r (Δ A D Q) + a r (A C D)$
$H e n c e a r (A B C D E) = a r (Δ A P Q)$

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A B C D E

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A C

D C

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

Δ

a r e a (A E D F) = a r e a (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. Which of the following is true.

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