In parallelogram ABCD of the accompanying diagram, the line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Line DP and CQ meet at O, show that area of triangle QPO is 9/8 of the area of the parallelogram ABCD.
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post the accompanying diagram
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》[QPO] = [QAM]+[PBM]+[AMONB]
》 = [AMONB]+[MDC]+[NCD]
》 = [AMONB]+[MDC]+[NOC]+[DOC]
》 = [ABCD]+[DOC] , [ABCD] = K
》 = K+[DOC]
HERE, [DCNM ] ALSO IS A PARALLELOGRAM.
SO,
》[DOC] = 1/4 [DCNM] , 2[ABCD] = [DCNM]
》 = 1/8 [ABCD]
》 = k/8
THERE FOR ,
》 [QPO] = K +K/8
》[ [QPO] = 9k/8 ]
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