Question

can anyone solve Q.30

Answer 1

hello,
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,
∴ Area (ΔABC) = 1/2 Area (∥gmABCD) ... (2)
Area (ΔQBP) =  1/2 Area (∥gmBPRQ) ... (3)
From equations (1), (2), and (3), we obtain
1/2 Area (∥gmABCD) = 1/2Area (∥gmBPRQ)
Area (∥gmABCD) = Area (∥gmBPRQ).

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