Question

please help to solve.maths question class ninth

Answer 1

First we have to do some Construction.
I'm Extending AD parallel to BQ also I'm extending line from Q parallel to AB say they Meet at K.
So ABCD is congruent to QCDK
Area(ABCD)=Area(QCDK).
AB=KQ
AD=BC
BC=CQ(Given)
In Tri.s ADP and QCP
AD=BC
angle DAP=angle CQP (AK||BQ)
angle ADC=angle QCD(ABCD and QCDK are congurent)
So By ASA congurency Tri.ADP is congruent to Tri. QCP
So Area(ADP)=Area(QCP)
So In tri APB and tri DKQ
the bases tri.s are Equal i.e AB=QK
also The height of The tri.s are also same (Height of ||grams are same)
So Area(APB)=Area(DKQ)

So,Area(ABCD)-Area(ADP)-Area(APB)=Area(QCDK)-Area(QCP)-Area(DKQ)
So Area(BCP)=Area(DPQ)


P.S Amazing question!