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Question

In the following figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, then:


  1. area of(BPC) = 13 area of (DPQ)  

  2. area of(BPC) = 38area of (DPQ)

  3. area of(BPC) = 14 area of (DPQ)

  4. area of(BPC) = area of (DPQ)


Solution

The correct option is D

area of(BPC) = area of (DPQ)


It is given that ABCD is a parallelogram

Join point A to point C.

Consider APC and BPC

APC and BPC are lying on the same base PC and between the same parallels PC and AB. Therefore,

Area (APC) = Area (BPC) ... (1)

In quadrilateral ACQD, it is given that

AD = CQ

Since ABCD is a parallelogram,

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 DCQ and ACQ

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

Area (DCQ) = Area (ACQ)

 Area (DCQ) - Area (PQC) = Area (ACQ) - Area (PQC)

 Area (DPQ) = Area (APC) ... (2)

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

Area (BPC) = Area (DPQ)

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