Question

The side AB of a pllgm ABCD is produced to any point P. A line through A pll to CP meets CB produced in Q and The pllgm PBQR is completed. show that,ar(llgm ABCD) =ar(llgm BPRQ)​

Answer 1

$\huge\bf\red{Shalom!}$ FOLLOW ME❤ & Mark Brainlist _____________Answer_____

Triangles on the same base and between the

same parallels are equal in area.

Diagonals of a parallelogram divides it into two Triangles of equal areas

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Given:

Two parallelograms ABCD and PBQR.

To show:

ar (ABCD) = ar (PBQR).

Proof:

Join AC & PQ .

Now  △ACQ & △APQ are on the same base AQ and between the

same parallel lines AQ and CP

ar(△ACQ) = ar(△APQ)

ar(△ACQ) – ar(△ABQ) = ar(△APQ) – ar(△ABQ)

[Subtracting ar(△ABQ) from both sides]

ar(△ABC) = ar(△QBP) — (i)

AC and QP are diagonals ABCD and PBQR.

Thus,

ar(ABC) = 1/2 ar(||gm ABCD) — (ii)

ar(QBP) = 1/2 ar(||gm PBQR) — (iii)

[Diagonals of a parallelogram divides it into

two Triangles of equal areas]

From (ii) and (ii),

1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR)

 

 ar(||gm ABCD) = ar(||gm PBQR)

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Hope

this will help you...