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प्रश्न
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 the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]

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उत्तर

Let us join AC and PQ.
ΔACQ and ΔAQP are on the same 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 are diagonals of parallelograms ABCD and PBQR respectively,
∴ Area (ΔABC) = 1/2Area (ABCD) ... (2)
Area (ΔQBP) = 1/2Area (PBQR) ... (3)
From equations (1), (2), and (3), we obtain
1/2Area (ABCD) = 1/2Area (PBQR)
Area (ABCD) = Area (PBQR)
संबंधित प्रश्न
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
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(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
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(i) ar(PRQ) = 1/2 ar(ARC)
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In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
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(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
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In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).

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