Question

Lines y = 3x, y = 6, y = 3x - 12 and y = 0 are plotted. The intersection of the lines forms a quadrilateral ABCD, where A, B is the point on the line y = 6; and C,D is the point on the line y = 0. A is (2,6) and D is (0,0). BC is extended such that BC = CQ. AQ intersects DC at P.
Then, area of $\Delta$BPC = area of $\Delta$ DPQ?

• A. True
• B. False

Answer 1

The correct option is A

True

ABCD is a quadrilateral with AB = CD and AB$\parallel$CD.A quadrilateral with opposite sides parallel and equal is a parallelogram.
Since BC = CQ, AD = BC. Thus, AD = CQ. And CB is extended to CQ, then CQ is parallel to AD.

Since CQ = AD, and CQ$\parallel$AD, then ADQC is a parallelogram.

AC is a diagonal of the parallelogram ABCD. DC and AQ are the diagonals of ADQC.
Triangles between the same base and same parallels are equal.
Area of $\Delta$ DCQ = Area of $\Delta$ QAC (between the parallels AD and CQ).

Area of $\Delta$DCQ + Area of $\Delta$ PCQ = Area of $\Delta$ QAC + Area of $\Delta$ PCQ
=> Area of $\Delta$ DPQ = Area of $\Delta$APC

But, Area of $\Delta$APC = Area of $\Delta$PBC (between the parallels PC and AB)

Hence, Area of $\Delta$BPC = Area of $\Delta$DPQ.