  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: area of(△BPC) = 13 area of (△DPQ)   area of(△BPC) = 38area of (△DPQ) area of(△BPC) = 14 area of (△DPQ) 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)  Suggest corrections   