The Total solution for NCERT class 6-12
The side AB of aparallelogram ABCD is produced to any point P. A line through A and parallel toCP meets CB produced at Q and then parallelogram PBQR is completed (see thefollowing figure). Show that
ar (ABCD) = ar (PBQR).
Let us join AC and PQ.
ΔACQ and ΔAQP are on thesame base AQ and between the same parallels AQ and CP.
∴Area (ΔACQ) = Area (ΔAPQ)
⇒Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)
⇒Area (ΔABC) = Area (ΔQBP) … (1)
Since AC and PQ arediagonals of parallelograms ABCD and PBQR respectively,
∴Area (ΔABC) = Area (ABCD) … (2)
Area (ΔQBP) = Area (PBQR) … (3)
From equations (1), (2),and (3), we obtain
Area (ABCD) = Area (PBQR)
Area(ABCD) = Area (PBQR)