what is the tricky way for solving the question of class 9^{th} ncert math of Areas of Parallelograms and Triangles chapter of ncert of exercise 9.3of math give me the best and simple way for solving this question in easy and simple way The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that ar(ABCD) = ar(PBQR). [Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

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# The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that ar(ABCD) = ar(PBQR). [Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).] Q.9

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AC and PQ are joined.

Ar(â–³ACQ) = ar(â–³A

PQ) (On the same base AQ and between the same parallel lines AQ and CP)

â‡’Â ar(â–³ACQ)-ar(â–³ABQ) = ar(â–³APQ)-ar(â–³ABQ)

â‡’Â ar(â–³ABC) = ar(â–³QBP) â€” (i)

AC and QP are diagonals ABCD and PBQR.

âˆ´,ar(ABC) = Â½ ar(ABCD) â€” (ii)

ar(QBP) = Â½ ar(PBQR) â€” (iii)

From (ii) and (ii),

Â½ ar(ABCD) = Â½ ar(PBQR)

â‡’Â ar(ABCD) = ar(PBQR)