Question

in fig.9.32 ,ABCD is a parallelogram and BC is produced to a point Q such that AD=CQ.If AQ intersects DC at P, Show that ar (BPC)=ar(DPQ).

Answer 1

>>>>>>>>>>here is your answer<<<<<<<<<<

It is given that ABCD is a parallelogram.

AD || BC and AB || DC (Opposite sides of a parallelogram are parallel to each

other)

Join point A to point C.

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC

are lying on the same base PC and between the same parallels PC and

AB. Therefore,

Area (ΔAPC) = Area (ΔBPC) ... (1)

In quadrilateral ACDQ, it is given that

AD = CQ

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced.

∴ AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram.

Consider ΔDCQ and ΔACQ

These are on the same base CQ and between the same parallels CQ and AD. Therefore,

Area (ΔDCQ) = Area (ΔACQ)

⇒ Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)

⇒ Area (ΔDPQ) = Area (ΔAPC) ... (2)

From equations (1) and (2), we obtain

Area (ΔBPC) = Area (ΔDPQ)

Answer 2

here is your answer

please like this answer