The side AB of a parallelogram ABCD is produced to any point P. A line through A parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Show that ar(parallelogram ABCD) = ar(parallelogram BPRQ).
Asked by Sthitaprajna Mishra | 4th Feb, 2014, 08:22: PM
Consider the above figure.
Triangles on the same base and between the same parallels have equal areas.
Subtract the area of triangle ABQ from both the sides, we have
Since AC and PQ are the diagonals of the parallelograms ABCD and PBQR, we have
Thus from equations (1), (2) and (3), we have
Answered by | 6th Feb, 2014, 10:01: AM
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