Q. The side AB of a IIgm ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then IIgm PBQR is completed.Show that ar(IIgm ABCD) = ar(IIgm PBQR)

Given, ABCD is a parallelogram and AC is the diagonal on the parallelogram.

We know that, the diagonal of a parallelogram divides it into two triangle having equal areas.

Now, PBQR is a parallelogram and PQ is the diagonal of the parallelogram.

ΔAQC and ΔAQP are on the same base AQ and between the same parallel AQ and CP.

∴ area (ΔACQ) = area (ΔAQP)   ( Triangles on the same base and between the same parallels have equal area)

⇒ area (ΔACQ) – (ΔABQ) = area (ΔAQP) – area (ΔABQ)

⇒ area (ΔABC) = area (ΔPBQ)

  • 67
What are you looking for?