Question

the side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. show that ar(ABCD) =ar(PBQR)

Answer 1

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Let us join AC and PQ. 
ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP. 
∴ Area (ΔACQ) = Area (ΔAPQ)
⇒ Area (ΔACQ) − Area (ΔABQ)
= Area (ΔAPQ) − Area (ΔABQ) ⇒ Area (ΔABC) = Area (ΔQBP) ...[ (1) Since AC and PQ are diagonals of parallelograms ABCD and PBQR
respectively,]
(1) ∴ Area (ΔABC) = 1/2 Area (∥gmABCD) ... (2) Area (ΔQBP) =  1/2 Area (∥gmBPRQ) ... (3) From equations (1), (2), and (3), we obtain1/2 Area (∥gmABCD) = 1/2Area (∥gmBPRQ) Area (∥gmABCD) = Area (∥gmBPRQ).





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