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Areas of Parallelograms and Triangles
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16/07/2021 4:23 pm
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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.). Show that: ar(ABCD) = ar(PBQR).
[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]
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16/07/2021 4:28 pm
AC and PQ are joined.
Ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) - ar(△ABQ) = ar(△APQ) - ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) — (i)
AC and QP are diagonals ABCD and PBQR.
∴ ar(ABC) = \(\frac{1}{2}\) ar(ABCD) — (ii)
ar(QBP) = \(\frac{1}{2}\) ar(PBQR) — (iii)
From (ii) and (ii),
\(\frac{1}{2}\) ar(ABCD) = \(\frac{1}{2}\) ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)