Math, asked by rubeena12, 1 year ago

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.


Alleei: post the accompanying diagram

Answers

Answered by Shaizakincsem
6

》[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 ]

Similar questions