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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. Show that ar (ABCD) = ar (PBQR).

Q: 9         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. \small 9.26). Show that  \small ar(ABCD)=ar(PBQR).

                 [Hint: Join AC and PQ. Now compare  \small ar(ACQ) and  \small ar(APQ).]

                    

Answers (1)
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M manish

Join the AC and PQ.

It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar(\DeltaABC) = ar(\DeltaADC) = 1/2 ar(||gm ABCD).............(i)

Also, ar(\DeltaPQR) = ar(\DeltaBPQ) = 1/2 ar(||gm PBQR).............(ii)

Since \DeltaAQC and \DeltaAPQ are on the same base AQ and between same parallels AQ and CP.
\therefore  ar(\DeltaAQC) = ar (\DeltaAPQ)

Now, subtracting \DeltaABQ from both sides we get,

ar(\DeltaAQC) - ar (\DeltaABQ) = ar (\DeltaAPQ) - ar (\DeltaABQ)
ar(\DeltaABC) = ar (\DeltaBPQ)............(iii)

From eq(i), (ii) and (iii) we get

\small ar(ABCD)=ar(PBQR)

Hence proved.

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