Question

In parallelogram ABCD of the accompanying diagram, 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. Lines DP and CQ meet at O. Show that the area of triangle QPO
is 9\8 of the area of the parallelogram ABCD.

Answer 1

Solution:-


》[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 ] PROVED


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